The Book of Integers - La Citadellela-citadelle.com/mathematics/the_book_of_integers.pdf · The Book of Integers 1. Understanding Integers 5 Iulia & Teodoru Gugoiu 1.1 (Definition)

The Book ofIntegers

Iulia Gugoiu

Copyright © 2007 by La Citadelle

www.la-citadelle.com

FOR HIGH SCHOOL STUDENTSFOR HIGH SCHOOL STUDENTS

MATHEMATICSMATHEMATICS

Teodoru GugoiuB.Sc. Hons., M.Sc., B.Ed.Mathematics TeacherWilliam Lyon Mackenzie CIToronto District School Board

B.Sc., B.Eng.Mathematics & Science Teacher

Iulia & Teodoru Gugoiu

The Book of Integers - Mathematics for High School Students

© 2007 by La Citadelle4950 Albina Way, Unit 160Mississauga, OntarioL4Z 4J6, [email protected]

All rights reserved. No part of this book may be reproduced, in any form or by any means, without permission in writting from the publisher.

ISBN 0-9781703-1-8

Edited by Rob Couvillon

“Simplicity is complexity resolved.”

“Don't look for obscure formulae or mysteries.It is pure joy that I am giving you.”

Constantin Brancusi, romanian sculptor

(see on the left side a drawing of “The Endless Column”)

Acknowledgements

We want to thank to all our students. Their feedback, comments, criticism, ideas, and requests help us to become better teachers.


everyday

Topic

1. Understanding integers2. Absolute value, sign, and opposite3. Number line4. Comparing integers5. Applications of integers6. Addition of integers. Axioms7. Additions of integers using the number line8. Addition of integers using rules9. Subtraction of integers10. Order of operations (I)11. Equalities and inequalities12. Equivalent equalities and inequalities13. Equations14. Inequations15. Multiplication of integers (I)16. Multiplication of integers (II)17. Order of operations (II)18. Division of integers (I)19. Division of integers (II)20. Order of operations (III)21. Equalities and equations22. Proportions and equations23. Inequalities and inequations24. Powers25. Exponent rules26. Order of operations (IV)27. Divisors28. Divisibility rules29. Prime factorization30. Set of divisors31. Highest Common Factor (HCF)32. Least Common Multiple (LCM)33. Square roots34. Cubic roots35. Roots of superior order36. Roots rules37. Equations with powers and radicals38. Link between radicals and powers39. Order of operations (V)40. Substitution

Answers

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Content

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The Book of Integers

1. Understanding Integers

5


1.1 (Definition) Integers are signed numbers made of:a) a sign, + for positive numbers and - for negative numbers b) an absolute value, that is a natural number Example: -5.The sign is - . The absolute value is 5.

1. Identify whether the following numbers are integers or not:

1.2 (No Sign Rule) If an integer does not have a sign, by default the sign is considered positive. So, the numbers 5 and +5 are identical.In general:

1.3 (Number Zero) 0 is considered neither positive nor negative. However, some generalizations require us to consider the following numbers: +0 and -0. Still, these two instances of the number zero are equal to 0. So:

1.4 (The Set of Integers) The set of all integers can be represented by using the set notation as:

1.5 (Boundless) The set of integers is unlimited or boundless. There is neither a biggest integer nor a smallest integer. The number of elements in this set (the number of integers) is infinite.

1.6 (Infinity) Although a largest integer does not exist as a regular number, the symbol (called plus infinity) is used to express an unlimited large positive number.Similarly, the symbol (called minus infinity) is used to express an unlimited large (as absolute value) negative number.These two symbols are not considered numbers because the arithmetic of these numbers is completely different from the arithmetic of regular numbers. For example:

¥+

¥-

2. For each of the following integers, identify the sign and the absolute value:

3. Use the short notation (by dropping the sign) to rewrite the following integers:

4. Rewrite the following numbers as integers (including the sign):

5. Classify the following numbers as positive, negative or neither:

6. Find a way to represent the set of negative integers (see 1.4 for the set of positive integers).

101)22)3)4)10)11)2)0)2)3) -----+ jihgfedcba

3)0)1)2)5) +++++ edcba

55)41)300)20)1) edcba

1000)123)11)7)5)0)10)2)5)1) -----+ jihgfedcba

10. Find the value of the following expressions that contain the infinity symbol:

aa +=

000 -=+=

,...}3,2,1,0,1,2,3{..., +++---=Z

,...}3,2,1{ +++=+Z

definednotis¥-¥

¥=+¥ 5¥

´¥¥

¥´¥

¥

¥¥¥´¥+¥¥+¥-¥

2)

1)10)

1000))

/))10))1)

3 jihgf

edcba

25,13,1,3 -+-+

)10)01.0)3)3

2)1)5)5.7)0)2)1) kjihgfedcba ---+ ¥-

7. Use the set notation (see 1.4) to represent all odd positive integers greater than 0 and less than 10.

8. Find the logical value (true or false) of each of the following sentences:

a) All integers are positive.b) There is a largest positive integer.c) The smallest integer is 0.d) Any natural number is also an integer.e) There are more positive integers than negative integers.f) is a positive integer.¥+

9. For each case, identify the integers that satisfy the given properties:

a) is positive and has an absolute value equal to 7b) is negative and has an absolute value equal to 11c) has an absolute value equal to 5d) has an absolute value equal to 0e) is positive and has an absolute value equal to ¥

The set of positive integers is:



2. Absolute value, sign, and opposite

6


1. Use absolute value notation (see 2.1), and find the absolute value of the following integers:

2.5 (Opposite Numbers) Two integers are called opposite if they have the same absolute value but opposite signs.So, +5 and -5 are opposite.+5 is the opposite of -5-5 is the opposite of +5

2.6 (General Definition) The opposite of any number a is -a. So, to get the opposite of a number, just place the - sign in front of the number. Examples:The opposite of 5 is -5.The opposite of -5 is -(-5) that is equal to 5.

2.7 (The Opposite of an Opposite) The opposite of an opposite of a number is equal to the number itself:

Example. Let’s consider the number 3. The opposite of this number is -3. The opposite of the number -3 is +3, which is equal to the original number 3.

2.8 (The Opposite of Zero) The opposite of 0 is 0:

2.1 (Absolute Value) If you drop the sign of an integer, you get its absolute value. The absolute value of an integer is a natural number. Use the function | | to express the absolute value. Examples:

(Read: the absolute value of -4 is 4)

2.2 (Number Zero) The absolute value of 0 is 0.

2.3 (Sign) If you drop the absolute value of an integer you get its sign. Use the function sign() to express the sign of an integer. Examples:

2.4 (Number Zero) The sign of 0 is not defined. 0 is neither a positive nor a negative number.

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4|4| =-

7|7||7|

6|6|

=+=

=+

0|0| =

+=+=

-=-

+=+

)10()10(

)7(

)5(

signsign

sign

sign

aa =-- )(

000 =+=-

2. For each of the following absolute values, find the possible value(s) of the integer(s):

3. For each case, find the sign of the integer (see 2.3):

4. For each case, find the opposite of the given integer:

5. Find the value of each expression. One case is solved for you as an example:

6. Find the value of each expression that contains the absolute value function. One case is solved for you as an example:

7)7)12)100)10)0)3)2)2)3) -----+ jihgfedcba

123)5)10)3)2)1)0) gfedcba

7)1)1)10)100)10)0)4)5)3) jihgfedcba -+---

77)7)2)11)0)20)3)3)20)12) -+---- jihgfedcba

)))32((()))35(())745())811())21()

)))20(())))4(())3())5())2()

+-+--+--+---+-

+-+----++---

jihgf

edcba

|2||8|)|3||5|)|2|2)|2||5|)|1||3|)

|)7(|)|3|)|)4(|)|25|)|)41(|)

-¸--´--´----+-

-----+--+-

jihgf

edcba

Example:

5)5())5(()))5((()))32((() =--=-+-=-+-=+-+-j

7. Find the value of each expression that contains the absolute value function. One case is solved for you as an example:

||)2|1(|)||8||)||55|5|)|)2(|)|)47(|)

||3||7||)|3||5|)3|2|1)2|3|))1|3(|2)

|2|5)|5||13|))1|5(|)|5|

|10|)|3||2||1|)

-+-----------

------+-++-+-´

---´-+---

--+-+-

onmlk

jihgf

edcba

Example:

325|2||5|) =-=---g

Example:

4|4||37|||3||7||) ==-=---j



3. Number Line

7


1. For each point represented on the following number line, find the corresponding integer:

3.1 (The Number Line) An intuitive representation of integers is based on the number line. There is one to one correspondence between the set of integers and a set of equidistant points on a line. This correspondence is illustrated in the following figure:

3.2 (Positive and Negative Integers) The positive integers are situated to the right side of the origin O. The negative integers are situated to the left side of the origin O.

3.3 (Number Zero) Number zero corresponds to the origin O.

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0 +1 +2 +3-1-2-3-4 +4

O

3.4 (Absolute value) The absolute value of an integer is equal to the distance between the origin O and the corresponding point on the number line. The greater the absolute value of an integer, the greater the distance between the corresponding point on the number line and the origin.

0 +1 +2 +3-1-2-3-4 +4

O M

3|3| =+

3.5 (Opposites) Two integers that are opposite to each other correspond to points on the number line that are symmetrically positioned relative to the origin O.

0 +1 +2 +3-1-2-3-4 +4

O NM

Example. The absolute value of the number +3 is 3. M is the point corresponding with +3 on the number line. The distance between M and O is equal to 3.

Example. The numbers -3 and +3 are opposite to each other. -3 corresponds to point M on the number line and +3 corresponds to N. The points M and N are symmetrically positioned relative to the origin O. The distances MO and NO are both equal to 3.

2. Plot the following points on the number line below:

3. A part of a number line is represented in the figure below. All the points are equidistant. Find the integers that correspond to the given points:

4. Find the corresponding integer and its absolute value, by calculating the distance between each point and the origin O.

5. For each of the points below (A, B, C, D, E, F), find the symmetrical point relative to the origin O (see 3.5):

+2

A

O

BC DE

+2+2+1 +3 +4 +5 +6 +7 +8 +9-8 -7 -6 -5 -4 -3 -2 -1 0

)3())4())5())7())2() --+- EeDdCcBbAa

-15

D EA

-35

B C

D EA B C

+10

+2+2+1 +3 +4 +5 +6 +7 +8 +9-8 -7 -6 -5 -4 -3 -2 -1 0

A B C D E

6. For each case, use the number line to identify an integer that is:

a) 2 units left of +7b) 3 units right of +2c) 2 units left of -5d) 4 units right of -2e) 3 units from +5f) 2 units from -6g) 3 units from the origin

+2+2+1 +3 +4 +5 +6 +7 +8 +9-8 -7 -6 -5 -4 -3 -2 -1 0



4. Comparing Integers

8


1. For each case identify the true statement:4.1 (Less, Equal, Greater) Any two integers a and b can be compared. The result of this comparison is that only one of the following statements is true:

4.4 (Comparing negative integers, 0, and positive integers) Any negative integer is less than 0 and 0 is less than any positive integer. Examples:

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4.2 (The Number Line) On the number line a smaller integer is placed at the left side of a bigger integer. So:

4.5 (Transitivity) The order (comparison) relation between two integers has the transitivity property:

bthatgreaterisaba

btoequaltisaba

bthatlessisaba

>

=

<

...3210123... <+<+<+<<-<-<-<

4.3 (Order) As you can see from the previous relation, the set of integers is an ordered set. Integers can be sorted (ordered) from smaller integers to bigger integers. There is neither a smallest integer nor a largest integer.

cathencbandbaif <<<

Example:

323002 +<-+<<- thenand

So, any negative integer is less than any positive integer.

4.6 (Comparing positive integers) When you compare two positive integers, the greatest integer is the one with the greatest absolute value. Example:

3553 +>++<+ or

4.7 (Comparing negative integers) When you compare two negative integers, the greatest integer is the one with the smallest absolute value. Example:

7557 ->--<- or

Indeed, -7 is at the left side of -5 and therefore is smaller.

70

02

+<

<-

07

20

>+

->

...3210123... >->->->>+>+>+

2. Sort the integers from the smallest to the greatest (use the < symbol):

3. Sort the integers from the greatest to the smallest (use the > symbol):

5. Place the symbols <, =, or > between each pair of integers to obtain true statements:

4. Use the transitivity property to combine the two true statements and get a third true statement (see 4.4):

6. Find the logical value (true or false) of each statement:

9. Find the set of values of the integer x so each statement is true:

8. Find the logical value (true or false) of each statement (Note: a<b<c is true if both a<b and b<c are true):

00;00;00)01;10;01)33;33;33)

50;50;50)35;35;35)21;21;21)

>=<>=<>=<

>=<>=<>=<

fed

cba

3;2;1;0;2;3)3;3;2;2;1;1)2;0;2;1;3)

0;2;1;3)2;1;0)1;2)

---++-+-+---+

---++

fed

cba

5;1;1;2;3)5;5;4;4;2;2)4;0;2;2;3)

0;1;2;3)2;1;0)1;2)

---++-+-+--++

------

fed

cba

1223)3110)4225)

1223)2002)2110)

-<--<-->-->+<<-

>><<-<<

andfandeandd

andcandbanda

57)55)42)24)73)

00)50)11)01)21)

+-+++----

---

jihgf

edcba

33)21)53)24)75)

11)50)11)22)23)

+>-->+<+->--<

+>-<-=+>->

jihgf

edcba

7. Two more comparison operators are (greater or equal operator) and (less or equal operator). Find the logical value (true or false) of each statement:

£³

03)20)033)45)32)

11)22)21)10)11)

³--³+³--£-+£

+£-³+-³-£³

jihgf

edcba

203)310)135)135)011)

113)021)321)101)210)

->>+->->-<-<-->->->->

+>->->>+>->-<<-<<

jihgf

edcba

22)30)40)14)31)

51)01)31)12)31)

->³->>+££->>->>-

£<³³->>+<<-<<

xjxixhxgxf

xexdxcxbxa



5. Applications of Integers

9


1. Use integers to describe the position of an object. The object is positioned:

A) 5 units right of the origin B) 4 units left of the originC) in the origin D) 1 unit rightE) 2 units left

5.1 (Position) Let’s suppose that an object is placed at point A on the number line.

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5.2 (Displacement) A change in position of the object is called displacement. The displacement of the object from the point A to the point B can be described:a) in words as: 5 units to the rightb) using an integer as: +5

The position of the object can be described:a) in words as: 2 units left of the origin Ob) using an integer as: -2If the object is moved from point A to point B, its new position can be described:a) in words: 3 units right of the origin Ob) using an integer: +3

0 +1 +2 +3-1-2-3-4 +4

OA B

0 +1 +2 +3-1-2-3-4 +4

OA B

+5

The displacement of the object from point B to point C can be described:a) in words: 2 units to the leftb) using an integer: -2

0 +1 +2 +3-1-2-3-4 +4

O C B

-2

5.3 (Temperature) Temperature can be expressed using integers. The temperature of 0 temperature of 20

0

0

°C corresponds to 0. A °C above °C is +20. A temperature 30°C below 0°C is -30. The smallest possible temperature is approximately 273°C below

°C and corresponds to -273.An increase in temperature of 10°C corresponds to +10 and a decrease in temperature of 40°C corresponds to -40.

5.4 (Altitude) Altitude can be expressed using integers. Sea level corresponds to 0. An altitude of 100m above the sea level corresponds to +100. A depth of 50m below sea level corresponds to -50. Rising 200m corresponds to +200 and falling 500m corresponds to -500.

5.5 (Bank Account) Bank account can be expressed using integers. A debit of $100 corresponds to +100. A credit of $25 corresponds to -25. A deposit of $50 corresponds to +50 and a withdraw of $75 corresponds to -75.

2. Use integers to describe the displacement of an object. The object is:

A) moved 2 units to the right B) moved 2 units to the leftC) fixed in the origin D) moved 3 units rightE) moved 5 units left

3. Use integers to describe the outside temperature. The temperature:

A) is 10 degrees above 0 B) is 15 degrees below 0C) for melting ice D) for boiling waterE) the lowest possible

4. Use integers to describe the change in the temperature. The temperature is:

A) increasing by 10 degrees B) decreasing by 5 degreesC) the same D) 3 degrees moreE) 10 degrees less

5. Use integers to describe the position of an object relative to sea level. The object is:

A) 1000m above B) 2m belowC) 20m under sea level D) at sea levelE) 7m higher than sea level

6. Use integers to describe the change in the altitude of an object. The object is:

A) raised 2m B) lowered 5mC) moved 7m upward D) moved 3m downwardE) not moved at all

7. Use integers to describe the balance of a bank account. The balance is:

A) a debit of $500 B) a credit of $150C) the initial value D) a debt of $31E) $125 cash

8. Use integers to describe the transaction in a bank account. The transaction is:

A) a deposit of $50 cash B) a deposit of a $550 chequeC) a withdraw of $40 cash D) a payment of a $45 billE) just updating the address



6. Addition of Integers. Axioms

10


1. For each case, identify the addends and the sum:6.1 (Addition) Addition is a binary operation (represented by the symbol +) between two operands (called terms or addends) to obtain a result called the sum. Example:

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The terms (addends) are 3 and 2.The sum is 5.

6.2 (Commutative Property) The order in which the addends are added is not important. This is the commutative property of addition. This property can be written as:

5)2()3( +=+++

abba +=+

Example:

6)2()4()4()2( +=+++=+++

6.3 (The Additive Identity) 0 is the additive identity of addition satisfying the property:

aaa =+=+ 00

Example:

5)5(00)5( -=-+=+-

6.4 (Opposites Cancellation) The sum of a number (integer) and its opposite is 0:

0)()( =+-=-+ aaaa

Example:0)3()3( =-++

6.5 (The Associative Property) If an expression contains more than 2 terms then the order in which the additions are done is not important. For 3 terms this property states that:

)()( cbacba ++=++

Example:

6333)21(

651)32(1

=+=++

=+=++

6.6 (Application) By combining the previous properties the sum of a complex addition (involving more terms) might be calculated very easily.Example:

7)7(00

)7()]4()4[()]2()2[(

)4()7()2()4()2(

-=-++

=-+-+++++-

=-+-+++++-

426)10)6(4)4)3(7)

4)2()2()1)1()2()321)

+==++=-+

+=++++=-++=+

fed

cba

2. Use the commutative property (see 6.2) to rewrite these additions:

)5()4()4)3()0)1()

05)32)10)

-+-+-+-

+++

fed

cba

3. Use the additive identity property (see 6.3) to find each sum:

)5(0)0)4()0)1()

05)10)00)

-++++-

+++

fed

cba

4. Use the opposites cancellation property (see 6.4) to find each sum:

)11(0)11(0))3(03)50)5()

)7()7())0()0())10()10()

)5(5))2()2())1(1)

+++-+-+++++-

-++-++++-

-++++--+

ihg

fed

cba

5. Use the associative property (see 6.5) to find each sum easily. One case is solved for you as an example. You might also need to apply the commutative property.

)11(511))7()7(3)5)5()3()

)37)17((17))316(14))75()5()

1)195())2())2(5()7)35()

+++-++-++-+-

+-+++++-

++++-+++

ihg

fed

cba

6. Group the terms conveniently (see 6.6) to find the sum. One case is solved for you as an example:

54)5(3)9(2)1313)4(17)19)3(11)

8)6(42)8642))1(39)3(1)

52)5(3)6)10(4))2()1(321)

)2()4()2()4()35)3())1()2(1)

++-++-++++-++-+

+-+++++-+++-+

++-++-+-+-+++

-+++++-++--+-+

lkj

ihg

fed

cba

Example:

000)9)9(()55(

))54()9(())5()32((54)5(3)9(2)

=+=+-+-=

=++-+-++=++-++-+l

Example:

550)5())11()11((

)511()11()11()511()11(511)

=+=++-++=

=+-++=+++-=+++-i



7. Addition of integers using the number line

11


1. Use the number line to find each sum:7.1 (Addition Redefined) The addition operation can be interpreted as:

old value + change = new value

Example. Let’s find the sum of:

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using this new definition of addition together with the number line:

+3 is the old value-5 is the change-2 is the new valueSo:

7.2 (More addends) If more terms (addends) are to be added, then the meaning of the addition operation is:

old+change+change+...+change=new

Example. Let’s find the sum of:

So:

)5()3( -++

0 +1 +2 +3-1-2-3-4 +4

O

-5

)1()5()3()6()4( -+++-+++-

0 +1 +2 +3-1-2-3-4 +4

+6-3

+5-1

7.3 (The hidden addition sign) When adding more numbers, usually the addition operator (+) between the numbers can be dropped without affecting the clarity of the expression. So, an expression as:

14532 -+-+-

is in fact equivalent of:

)1()4()5()3()2( -+++-+++-

Let’s use number line to find the sum:

0 +1 +2 +3-1-2-3-4 +4

+3-5

+4

-1

So:

2)5()3( -=-++

3)1()5()3()6()4( +=-+++-+++-

114532 -=-+-+-

2. Find the addition operation that corresponds to each arrow on the number line:

3. Use the number line to find each sum:

4. Find the addition operation that corresponds to the sequence of arrows on the number line:

5. Simplify the following expressions by hiding the addition operators between two consecutive terms. Do not calculate the sum.

6. Rewrite each expression by revealing the hidden addition operators. Do not calculate the sum.

7. Use the number line to calculate each sum:

)5()3()0)2())6()2())5()3()

)1()2())1()4())3()2())3()4()

)2()1())2()3())3(0))2(0)

-+++-++--++

-+-++--+--++

++-+++-+++

lkji

hgfe

dcba

0 +1 +2 +3-1-2-3-4 +4

a)

b)c)

d) e)

)3()3()2()2())3()1()2()5())1()2()1()

)6()4()2())3()2()1(0))3()2()1()

)3()5()2())3()1()3())3()2()1()

-+-++++-+-+-++-+-+-

-+++--+++-+++-++

-+-++-+++--+++

ihg

fed

cba

0 +1 +2 +3-1-2-3-4 +4

)5()4()2()1())1()3()5())3()2()2()

)5()2())2()3())2()1()

-+++++--+-++++-+-

-+-++--++

fed

cba

212345)3322)4321)

53)32)21)

---+-++--+-+-

--+--

fed

cba

642531)10631)8421)

7654321)5342)5432)

5420)531)222)

2345)321)531)

+++---+-+--+-

+-+-+-++---+-

-+-+-----

-+----+-

lkj

ihg

fed

cba


|2||6|5)|5||2|)|3|2)|52|) ---------- dcba

Example:

321265|2||6|5) -=--=--=----d



8. Addition of integers using rules

12


1. For each case, calculate the sum of the positive integers (see 8.2):8.1 (Addition Rules) The method of using the number line to add two or more integers is not useful for adding large numbers (in absolute value). In these cases the addition rules presented bellow must be used.

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Examples:

8.2 (The Addition of Positive Integers) If you add two or more positive integers:a) the sign of the result is positiveb) the absolute value of the result is the sum of the absolute values of each of the terms

...)(...)()()( ++++=++++++ cbacba

9)423()4()2()3(

7)52()5()2(

+=+++=+++++

+=++=+++

8.3 (The Addition of Negative Integers) If you add two or more negative integers:a) the sign of the result is negativeb) the absolute value of the result is the sum of the absolute value of each of the terms

...)(..)()()( +++-=+-+-+- cbacba

Examples:

8.4 (The Addition of Integers with Opposite Signs) If you add two integers with opposite signs:a) the sign of the sum is the sign of the integer with the biggest absolute valueb) the absolute value is the difference between the greatest absolute and the smallest absolute valueCase 1. If the absolute value of the positive number is greater than the absolute value of the negative number:

10)4321(4321

8)413()4()1()3(

4)31()3()1(

-=+++-=----

-=++-=-+-+-

-=+-=-+-

)()()( bAbA -+=-++

Case 2. If the absolute value of the negative number is greater than the absolute value of the positive number:

)()()( aBBa --=-++

Examples:

Examples:

3)36(63

2)35()3()5(

+=-+=+-

+=-+=-++

3)14(14

5)27()7()2(

-=--=+-

-=--=-++

2. For each case, calculate the sum of the negative integers (see 8.3):

3. In each of the following cases, the absolute value of the positive integer is greater than the absolute value of the negative integer. Find the sum:

4. In each of the following cases, the absolute value of the negative integer is greater than the absolute value of the positive integer. Find the sum:

5. For each case, calculate the sum:

333221)300020010)302510)321)

5432112345)3525550)3251250)1351000)

)375()25())35(200)123321)24101)

3321))17(5))7()3()31)

++++++++

++++

++++

+++++++

ponm

lkji

hgfe

dcba

5150015050200)552510)432)

1001234)33332222)35001000)

)350(450))125()125())25(200)

)3()3())3()2()21)

-----------

------

-+--+--+-

-+--+---

lkj

ihg

fed

cba

)50()150())3333()1111()300250)12502500)

)125()550())300()100()12525)50100)

)15()55())55()20()2015)510)

75))6()3())2()4()35)

-++++-+--

-++++-+--

-++++-+--

+-++--++-

ponm

lkji

hgfe

dcba

)350()125())2200()4400()4005500)2000150)

)250()150())150()600()125150)20050)

)40()15())25()30()1025)103)

49)85))3()7())4()3()

-++++-+--

-++++-+--

-++++-+--

+--++--++

ponm

lkji

hgfe

dcba

)2()3()1()5()3()1()1550)0325125)

)300()125()125()1505)220200)

1001000100)100101)502010)

655)420)4321)

++++-+-+-++----

-+++--+++

--------

+-+-----

lkj

ihg

fed

cba

6. Group the terms conveniently to find more easily each sum. One case is solved for you as an example:

252015105)15662534)1523105)

46545)7283)16743)

-+-+---+-+-+--

-+-++--++-+-

fed

cba

Example:

90)10100(10100)1525()6634(15662534) -=--=+-=-+--=--+-e



9. Subtraction of integers

13


9.1 (Subtraction) Subtraction is a binary operation between an operand called the minuend and an operand called the subtrahend to get a result called the difference:

© La Citadelle

differencesubtrahendminuend =-

9.2 (Number Line) You can use the number line to calculate the difference between two integers. To do that use this rule:

changevalueoldvaluenew =-

Example:

0 +1 +2 +3-1-2-3-4 +4

O

+5

)2()3( --+

where:-2 is the old value+3 is the new value+5 is the change from the old value to the new valueSo:

5)2()3( +=--+

9.3 (Subtraction Redefined) Subtraction between a minuend and a subtrahend can be redefined as an addition between the minuend and the opposite of the subtrahend:

)( baba -+=-

Example:

9)5()4()5()4( -=-+-=+--

9.4 (Anti-commutative Property) The subtraction operation has the anti-comutative property:

)( abba --=-

Example:

2)3()5(

2)5()3()5()3(

+=+-+

-=-++=+-+

9.5 (Sign Rules) The following sign rules can be useful when dealing with addition and subtraction of integers:

aaa

aa

aa

aaa

=+=--

-=+-

-=-+

=+=++

)(

)(

)(

)(

Examples:

23553)5()3(

945)4()5(

3)3())3((

=-=+-=--+-

-=--=-++-

=--=+--

1. Use the number line to find each difference (see 9.2):

2. Find the subtraction operation that corresponds to each arrow on the number line (see 9.2):

)5()7())5(2)52))5()3()

)2(0))1()4())3()2())1()5()

)2()2())2()3())1()3())1()2()

--------+

+-+-+---+--

---+-++----+

lkji

hgfe

dcba

0 +1 +2 +3-1-2-3-4 +4

a) b)c)

d) e)

3. Rewrite each subtraction as an addition with the opposite, then find the result:

)7(5))2()7())3()2())5()3()

)1(3))1()4())4()2())2()3()

)2(5)35)40)51)

+----------+

----++-++-+

-----

lkji

hgfe

dcba

4. Use the anti-commutative property to calculate the differences (see 9.3):

143123))5(0))12()6()107)

144))8()3()40)73)

-+-+-+-

-+-+--

hgfe

dcba

5. Use the sign rules to simplify the following expressions (see 9.4):

)))5((()))5(()))6(())4()

))2(())3())3())5()

-+---+------

+--+--+++

hgfe

dcba


))1(())1(()))3(())5(()))1(())4(()

))6(())2(())))4((())6(()))6(())4(()

))5(()3()))2(()3())3()5()

+--+-+---+---+----

-+--+--+-+-+-------

+--+---+--+-++

ihg

fed

cba

+2+2+1 +3 +4 +5 +6 +7 +8 +9-8 -7 -6 -5 -4 -3 -2 -1 0

Example:

1046)4(()6()))4((())6(() =+=--+--=-+-+-+-e


|3||2||1|)|7|5||)|3||4|)|5||2|) -+---+--------+ dcba

Example:

2|2||75||7|5||) =+=+-=+--c



10. Order of operations (I)

14


1. Use the “left to right” rule (see 10.1) to find the result of each sequence of operations:

10.1 (Priority) The addition and the subtraction are considered operations of the same priority. Therefore, if more operations appear within an expression then the operations must be done in the order they occur (from left to right). Example:

© La Citadelle

330344

3413

34152

-=-=--=

=--++=

=--++-

10.2 (Grouping) Sometimes it is useful to group all the positive terms and negative terms separately:

51512

)753()246(

752436

-=-=

=++-++=

=--++-

10.3 (Brackets) To change the order of operations, you can use brackets. If brackets appear within an expression then the operations inside of the brackets must be done first. Example:

)()( dbecaedcba +-++=+-+-

Example:

51311)3(1

)43()52(1

=++=+--=

=+-+--

10.4 (Nested Brackets) In complex expressions the brackets can be nested one inside of another. In this case start with the inner-most operations . This rule might be applied more than once if necessary. Example:

4)1(12

)232()12(2

))2(32())1(2(2

))64(32()3)21(2(2

-=---=

=+--+-+-=

=-------+-=

=----+---+-

10.5 (Standard Grouping Symbols) To differentiate nested grouping symbols, three types of standard grouping symbols are used: ( ) called parentheses, [ ] called square brackets, and { } called braces. By convention the order of nesting is:

{ [ ( ) ] }Example:

10}12{2}2]5[5{2

}2]1)0()1(3[5{2

}2]1)165()32(3[5{2

=+-=+--+-=

=+-----+-=

=+-+--+----+-

52015105)65432)28765)

2345)5432)4321)

432)143)321)

-+---+-+-+-+-

-+-++--+-+-

-+----+

ihg

fed

cba

2. Use grouping of positive and negative terms (see 10.2) to find the result of each sequence of operations:

5030252015)246810)654321)

3145)3251)4143)

432)142)523)

-+--+-+--+-+-

---++--+---

++-+---+

ihg

fed

cba

3. For each case, first do the operations inside of the brackets, then find the result (see 10.3):

2)54()32(1))75()12(0))52()25()31()

)74()53(1))43()52())42()42()

)52(0))53(2))42(1)

--+--+--+-+-----

----------+

---+-+--

ihg

fed

cba

4. For each case, complete the operations starting with the inner-most brackets (see

4)3)2)1)65(4(3(2(1))1)24()75(()43()

)))14((()))14((()))85(3(2)

)1)1)3((())4)53(()25()

2)1)53(2()))21(3(2)

----------+------

--+--+--+---+-

-+-----+---+-

+------

hg

fe

dc

ba

5. For each case, find the result using the order of operation for “standard grouping symbols” (see 10.5):

1}2]1)97(2[3{2))]}13()95[()]53()21[({1)

)]}32(3[2{1))]21(3[)}32()]21(1{[)

)]}26(5[)75({2))]43(3[)]43(3[)

]4)29(3[)75()]1)74(2[1)

----+-+-+---------

+-------+-----

+------++---

+---+-------

hg

fe

dc

ba


|4)54(3|)|73|)73()|)74(3|)

|)3||2(||2|)|)3|2(|)7|5()|)5|3(2)

----------

--------------

fed

cba

Example:

112)1(2)32(2|)3||2(||2|) -=+-=---=---=------c



11. Equalities and Inequalities

15


1. Find the logical value (true or false) of each statement (see 11.1):11.1 (True/False Statements) In mathematics ideas are expressed as statements. A statement can only be true or false.Example: “3<4” is a true statement.Example: “3 is a divisor of 10” is a false statement.

© La Citadelle

11.2 (Relational Operators) Some of the most common types of statements are binary relations involving the following relational operators:

Examples:

11.4 (Symmetry) Equality has the symmetry property that states:

bababa

bababa

¹³>

=£<

truetrue

falsetrue

22;11

20;03

-³--¹

+=<-

11.3 (Equality) If the relational operator is “=” then the binary relation is called equality. Examples:

false

true

4532

00

-=-

=

or:

“a=b“ is true if and only if “b=a“ is true.“a=b“ is false if and only if “b=a“ is false.Example:

523352 +-=+Û+=+-

abba =Û=

abifonlyandifba ==

11.5 (Inequality) If the relational operator is not “=” then the binary relation is called inequality. Examples of inequalities:

truetrue

truefalse

375;221

321;321

-³--<--

=+¹+

11.6 (Mirroring) For any inequality there is an equivalent inequality obtained by mirroring of the original:

abba

abbaabba

abbaabba

¹Û¹

£Û³<Û>

³Û£>Û<

;

;

The original inequality is true if and only if the mirrored inequality is true. The original inequality is false if and only if the mirrored inequality is false. Examples:

2222

533353

1001

-¹+Û+¹-

+->Û<+-

<Û>

121)2)93)

215)10)21)

=--

>=

fintegeraniseofdivisoraisd

bydivisibleiscba

2. Find the logical value (true or false) of each statement containing operational operators (see 11.2):

22)2515)52)22)

11)30)10)22)

10)12)11)11)

->-->+-³-+£

+>-->+<-¹

->>-+=-=+

lkji

hgfe

dcba

3. Do the required operations on both sides and find the logical value (true or false) of each of the following equalities (see 11.3):

50155)143532)123321)

3652)862)213)

542)121)321)

-=-+-=-++-=-+

-=--+=---=-

-=--=-=+

ihg

fed

cba

4. Use the symmetry property (see 11.4) to rewrite each equality and then find the value (true or false) of each statement:

)74(30)317531)2453)

753)742)231)

---=--=-+--=+-

-=---=--=

fed

cba

5. Do the required operations on both sides and find the logical value (true or false) of each of the following inequalities (see 11.5):

0)52()74(1))62(153)

)2010(15)2015(10))83(31)23(2)

228642))41(2)73(1)

2251)5162)

>--+---->-

--£--+--³+--

--¹-+----£--

--³---<--

hg

fe

dc

ba

6. Use mirroring (see 11.6) to rewrite the following inequalities and then find the value (true or false) of each statement:

432321))22()85(1)364532)

22)21)10)

+-¹+----£--+-³+-

+<-->->

fed

cba

7. Find the logical value (true or false) of each statement containing operational operators and absolute value function. One case is solved for you as an example:

|1||1|)|3|0)|1|0)|5|5)

|1|0)|1||2|)|3||4|)|3||3|)

+>-->+<-¹

-<-<-+>++=-

hgfe

dcba

Example:

)(12|1||2|) falsec <Û-<-



12. Equivalent Equalities and Inequalities

16


1. Do the required operation, then simplify, and finally find the logical value (true or false) of each statement:

12.1 (Adding or Subtracting) By adding (or subtracting) the same number to (from) the left side and to (from) the right side of an equality (inequality) you get an equivalent equality (inequality):

© La Citadelle

Examples:

.etc

cbcaba

cbcaba

+<+Û<

+=+Û=

)(

42442242

falseareboth

++=+-+Û+=-+

)(

2422314231

trueareboth

-+<-+Û+<+

12.2 (Moving Terms) Moving a term from one side of an equality (inequality) to the other side requires changing the term to its opposite in order to get an equivalent equality (inequality):

bcacba

bcacba

+=Û=-

-=Û=+

Examples (true statements):

12532513

22312321

-+-¹-Û+-¹+-

+=+Û=+-

12.3 (Empty side) After removing the last term from one side of an equality (inequality) to the other side replace the empty side with 0. Indeed:

abbaba -=Û=+Û= 00

03213210

321202321

¹---Û++¹

+-+-=Û-=-+-

12.4 (Positive Terms) By moving terms from one side of an equality (inequality) to the other side, it is always possible to have an equivalent equality (inequality) with positive terms on both sides. Example:

522144142542 ++=++Û--=-+-

12.5 (Simplifying) If both sides of an equality (inequality) contain the same term then this term can be cancelled out:

.etc

bacbca

bacbca

<Û-<-

=Û+=+


42134321

765765

-¹-Û+-¹+-

<Û-<+-


6;263)5;532)1;217)

4;042)3;132)2;22)

addfaddeaddd

addcaddbadda

-£--³----<+-

->+-+¹-+-=+

2. By moving terms from one side to the other side (see 12.2), find 3 equivalent equalities (inequalities):

101551510)23312)1235)

164)432)132)

-+-£-+-³+--¹+-

->+--<---=-

fed

cba

3. Find an equivalent equality (inequality) by moving all terms from the right side to the left side:

151051510)3221)223)

210)413)422)

+-£---³--+¹

+->--<--=-

fed

cba

4. Find an equivalent equality (inequality) by moving all terms from the left side to the right side:

23573)1510105)311)

2121)231)374)

-+£+-+--³--¹+

->+-<---=-

fed

cba

5. Find an equivalent equality (inequality) by moving terms from one side to the other side so finally each side to have only positive terms (see 12.4):

321432)52421)43)

23)12)121)

---£----³+--¹-

->--<--=-

fed

cba

6. Find an equivalent equality (inequality) by canceling out the identical terms on both sides (see 12.5):

5323312)114421)155)

2232)133)11121)

+-+-£-+-+-³-+-+-¹-

+->-+-<-+-=+-

fed

cba

7. Find an equivalent equality (inequality) so one side is 0 and the other side is an integer. One case is solved for you as an example:

2)85()53()62)1)84(2())23()32(1)

2135)2142)213)

+£-----³+-----¹--

-->+--<---=-

fed

cba

Example:

)(010878762)1)84(2() truee ³Û³+-Û-³-Û--³+---

8. Find the logical value (true or false) of each statement. One case is solved for you as an example:

21|2||1|)|3||1|)|8||8|)|7||7|)

0|1|1)|2||5|)|3||0|)|2||2|)

-->-+-+£-+³-+¹-

=-+->--<+=-

hgfe

dcba

Example:

)(3332121|2||1|) trueh ->Û->+Û-->-+-



13. Equations

17


1. Find if the given value of x is a solution of the following equations (use substitution as explained in 13.2).

13.1 (Equation) An equation is an equality containing an unknown number represented by a letter and called variable. Example:

© La Citadelle

521 -=+- x

In this equation x is the unknown number or variable.

13.2 (Solution) A solution of an equation is the value (number) of the variable that makes the equation a true statement. If an equation has more solutions, these solutions together form a set solution. Example:

31 -=+x

A solution of this equation is x = -4. Indeed, if you replace x by -4 in the equation, you get a true statement:

314 -=+-

13.3 (Solving) To solve an equation means to write a sequence of equivalent equations until you isolate the variable. Example:

6432 -=-+ xa) regroup the left side (LS) and simplify the right side (RS): 2)32( -=-+x

b) simplify LS: 21 -=-x

c) move -1 from LS to RS: 12 +-=x

d) simplify RS: 1-=x

13.4 (Variable on RS) If the variable appears on the RS of an equation, then use the symmetry property of an equality. Example:

5523

233152

=Û=Û=+Û

Û-=Û-+=+-

xxx

xx

13.5 (Opposite Variable) If -x appears instead of x in the equation, then move -x to the other side. Example:

11

212121

-=Û=-Û

Û=-Û+=Û=-

xx

xxx

13.6 (Brackets) If the equation contains brackets, then simplify and expand (remove) the brackets until you succeed in isolating the variable. Example:

4:4:

51:

51:

512

5)1(2:

5)23(2

5))2(3(2

-==-

=-

=-

=--

=+-

=+--

=---

xmirrorxsimplify

xRStoxmove

xsimplify

xexpand:

xsimplify

xexpand:

x

4;5)2(1)2;11)2;132)

0;32)1;21)0;0)

-=-=---=+=--=+=-

=+=---==+==-

xxfxxxexxd

xxcxxbxxa

2. Solve for x by isolating the variable on the left side and moving all the other terms to the right side (see 13.3):

530)2153)12)

40302010)3132)53)

51)13)21)

+--=--=--=-

-=-+-+=-+++=+

+=+--=-=+

xixhxg

xfxexd

xcxbxa

3. Solve for x by isolating the variable on the right side and moving all the other terms to the left side (see 13.4):

15105)321)32)

201015)322)211)

55)32)21)

-=---=----=-

-+=-+--=+-+=-

+=-+-=-=-

xixhxg

xfxexd

xcxbxa

4. Use the “simplify and expand” method explained in 13.6 to solve the following equations for x:

)]15(5[105))32(1)]3(4[2)

0)3()23())2(53)

1)2(2)2)1(1)

+---=---=---

=--++-=-

-=--+=--

xfxe

xdxc

xbxa

5. Solve for x using the “working backward” method (the first case is solved for you as an example):

))1)(4(53(2)74(1)52)1)3(2(1)

1)1)5(2(1)01)5()52()

)15(510)3)1(2)

------=--=-----

-=+---=+---

--=--=--

xfxe

xdxc

xbxa

4

51155)1(23)1(3)1(2)

-=Û

Û-=Û-=Û-=--Û--=--Û-=--

x

xxxxxa

Example:

6. Solve for x. One case is solved for you as an example:

|3||2||1|||)|2||1|||)12||)

|5|||)6||1)5||)

02||)0||)1||)

-=-+---=-+-=-

-==+=-

=-==

xixhxg

xfxexd

xcxbxa

Example:

22||213||321|||3||2||1|||) ±=Û=Û-+=Û=+-Û-=-+-- xxxxxi



14. Inequations

18


14.1 (Inequation) An inequation is an inequality containing an unknown number represented by a letter and called variable. Example:

© La Citadelle

52 -³+x

14.2 (Solution) A specific value of the variable that makes the inequation a true statement is called a solution of the inequation. Example:

51 <- x

x = 1 is a solution of this inequation. Indeed, by replacing x with 1, the previous inequation become a true statement:

50511 <Û<-

14.3 (Set Solution) In general, an inequation has more than one solutions. The set solution is the set of all solutions of a given inequation. Example:

Zxx Î-³ ;1

The set solution is:

,...}4,3,2,1,0,1{ ++++-=x

and has the following graphical representation:

0 +1 +2 +3-1-2-3-4 +4

14.4 (Solving) Solving an inequation implies finding a sequence of equivalent inequations until the variable is isolated. Example:

2

2

123

;132

£

³

³+-

Î-³+--

x

x

x

Zxx

0 +1 +2 +3-1-2-3-4 +4

14.5 (Brackets) If an inequation contains brackets, then simplify and expand (remove) the brackets until the variable is isolated. Example:

0

0

415

415

4]1[5

4]23[5

4)]2(3[5

>

<

<--

<--

<+-

<+--

<---

x

x

x

x

x

x

x

0 +1 +2 +3-1-2-3-4 +4

1. Find if the given value of x is a solution of the given inequation (use substitution as explained in 14.2).

1);2(42)3;25)4;03)

2;23)2;12)2;01)

+=--->-+=-³-+=£+-

-=-<+-+=->--=¹+

xxfxxexxd

xxcxxbxxa

2. On a number line, graphically represent the set solution of each inequality:

12)32)0)1)2)

33)31)1)3)2)

>-£-³+£³+£-³

+<<-+<->-<->-¹

xandxjxandxixhxgxf

xexandxdxcxbxa

3. Solve for x by isolating the variable on the left side:

15102030)5321)5231)

23)12)02)

-³-++--£---³-+

-<-+>+¹-

xfxexd

xcxbxa

4. Solve for x by mirroring the inequation:

xhxgxfxe

xdxcxbxa

³-³£+³-

<<->-¹-

3)0)5)7)

0)5)1)1)

5. Solve for x by isolating the variable on the right side, simplifying, and then mirroring the inequation (the first case is solved for you as an example):

10515105)5385)3274)

321)53)40)

++-³+-+-£+--+³-

+-<-->-+¹

xfxexd

xcxbxa

444040) -¹Û¹-Û¹-Û+¹ xxxxaExample:

6. Solve for x by moving the variable to the other side, simplifying and eventually mirroring the inequality:

1510105)4219)1351)

531)415)30)

+--³+-+-£+---³-

+-<--->+---¹

xfxexd

xcxbxa

7. Solve for x using the “expand the brackets and simplify” method (see 14.5):

)]25(10[155))25(3)]1(2[1)

2)5()23())5(55)

3)1(5)5)2(3)

---->-+---³-----

-£--+--+-<-

->---¹--

xfxe

xdxc

xbxa

8. Solve for x using the “working backward” method (the first case is solved for you as an example):

]1)43([)]1(3[)]52(1[)0]2)3([)75()

123)2())52(41)

72)3(5))3(23)

+------>---³+-----

+-£+---+--<-

->----¹-

xfxe

xdxc

xbxa

835

53)3(5)3(23)3(23)

¹Û+¹Û

Û¹-Û--¹-Û--¹--Û--¹-

xx

xxxxa

Example:



15. Multiplication of Integers (I)

19


1. Use the short notation for the following additions:15.1 (Shortcut) Multiplication was initially designed as a shortcut for addition of like terms:

© La Citadelle

anaaaatimesn

´=++++ 44 344 21 ...

n is called the multiplier, a is called the multiplicand, and the result of the multiplication operation is called the product. Example:

1025222225

=´=++++ 44 344 21times

15.2 (Commutativity) The multiplication operation is commutative (the multiplier and the multiplicand can be interchanged without affecting the product):

abba ´=´

Example:

1555553

153333335

3

5

=++=´

=++++=´

43421

44 344 21

times

times

15.4 (Multiplying integers with opposite signs) The product of two integers with opposite signs is negative:

)()()( baabba ´-=´-=-´

Example:

)32(6)3(2:

)3(2)3()3(

6)3()3(

´-=-=-´

-´=-+-

-=-+-

So

15.5 (Multiplying negative integers) The product of two negative integers is positive:

baabba ´=-´-=-´- )()()()(

Example: 12)43()4()3( +=´+=-´-

15.6 (Multiplying by 0) The product of any number and 0 is 0:

000 =´=´ aa

Example: 0)7(00)7( =-´=´-

15.7 (Multiplicative Identity) The product between a number and 1 is equal to that number (1 is called the multiplicative identity):

aaa =´=´ 11

Example: 3)3(11)3( +=+´=´+

15.3 (Multiplying positive) The product of two positive integers is positive:

)()()()()( baabba ´+=+´+=+´+

7777777)111111))2()2()2()

33333)555)111)

--------------+-+-

-----++++

fed

cba

2. Use the expanded notation for the following multiplications:

)18(3))1(1))10(5))4(3))3(5)

03))5(5))1(2))5(4)23)

+´-´-´-´-´

´+´-´-´´

jihgf

edcba

3. Use the commutativity property of multiplication to expand the following multiplications in two different ways (see 15.2):

20)11)33)43)25)

35)26)24)32)51)

´´´´´

´´´´´

jihgf

edcba

5. Multiply the following integers with opposite signs (see 15.4):

)6()5()2)10()3)4())3(3))1(1)

)2(6))2()2()1)3())5()1())3(2)

+´-´-´--´-´

-´+´-´--´+-´

jihgf

edcba

6. Multiply the following negative integers (see 15.5):

4. Multiply the following positive integers (see 15.3):

)5()6())10()5()1010))2()3()

)10(3)5)3())5(2)11)

+´++´+´+´+

+´´++´´

hgfe

dcba

)5()5())2()10())9()1())4()3())4(5)

)4()2())2()5())1()3())3()2())1(1)

-´--´--´--´--´-

-´--´--´--´--´-

jihgf

edcba

7. Multiply the following integers (see 15.6 and 15.7):

)10(1)01)110))1(1)

11)00))5(0))1(0)

-´´´--´

´´+´-´

hgfe

dcba

8. Multiply the following integers:

1)111())2468(0))1()531())1()123()

)50()6())50()6())20()6()66)

)25()5())40()5())2()30())20()5()

)10()10()10)10())10(10)1010)

)4()5()4)5())4(5)45)

´--´-´+-´-

-´+-´-´-´

+´--´++´--´-

-´-´--´´

-´-´--´´

tsrq

ponm

lkji

hgfe

dcba

9. Find the value of each expression:

|0|)5432()0|5|)|3||3|)|2|1)

|)1()333(|)|)5(0|)|)4()2(|)|)3(2|)

´-´--´--´

-´--´+´--´

hgfe

dcba



16. Multiplication of Integers (II)

20


1. Complete the following multiplication statements:16.1 (Divisors) The operation of multiplication has a special meaning when the multiplicand and the multiplier are integers:

© La Citadelle

cba =´

a and b are called factors or divisors of the product c.c is a multiple of a and b.Example:

16.2 (Associativity) The operation of multiplication has the property of associativity (the order in which 3 numbers are multiplied is not important):

15)5()3( -=+´-

-3 and +5 are factors or divisors of -15.-15 is a multiple of -3 and +5.

cbacbacba ´´=´´=´´ )()(

By convention, if more than one multiplication operations appear, the operations are done in the order they occur, from left to right. Example:

24)4()6()4()2()3( +=-´-=-´+´-

16.3 (Distributivity) The multiplication operation is distributive over the addition operation:

cabacba ´+´=+´ )(Example. The expression:

)]4()5[()3( -++´-

can be evaluated in two different ways:

îíì

-=+-=-´-++´-

-=+´-

31215)4()3()5()3(

3]1[)3(

16.4 (Positive Factor) If the number in front of the bracket is positive, the signs of the final terms are coincident with the signs of the terms inside of the brackets:

eadacabaedcba ´-´+´-´=-+-´ )(

16.5 (Negative Factor) If the number in front of the bracket is negative, the signs of the final terms are opposite to the signs of the terms inside of the brackets:

dacabadcba ´+´+´-=--´- )(

îíì

=++-=´+´+´-

=-´-=

=--´-

93126134323

9)3(3

)142(3Example:

îíì

=+-=´+´-´

=´=

=+-´

81086524232

8)4(2

)543(2Example:

)()12(36))()15(60))()10(100)

0)()1())(11)12)()3()

)()5(10))()4(16))(420)

´-=-´-=+´-=-

=´-´=--=´-

´-=+´-=´=-

ihg

fed

cba

2. Use the associativity property (see 16.2) to calculate easily the product:

2)15()75()4())4()5(6)25())4()5(15)12()

4)5(3)25())5()10(2)11()2)15(4)2()

5)2(9))5()7()4()5)3(2)

´-´-´--´-´´--´-´´-

´-´´--´-´´-´-´´-

´-´-´+´-´-´

ihg

fed

cba

3. Use the distributivity property (see 16.3) to rewrite the following expressions by expanding the brackets. Then evaluate the final expression.

)152()4())53()2())25(1)

)423(5))52()3())43(2)

)]4()1()3[()2())]5()2[()3())43(2)

+--´--´-++´-

-+-´++´+--´

-++--´+-++´-+´

ihg

fed

cba

4. Use the distributivity property (see 16.3) to factor out the common factors, then evaluate the final expression. One case is solved for you as an example:

42532522525)315215)410310210)

113112511111)243221)223212)

275717)105310)5515)

5443)5232)3121)

´+´-´+-´-´-´+´-´

´-´+´-´´-´+´-´+´-´-

´+´-´´-´´+´-

´-´-´-´´+´

lkj

ihg

fed

cba

Example:

4)2(2)431(2423212243221) -=-´=-+-´=´-´+´-=´-´+´-h

5. Evaluate each of the following expressions in two different ways (see 16.3, 16.4 and 16.5):

)321()5())152()3())51(2)

)213(2))31()2())43(2)

)]3()2()1[()4())]3()2[()1())32(1)

+--´-+-´-++´-

-+-´+-´--´-

--++-´+-++´-+´

ihg

fed

cba

6. Rewrite one of the factors of multiplication to calculate the product more easily. One case is solved for you as an example:

)51()19())21()19())4(51)

)6(102)5)39()993)

+´--´--´

-´´-´

fed

cba

Example:

969311000

5120100051)20(1)20(5051)20()150(

151)20(51)120(5151)120()51()19()

-=+-=

=+--=+-´+-´=+-´+=

=´+-´=+-´=´+-=+´-f



17. Order of Operations (II)

21


1. For each of the following expressions, analyze each operator and decide if it is an unary or binary operator:

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17.1 (Unary & Binary Operators) In mathematics, the symbols + and - are used as unary operators to express the sign (of a number or of opposite) and as binary operators to express the addition and the subtraction operations. Example:

)4()3()2(1 --+´-+-

First, second and forth - operator are unary. Third - operator is binary.First + operator is binary and second + operator is unary.The x operator is always a binary operator.The brackets are used here to express the sign of numbers, not to change the order of operations.

17.2 (Default Order of Operations) If an expression contains additions, subtractions and multiplications but not grouping symbols then the default order of operations is:a) do all the multiplication operations in the order they appear (from left to right)b) do all the additions and subtractions in the order they appear (from left to right)Example:

347

)4()7()4()6(1

)4()3()2(1

-=+-=

=---=---+-=

=--+´-+-

17.3 (Order of Operations Algorithm) If an expression contains grouping symbols then use the following algorithm to find the value of the expression:1. Identify the innermost bracket(s)2. Perform the operations inside the innermost bracket(s) using the default order of operations3. Replace the innermost bracket(s) with the result you got on step 24. If there are still more grouping symbols then go back to step 15. If there are no more grouping symbols, then perform the operations of the remaining expression using the default order of operations.Example:

75935)1(]9[3

5)1(]9162[3

5)1(]9)4(42[3

5)43(]9)51(42[3

-=+--=+-´+-=

=+-´-++-=

=+-´--´-+-=

=+-´--´-+-

3655)1)6()5()5(6)

67))5(())5()3()2()1()

+´--+-´+--´

´-+--+--´++-

dc

ba

2. Use the “left to right” rule (see 17.2) to calculate the value of the following expressions containing more than one multiplications:

)4()3(5)2(10)2)1()2(4)6()5)4(3)2(1)

)10(0)5(10)1003)2(0)0)4()5(1)

)2()2()2()2())2()3()1(2))3()2()1()

)3()2(1)3)2(1)321)

-´-´´-´´-´-´´-´-´´-´

-´´-´´´-´´-´-´

-´-´-´--´-´-´-´-´-

-´-´´-´´´

lkj

ihg

fed

cba

3. Use the default order of operations (see 17.2) to calculate the value of the following expressions (brackets are used here to express the sign of integers):

254321)2)4(3)2()1())4()3(21)

4321)43)2(1)4321)

4321)4321)4321)

3)2(1)321)321)

321)321)321)

´+´-´´-´--´--´-´+-

´-+--+-´+´´-

+-´-´´-´-´

+-´-+´-´--

-´+´´+

onm

lkj

ihg

fed

cba

4. Use the order of operations algorithm (see 17.3) to calculate the value of the following expressions:

2)54)32((1)2)4())32()1(())43()21()

4)32(1))4()32(1))43()21()

4)32(1)43)21()4)32(1)

)3)2(1())32(1)3)21()

)32(1))32(1)3)21()

´+´-´´-+--´---´+-

´-+--´+-´+-´--

+-´-´´-´-´

+-´-+´-´--

-´+´´+

onm

lkj

ihg

fed

cba

)))223(53(23(32)4)5()2)1()2)5)3(2(((()

1)1(}3]13)212()1()2[(4{]2)1()34[(2)

)1()3)2(2()24(2)21()2()121()

42)12()4)4(3)3(2()

´-´--´-´-´--´--´-+-´-

--´+-´´-´-´-´---´+-´-

-´+-´´+-´´+-´-´´-

-´--´+-´--´

s

r

q

p

)1()2(3)2(3212)1()2(4)2()1(3)4(5)

)5(2)2(2)2()4(3)2()1()2(321)

)4(21)5(4)4(3)3(2)

-´-´--´´´+´-´-´-+´-´+-´-

-´+-´´---´´-+-´-´´-

-´´--´+-´--´

r

q

p

5. Calculate the value of the following expressions:

|)2(3||655|)|3|2|)2()5(|))53(|53|)

|5||3||2||1|)1|)6()5(||)5(6|)|32|2)

-´-´---´--´+--´-

+--´++-+-´+--´-´

fed

cba



18. Division of Integers (I)

22


1. Change each of the multiplication statements to an equivalent division statement:

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18.1 (Division) The division operation was initially designed as the opposite operation of multiplication. So, a multiplication statement like:

cba =´

is considered equivalent to any one of the following division statements:

a

cbor

b

ca ==

Example:

ïî

ïí

ì

-

-=+

+

-=-

Û-=+´-

3

124

4

123

12)4()3(

18.2 (Definitions) The division operation is an operation between a dividend and a divisor to obtain a result called quotient:

quotientdivisor

divident=

18.3 (Sign Rules) The sign rules for division are similar to the sign rules for multiplication:

b

a

b

a

b

a

b

a

b

a

b

a

b

a

b

a

+=-

--=

+

-

-=-

++=

+

+

Examples:

76

42

6

424

5

20

5

20

53

15

3

152

5

10

5

10

=+=-

--=-=

+

-

-=-=-

+=+=

+

+

18.4 (Division Operators) There are four operators used to express the division operation:

bababab

a:/ ¸

Examples:

18.5 (Division by 1) Any number divided by 1 is equal to the initial number:

aa

=1

18.6 (Division by 0) Division by 0 is not defined.

definednota

=0

6)3(:)18(5)2()10(

3)2/()6(24

8

-=-++=-¸-

-=+--=+

-

000)00)1()5)3(15)1025)

11)1()4)2(8))3()3(9))1(12)

=´=´-´-=--=´-

-=´-´-=--´-=-´=-

hgfe

dcba

2. Change the division statements to equivalent multiplication statements:

1010

100)0

2

0)2

3

6)

5

102)2

4

8)

45

20)10

1

10)0

1

0)4

3

12)1

1

2)

=-

-=

-=

-

--=--=

-

-=-

-=-

=-=-

-=-

jihgf

edcba

3. Find the value of each expression (see 18.3):

5

25)

6

12)

5

15)

3

9)

4

16)

3

9)

6

18)

5

20)

4

4)

2

0)

5

10)

2

2)

-

-

-

-

-

-

-

-

-

----

-

lkjihg

fedcba

4. Find the value of each of the following expressions containing different types of division operators (see 18.4):

)5(:25))2(8)10/)20()3

9)

)5(:5))4()8())2/(6)1

3)

--¸-+

-

-+¸---

+

hgfe

dcba

5. Find the value of the following special cases of integer division (see 18.5 and 18.6):

0

0)

7

0)

1

6)

0

2)

1

5)

11

11)

10

10)

1

1)

5

5)

5

5)

1

3)

3

3)

lkjihg

fedcba

--

+

-

+

-

----

-

-

++


25

625)

125

1000)

150

600)

40

200)

50

250)

11

121)

50

100)

25

75)

10

50)

5

45)

12

36)

7

28)

---

+

-

+

-

----

-

-

+

-

+

-

lkjihg

fedcba

7. Find the value of each expression containing the absolute value function. One case is solved for you as an example:

5

15)

11

0)

|5|

|0|)

5

|10|)

4

20)

|3|

18)

4

|12|)

|5|

|15|)

|5|

5)

3

|3|)

----

---

-

-

-

--

-

-

-

--

jihgf

edcba

Example:

225

10

5

|10|) =-=

-=

-

-g



19. Division of Integers (II)

23


1. Write the following ratios of integers as rational numbers in lowest terms:

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19.1 (Rational Numbers) The set of integers Z is not closed under the operation of division. This means that the quotient of two integers might not be an integer. In these cases the quotients of two integers define more complex numbers in mathematics than integers called rational numbers. Example:

3

7

+

-

19.2 (Lowest Terms) The quotient of two integers can be simplified by dividing both the dividend and the divisor by common factors until you get the quotient in lowest terms. Example:

3

4

2)6(

2)8(

6

8

3)18(

3)24(

18

24

+

-=

¸+

¸-=

+

-=

¸+

¸-=

+

-

19.3 (Signed Decimals) If the quotient of two integers is not an integer, you can express the quotient using a signed terminating or non-terminating decimal.

6.2...666.23

56.2

5

13-=-=

+

--=

-

+

19.4 (Multiples) Given an integer, you can find its multiples by multiplying the integer by +1, -1, +2, -2, +3, -3, ....So, the multiples of -3 are:

,...12,12,9,9,6,6,3,3 +-+-+-+-

19.5 (Exact Division) If the dividend is a multiple of the divisor, the quotient is an integer. In this case the dividend is divided exactly by the divisor. Example:

43

12-=

-

+

19.6 (Remainder) If the dividend is not a multiple of the divisor, the division has a quotient and a remainder. The quotient is an integer and the remainder is a natural number less than the absolute value of the divisor. Here is the division statement and some examples:

||0; drrdqDord

rq

d

D<<+´=+=

1)2(59152

9

12)5(9152

9

1)2()4(9142

9

1249142

9

+-´=-=-

-

+´-=--=-

+-´-=-=-

+´==

becauseR

becauseR

becauseR

becauseR

16

18)

16

24)

10

25)

12

30)

12

20)

6

8)

18

32)

15

20)

4

6)

6

16)

4

14)

4

3)

-

-

-

-

-

-

-

-----

-

lkjihg

fedcba

2. Express the following ratios of integers as signed terminating decimals (see 19.3):

20

5)

16

24)

10

35)

24

6)

25

10)

15

3)

10

2)

10

7)

8

5)

12

3)

5

2)

4

1)

-

--

-

-

--

-

-

-

-

--

-

lkjihg

fedcba

3. Express the following ratios of integers as signed non-terminating repeating decimals (see 19.3):

90

15)

24

8)

15

5)

6

2)

44

4)

12

2)

15

20)

12

4)

6

4)

3

1)

---

-

---

-

--

-

-

-

jihgf

edcba

4. For each given integer, find 5 positive and 5 negative multiples:

10)8)4)5)2)

7)1)6)5)2)

+++-+

---+-

jihgf

edcba

8. Find the quotient and the remainder for each case (see 19.6):

4

15)

4

15)

4

15)

4

15)

3

7)

3

7)

3

7)

3

7)

-

--

-

-

--

-

hgfe

dcba

7. Rewrite each of the following division statements (see 19.6):

345

17)34

5

17)23

5

17)23

5

17)

334

9)33

4

9)12

4

9)12

4

9)

RhRgRfRe

RdRcRbRa

=-

--=

--=

-=

=-

--=

--=

-=

6. Find the quotient and the remainder for each case:

2

15)

3

15)

4

15)

7

15)

3

11)

5

9)

4

7)

2

5)

hgfe

dcba

5. Rewrite each of the following division statements:

1102

21)43

5

19)25

3

17)13

3

10)

324

11)03

3

9)12

4

9)12

3

7)

RhRgRfRe

RdRcRbRa

====

====



20. Order of Operations (III)

24


1. Use the “left to right” order of operations to find the value of each of the following expressions (see 20.1):

© La Citadelle

20.1 (Consecutive Divisions) If more divisions occur consecutively within an expression, the divisions must be done one by one in the order that they occur (from left to right). Example:

20.2 (Brackets) Use brackets to specify a particular order of operations. Example:

2)2()4()2()6()24( -=-¸+=-¸-¸-

8)3()24()]2()6[()24( -=+¸-=-¸-¸-

20.3 (Consecutive Multiplications and Divisions) Multiplications and divisions are considered operations of equal priority. If multiplications and divisions appear consecutively, perform the operations in the order that they occur (from left to right). Brackets can be used to specify a different order. Examples:

6323)12()24(

3)12()4()6(

=´=´-¸-=

=´-¸+´-

4222)3()6(

2)]4()12[()6(

=´=´-¸-=

=´-¸+¸-

20.4 (Default Order of Operations, Updated) Multiplication and division are considered to have higher priority than addition and subtraction. If an expression contains addition, subtraction, multiplication and division but not grouping symbols, then the default order of operations is:a) do all the multiplication and division operations in the order that they appear (from left to right)b) do all the addition and subtraction operations in the order that they appear (from left to right). Example:

20.5 (Order of Operations Algorithm) If an expression contains grouping symbols, use the Order of Operations Algorithm presented in 17.3 and the Default Order of Operations presented above (20.4). Example:

2)6()2(6

)2(33)6(32

-=---+-=

=-´-¸-+´-

244284]14[2

59]12)14(3[2

)5())3(6(

]1)2()2()7(3[2

2)10())4(1232(

]1)3(6)24()321(3[2

-=+-=+´-=

=-+----´-=

=-+--+

+--+-´-´-=

=¸-+-¸-´+

+--¸+¸-´´+-´-

)1()2()3(236)5)1()3(75))5()2()1(40)

)1()2()2(32)2)2(264)2)3(224)

2)5(50)5)2(20)2)2(12)

-¸-¸-¸¸-¸-¸-¸-¸-¸-¸

-¸-¸-¸¸-¸¸¸-¸¸-

¸-¸-¸-¸-¸-¸

ihg

fed

cba

2. Use the “left to right” order of operations to find the value of each of the following expressions (see 20.3):

)2()8()6()4(1)3)2()3(2)2)3()6(3)

)1()1()1(5)2)2(216)3)1(212)

2)5()1(5)2)5(10))2()3(4)

-´-¸-´-´´-¸-´¸-¸-´-

-¸-´-¸¸-´¸´-¸¸-

´-¸-´´-¸--¸-´

ihg

fed

cba

3. Start with the inner-most brackets and find the value of each expression (see 20.2):

)5(25

)5(10

)12(24

)3()12()

)33(94

)2(2)3()

)]2(5[20

6)4()3()

4

)4(16

)2(2

)12()1()

)2(4

1612))2(

4

82)

)]2()4[()]3(8[)]4)5[(20))]2()8[(16)

-¸

-¸¸

-¸

-¸-

´-¸´

-´´-

-´¸

¸-´--

-¸¸

-´

-´-

-´

-¸--¸

-

´-

-´-¸-´´-¸--¸-¸

ihg

fed

cba

4. Use the “default order of operations” to find the value of each of the following expressions (see 20.4):

1)1()1()1(1))3(24)8(2))1(3)12(24)

34)5(2054)5)2(3)5(15)13)2(101)

5)1()3(210))2(3412))2(4321)

--¸-+-´--´-¸-´-´+-¸

´+-¸+´-+-´+-¸+--¸´-

+-¸-´--´+¸--¸-´+

ihg

fed

cba

5. Use the “order of operations algorithm” (see 20.5) to find the value of each of the following expressions:

)2(23

)4(163

1)2(2

)6(22)

1)2()1(

1)3(122))2(

5

4231)

)1)2(2(]4)2(3[)2)333(12)1)5()321()

-´+

-¸+-

--´

-´+-

+-´-

+-¸+--´

-

´--

--´¸--´--´+-¸-+-¸´-

fed

cba

)2(232

)5(204

53

8))2104()]2(45[)]27(15[)

32)5(15

6)3(21)2()]5)2)6(2[()53()]43(2[1)

-´+´-

-¸+-+

-

-¸-´-´+¸+-¸--

-÷÷ø

öççè

æ

+-¸

¸-´+-´-¸´-+¸+-´---

ji

hg

)]3(3[5

15

3

152

244

)2(23

3)5(3

2

10

)2(2

153

45

2102

24)3(

)

3)5(2

2852

21

3)1315(

)3(91

3)554(

)3()9(

1)1322(

2

254)

5)254)5(3()1)3(9()22210()4832(2)

-´¸-

´+´-

-´-+

-´-

+-´

--

¸

-´

-´-

´-´

¸--

+´-

+-´

¸+´¸

--

´+¸-

-¸+-

--´´

-¸-

-+´+--+

-

¸´

¸¸´--´++-¸´´+¸-+¸+---

m

l

k



21. Equalities and equations

25


1. All the following equalities are true statements. Multiply or divide both sides of each equality by the given integer and check if the statement remains true (see 21.1).

© La Citadelle

21.1 (Multiplying and Dividing) By multiplying (or dividing) the left side and the right side of an equality by the same number, you get an equivalent equality:

0;

0;

¹=Û=

¹´=´Û=

cc

b

c

aba

ccbcaba

true

false

;113

69

3

3693

;126

)3()4()3()2(42

-=-Û-

-=

-Û-=

+=-Û

Û-´-=-´+Û-=+Examples:

21.2 (Unknown Factor) Solution 1. Rewrite the multiplication as an equivalent division:

a

bxbxa =Û=´

42

88)2(82 +=

-

-=Û-=´-Û-=´- xxx

21.3 (Unknown Dividend) Solution 1. Rewrite the division as an equivalent multiplication:

baxba

x´=Û=

12)3()4(34

-=+´-=Û+=-

xx

21.4 (Unknown Divisor) Solution 1. Rewrite the division as an equivalent multiplication and then back as an equivalent division:

b

axxbab

x

a=Û´=Û=

46

24)6(246

24-=

+

-=Û´+=-Û+=

-xx

x

Solution 2. Divide the equality by the second factor:

a

bx

a

b

a

xabxa =Û=

´Û=´

Solution 2. Multiply the equation by the divisor:

baxbaa

xab

a

x´=Û´=´Û=

Solution 2. Multiply the equation by the divisor and divide the equation by the initial quotient:

63

18;

3

18

)3(

)3(;18)3( -=

-

+=

-

+=

-

´-+=´- x

xx

18)9()2(

);9()2()2(

)2(;92

-=+´-=

+´-=-

´-+=-

x

xx

b

ax

b

xb

b

axbaxbx

x

ab

x

a=

´=´=´=´= ;;;;

2;4

4

4

8;48;4

8;4

8=

´=´=´=´= x

xxxx

xx

6;3)3(5)3(4)1;3)5()105(4)

2;4)3(22)3;321)

-+-´=+´--+-¸-=

-+-´=---=-

bydividedbymultiplyc

bydividebbymultiplya

2. Solve each of the following equations for x by replacing each multiplication statement with an equivalent division statement (see 21.2):

)3(0)520))3(9)210)

147)124)8)4()153)

-´=´-=--´=´-=-

-=´=´-=-´-=´

xhxgxfxe

xdxcxbxa

3. Solve each of the following equations for x by dividing both sides by a convenient integer (see 21.2):

)5(0)525))3(18)321)

16)4()153)12)2()41)

-´=´=--´=-´-=-

-=-´-=´--=-´-=´-

xhxgxfxe

xdxcxbxa

4. Solve each of the following equations for x by replacing the division statement with an equivalent multiplication statement (see 21.3):

11)

37)

71)

65)

010

)44

)45

)43

)

-

-=-

-=

+=-

-=

=+=-

-=-=-

xh

xg

xf

xe

xd

xc

xb

xa

5. Solve each of the following equations for x by multiplying both sides by a convenient integer (see 21.3):

33)

25)

34)

23)

03

)62

)43

)31

)

-

-=-

-

-=-

-=

-=-

=-

-=-

-=-

+=-

xh

xg

xf

xe

xd

xc

xb

xa

6. In order to solve each of the following equations for x, rewrite the division statement as a multiplication statement and then back as a division statement (see 21.4):

xl

xk

xj

xi

xh

xg

xf

xe

xd

xc

xb

xa

-==-

-=

-=-

-=-

-

-=+

-=-=-

+=-

-=-

--=+=

-

287)

255)

82)

11)

369)

505)

217)

183)

624

)525

)315

)15

)

7. In order to solve each of the following equations for x, multiply and then divide both sides by convenient integers (see 21.4):

xl

xk

xj

xi

xh

xg

xf

xe

xd

xc

xb

xa

-=-

-==-

-=-

-=-

-

-=-

-=-=-

-=-

+=-

-+=

-+=

-

131)

3003)

369)

306)

408)

217)

3311)

162)

530

)20100

)735

)11

)



22. Proportions and Equations

26


1. Use cross-multiplication and division to solve for x the following equations (see 22.2):

© La Citadelle

22.1 (Proportions) A proportion is an equation with a ratio on each side:

d

c

b

a=

22.2 (Solving a proportion) To solve a proportion means to find one term of the proportion when three other terms are given. Use cross-multiplication and division to isolate the unknown term:

b

cax

b

xb

b

caxbca

c

b

x

a ´=

´=

´´=´= ;;;

Example:

14;5

90

5

5;905;

30

3

5-=

-=

´-=´=

-x

xx

x

22.3 (Shortcut for a proportion) To find a term of a proportion multiply the adjacent terms and divide by the opposite term:

a

cbd

b

dac

c

dab

d

cba

d

c

b

a

´=

´=

´=

´=

Û=

;;;

Example:

203

60

3

5)12(

53

12-=

-=

´-=Û=

-x

x

22.4 (Other equations) The main idea in solving an equation is to isolate the variable using equivalent equations.Example 1:

95445

2

858)5(2

311)5(211)5(23

=+=Û=-Û

-

-=-Û-=-´-Û

+-=-´-Û-=-´--

xx

xx

xx

Example 2:

52

10

1

210210

2

4104)2(10

354)2(10

43)2(105

-=-

=Û=-

Û=¸-

Û-

-=¸-Û-=-´¸-

Û--+=-´¸-

Û+=+-´¸-

xx

x

xx

x

x

Example 3:

42

8

1

28

248

3

122

8

1

3

28

12

3

9

28

153

-=-

=Û=-

Û-=-

Û=+-

Û=

+-

Û-

-=

+-

+-

xx

xx

xx

210

0)

812

3)

4

25)

8

4

2)

-=

--=

--=

-

-=

-

xd

xc

xb

xa

2. Use cross-multiplication and division (the shortcut method explained at 22.3) to solve for x the following equations:

51

0)

63

2)

2

315)

5

5

1)

+=

--=

--=

-

-=

-

xd

xc

xb

xa

3. Use equivalent equations to solve for x the following equations:

1)32(410

1)5(3

)2(102

2)4(3)

3)2313(2]2412[)]3()32(5[)

)32(1

18

1

3)

10

5

6

)2(25)

35)3(2)21)1(3)

25

8

5

20)

3

6

16

4)

9

6

3

13)

5

53

10

6)

4

2

42

5)

2

4

2

3)

)5(25)146)222)

2512)534)932)

++-´--

--´-=

-¸--

¸-´-

´+´---=+¸-¸-´+´-

-¸-

-=

-

-=

-

-´-

-=-+¸-+=--´-

+¸

-=

--=

+¸

-

-

-=

-¸-

´-=

--=

+´

-

-

-=

-

-¸-=+-=-¸--=+¸

+´-=-=´--=-´

x

r

xq

xp

xo

xnxm

xl

xk

xj

xi

xh

xg

xfxexd

xcxbxa

4. Solve for x using the “working backward” method. One case is solved for you as an example.

3

3

3510

2)1)2()2(36)

06)7()318()3()32)4(2)1()

0415)4(5)054512)

5)4(5)8)4(10)

-=

+¸¸

--=-¸¸-¸

=+-¸¸-´¸-=-´´´-

=-¸´-´=+¸´¸-

-=-¸´-+=-´¸

xhxg

xfxe

xdxc

xbxa

Example:

52

110

1

210210)4(8108)4(10)

-=-

´=Û

Û-

=Û-=¸Û-¸+=¸Û+=-´¸

x

xxxxa

xh

xg

xf

xe

1

1

1)

82

4)

6

93)

8

10

4)

-=

+

-=

++

-=

--=

-

-

xh

xg

xf

xe

3

10

15)

204

5)

3

28)

14

4

7)

-=

+

-=

+

-

+

-=

-

-

-=

+

-



23. Inequalities and Inequations (II)

27


© La Citadelle

23.1 (Positive Numbers) By multiplying (or dividing) an inequality (or inequation) by a positive number you get an equivalent inequality (or inequation).Example 1 (all statements are false):

215

10

5

5105 +>-Û

+

+>

+

-Û+>-

Example 2 (all statements are true):

86);4()2()3()2(;43 -³--´+³-´+-³-

23.2 (Negative Numbers) By multiplying (or dividing) an inequality (or inequation) by a negative number you get an equivalent inequality (or inequation) only if you change its sense. Examples (all are true):

433

12

3

9129

632)3(1)3(21

£Û-

-£

-

-Û-³-

->-Û´->´-Û<

23.3 (Inequations) The main idea in solving an inequation is to write a sequence of equivalent inequations until you can isolate the variable. Examples:

23.4 (Inequations with Ratios) If you multiply an inequation by an expression containing the variable, you have to separately analyze the cases when the expression is positive or negative.

15);15()1()()1(;15

);3()5(5

)5(;35

.2

42

8

2

282.1

-³+´-³-´-+£-

-´-£-

-´--³

-

-

+<Û-

-<

-

´-Û->´-

xxx

xx

xx

x

1

62.2

+-

+>-

xEx

101.1 <Û>+- xxCase

solutionnoxandx

xxx

xxx

Þ><

>-<+-+>-´+-

+-

+´+->-´+-

41

4;31;6)2()1(

;1

6)1()2()1(

101.2 >Û<+- xxCase

}3,2{41

4;31;6)2()1(

;1

6)1()2()1(

++=Þ<>

<->+-+<-´+-

+-

+´+-<-´+-

xxandx

xxx

xxx

}1,2,3,4,5,6{60

;6;60

)(6;60

16

;3

318

3

1;3

18

;9

3)9(

2)9(;

9

32.1.

------=Þ-³<

-³£-<

-£³->

³-

-

-³´

--£

-

-´-£

-´-

-

-³

-

xxandx

xxthenxIf

solutionnoxxthenxIf

xxx

xxEx

1. All of the following inequalities are true statements. Multiply or divide both sides of each inequality by the given positive integer and check if the statement remains true (see 23.1).

2;4)6()3(4)3;3)4(120)

3;5)1(26)2;321)



-³-¸-´-+-¸-£

+-´>++-<-

2. All of the following inequalities are true statements. Multiply or divide both sides of each inequality by the given negative integer and change the sense of the inequality. Check if the statement remains true (see 23.2).

2;2)3()3(2)3;1)5(100)

5;15)10(25)2;432)

--³-¸-´-+-¸-£

-+-´>+--<-



3. Solve the following inequations for x by multiplying or dividing both sides by a convenient integer (see 23.3). Look only for integral solutions of x (that is , solutions where x is an integer).

110)

51)

10)

11)

30)

25)2

5)1

2)

48)36)102)93)

-£

->-<

-³

-³

->--£

--<

-

´<-´-£-³´-->´

xl

xk

xj

xi

xh

xg

xf

xe

xdxcxbxa

4. Solve the following inequations for x. Find only the integral solutions of x (x is an integer). An analysis of the sign of x might be necessary (see 23.4).

xl

xk

xj

xi

xh

xg

xf

xe

xd

xc

xb

xa

10025)

71)

62)

11)

105)

62)1

2)2

6)

51)2

2)0

2)0

1)

-³-

-³

-³-

-³-

-³-

->--£-<

-

-

->--³³

-<

5. Solve the following inequations for x by isolating the variable (see 23.3). Look only for integral solutions of x (x is an integer).

)62(3

10)

13

102)

3

82)

3

32)1(5)4

3

)1(2)2)1(38)

)2(510)9)32(3)8)2(2)

-´´-

->

+´->-

-

->-

-

+¸->--£

+´--+´<-

+-´-£-³-´´-->-´-

xi

xh

xg

xf

xexd

xcxbxa

6. Solve the following inequations for x. Look only for integral solutions of x (x is an integer). One case is solved for you as an example:

3|12

|)3|1|3)51||)0||)

0|2|)10|2|)0|1|)2||)

2|2|)10|2|)0|1|)2||)

<--->£+<

³-³´>-<

=-=´=-=

xlxkxjxi

xhxgxfxe

xdxcxbxa

Example:

002222202.2

42222202.1

2|2|)

=Û=Û=-Û=-Þ<Û<-

=Û+=Û=-Þ>Û>-

=-

xxxxxxCase

xxxxxCase

xd



24. Powers

28


1. Change the following repetitive multiplications to powers (see 24.1):

© La Citadelle

24.1 (Shortcut) Powers were introduced as shortcuts to repetitive multiplications of a number by itself:

44 344 21timesn

n aaaaa ´´´´= ...

aaaa ´´´´ ...

Example: 3)2()2()2()2( -=-´-´-

24.2 (Expanded and exponential notations) The expanded notation is:The exponential notation is:where a is called the base and n is called the exponent.The full expression represents a power.Examples:

na

na

2

4

)5()5()5(.2

)3()3()3()3()3(.1

-=-´-

-´-´-´-=-

24.3 (Negative exponents) A power with a negative exponent is defined as:

44 344 21timesn

n

n

aaaaaa

´´´´==-

...

11

Example:

8

1

)2()2()2(

1

)2(

1)2(

3

3

-=

-´-´-=

-=- -

As you can see from the previous example, an integer raised to a negative integral exponent is a rational number.

24.4 (Base 0 and 1) 0 cannot be a base for powers having integral negative exponents because division by 0 is not defined. Also 1 can be a base for any integral exponents. This operation doesn’t have an inverse operation (see sections about Roots).

11111

000000

3

4

=´´=

=´´´=

24.5 (Power Operator) The power is a binary operation between a base and an exponent. Usually the base is written normally and the exponent is written as superscript. Sometimes the ^ operator is used to denote a power operation, especially on calculators and in programing languages:

naan ^Û

Example:

9

1

)3()3(

1)2()^3( =

-´-=--

)1()1()1()1()1()1())2()2()2()

)10()10()10()10()10())4()

)5()5()5()5()3333333)

-´-´-´-´-´-+´+´+

-´-´-´-´-+

-´-´-´-´´´´´´

fe

dc

ba

2. Use expanded notation to express the following powers (see 24.2):

325323

243234

)6())7()1))5())10()10)

)8())2())3())1())2()3)

----

----+

lkjihg

fedcba

3. Use expanded notation to express the following powers (see 24.3):

45232

23312

)5()1))4())10()10)

)3()5))1())2()4)

-----

-----

---

--+

jihgf

edcba

4. Use expanded notation to find the value of the following expressions. One case is solved for you as an example.

023232

023232

124533

)10()5))2())2()0)0)

7)10))10())10())10())10()

)5()10))1())1()2)2)

---

----

---

---

---

--

srponm

lkjihg

fedcba

Example:

100

1

)10()10(

1

)10(

1)10()

2

2 =-´-

=-

=- -i

5. Use the ^ operator to express the following powers (see 24.4):

324312 )10()10))5())2()2)3) --- --- fedcba

6. Convert the ^ notation to the regular exponential notation (see 24.4):

)1()^10())2()^5()2)^5())2(^5)2^5) ------ edcba

7. Use the expanded notation to find the value of the following expressions. One case is solved for you as an example.

)3()^2())3()^10()2)^5())2(^10)2^3) ------ edcba

Example:

8

1

)2()2()2(

1

)2(

1)2()3()^2()

3

3

-=

-´-´-=

-=-=-- -e

8. Find the value of each expression containing powers and absolute value functions. One case is solved for you as an example.

|2||2||3|12 |2|)|5|)10)|10|)|2|) ------ ---- edcba

Example:

4

1

2

12|2|)

2

2|2| ===- ---e

24.5 (Exponent 0 and 1) Any number (excluding 0) raised to power of 0 is 1. Any number raised to power of 1 is equal to itself:

aaa == 10 1



25. Exponent Rules

29


1. Simplify the following expressions by adding the exponents (see 25.1):

© La Citadelle

25.1 (Adding Exponents) The product of two or more powers having a common base is a power with the same base and an exponent equal to the sum of exponents:

nmnm aaa +=´

Example:85353 )2()2()2()2( -=-=-´- +

25.2 (Subtracting Exponents) The division of two powers having a common base is a power with the same base and an exponent equal to the difference of exponents:

nm

n

m

aa

a -=

Example:235

3

5

)3()3()3(

)3(-=-=

-

- -

25.3 (Multiplying Exponents) A power raised to another power is a power having the base of the first power and the exponent equal to the product of exponents:

nmnm aa ´=)(

Example:63232 )5()5())5(( -=-=- ´

25.4 (Multiplying Bases) Multiplying two or more numbers and then raising the product to a power is equivalent to raising each number to that power and then multiplying all those powers:

nnnn xbaxba ´´´=´´´ ...)...(

Example:444 )3()2()]3()2[( +´-=+´-

25.5 (Dividing Bases) Dividing two numbers and then raising the ratio to a power is equivalent to raising each number to that power and then dividing the powers:

n

nn

b

a

b

a=÷

ø

öçè

æ

Example:

5

55

)3(

)2(

3

2

+

-=÷

ø

öçè

æ

+

-

25.6 (Simplification) Use exponent rules to simplify an expression containing powers. Example:

2)4(6

4

4

4

6

4

)2(2)2(3

4

223

)2()2()3(

)3(

)2(

)2(

)]3()2[(

)3()2(

6

])3()2[(

----

-

-

-

-

-

-´-´

-

-

-=-=-

-´

-

-=

=-´-

-´-=

-´-

325315223 00)55))10()10())3()3()22) ´´-´--´-´ edcba

2. Rewrite the following expressions as a product of two powers:

132213522123 )1()0)4))10())2()4) ++++++ --- fedcba

3. Simplify the following expressions by subtracting the exponents (see 25.2):

2

2

5

3

4

4

3

5

1

4

2

5

3

3)

)4(

)4()

)1(

)1()

)3(

)3()

)10(

)10()

3

3)

2

2) gfedcba

-

-

-

-

-

-

-

-

4. Rewrite the following expressions as a division of two powers:

122153634132 )1()4)1))10())5()5) ---+--- --- fedcba

6. Expand the brackets (write the expression as a product of two powers as explained at 25.4):

23234 )02())]3(2[))]2()5[())14())32() ´--´-´-´-´ edcba

7. Simplify (write each expression as a single power):

21524322 )5()5()00))10()10())1()1()22) -´-´-´--´-´ edcba

8. Expand the brackets (write the expression as a ratio of two powers, as explained at 25.5):

323524

3

5)

2

2)

3

0)

2

1)

3

2)

3

2) ÷

ø

öçè

æ

-÷ø

öçè

æ

+

-÷ø

öçè

æ÷ø

öçè

æ

-

-÷ø

öçè

æ -÷ø

öçè

æfedcba

9. Simplify (write each expression as a single power):

5

5

4

4

7

7

2

2

5

5

3

3

)3(

2)

)1(

0)

)2(

)6()

)4(

)1()

4

)2()

4

2)

---

-

-

--fedcba

10. Use exponents rules to simplify, then evaluate:

55

7

25

2

23

4

532

2

32

4

32

)5()2(

)10()

)5()5(

)5()

55

5)

)10()10()10())3()3(

)3()3()

2

22)

-´-

-

-´-

-

´

-´-´--´-

-´-´

-

-

fed

cba

5. Simplify (write the expression as one single power as explained at 25.3):

211123232323 ))7(())3()])2[())2())2())2() ------- -- fedcba

32

232

4

23

322

222

)2(16

)8()4()

)2(

)2()2()

])2()6[(

])3()4[()

-´

-´-

úúû

ù

êêë

é

-

-´-

-´-

-´--

--

-

ihg

12

22

222

3242

12

23

324

32

))4((

)2(2

])2()8[(

])4()4[()

)2()2(

)2()2(

))2(()2(

)2()2()2()

----

-

--

-

-

-´´

+¸-

-¸-÷÷

ø

ö

çç

è

æ

-´-

-¸-¸

-´-

-´-´-kj



26. Order of Operations (IV)

30


1. Evaluate the following expressions (see 16.1):

© La Citadelle

26.1 (Negation versus Power) The operation of raising a number to a power has a higher priority than the negation operation (changing a number to its opposite). Examples:

16)2(

16)2(2

4

44

+=-

-=-=-

26.2 (Right to Left Order for Exponents) If an expression contains consecutive raising operations to a power, perform the raising operations from right to left:

cbcbbb aabutaacc ´=Û )()(

Examples:

6422)2(

512222

62323

9)3(3 22

===

===

´

26.3 (Bases or exponents as expressions) If the base or the exponent of a power is an expression itself, first evaluate the base and the exponent separately, then evaluate the power:

)()^( baab ÛExample:

16)2()46()432( 462232 =-=+-=+´- +-´+-

26.4 (Default Order of Operations Updated) If an expression contains additions, subtractions, multiplications, divisions and powers but not grouping symbols, then the default order of operations is:a) do all the powers operations from right to leftb) do all the multiplication and division operations in the order they appear (from left to right)c) do all the addition and subtraction operations in the order they appear (from left to right). Example:

26.5 (Order of Operations Algorithm) If an expression contains grouping symbols, use the Order of Operations Algorithm (17.3) and the Default Order of Operations (26.4). Example:

213618)3()3(218

9)27()3(4892

3)3()3(2832 2322

-=+--=---´+-=

=¸---´¸+´-=

=¸---´¸+´-

4)11(

)11616()116)2(8(

)116)24(8(

)116)484(8(

)1)4()284(2(

2

22

2

)1()2(

)1()53(223

=+=

=+¸=+¸-´-=

=+¸+-´-=

=+¸¸+-´-=

=+-¸¸+-´-

-´-

-´-

111021

004422

10)10))2())123())2())2()

)10()10))2()2))2()2)

----

--

-------

------

lkjihg

fedcba


22222)3(3)3(

333)3(233

)2())2()2))2())2()2)

2))2()2)2))2()2)2222222

22222

lkjihg

fedcba

---

-

---

---


2

2

2

2

39212

331

4322

2]4)32[(2232132

)24(

)48()510()

221

432))2()

])2()31[())228())321()

-

¸-´-÷ø

öçè

æ

-

´-

-´-

¸´--+´-+-

¸-

¸´¸-÷ø

öçè

æ

´--

´+-

-¸--+¸--´+

fed

cba

4. Use the default order of operations (see 16.4) to evaluate the following expressions:

2323422222

2222323222

2)5()2(3)3()241642)3321)

)6()3()2()32432)321)

¸-´--¸-¸-¸´-¸-´-

-¸-´-+¸---+-

fed

cba

6. Use the order of operations algorithm (see 16.4) to evaluate the following expressions:

2

2

23

32

22

3262

332

)22()42(2

2

23

224

323222)3()23(

23222232

32)33(

1)322()43(

222

123)

)2()2(22

2)2()2(2)

])22[()12)4(2

)12()12(2)

3]1)2()2[()]21()53(1[))423(2)

)2323()2510()523()5)3431()

2)32(

÷÷

ø

ö

çç

è

æ

+¸-

-´-´+-´

÷÷

ø

ö

çç

è

æ

+-

--

-´-¸-

+-´--

-÷÷

ø

ö

çç

è

æ

-+-¸

-´+-

--¸¸+-´+-+-+´--

-´+-´+¸-¸+--+´--

-

-

--

-´+-

-

hg

fe

dc

ba

2)13(1

22

4

12)3(

21

53

)2(

)2(2

12

3)

2)223()3()5225()23(2)2()

2

2

422

00

2

3

2

4

221

524

2

2

524322252332223

22

22

-÷÷

ø

ö

çç

è

æ

--

-´

-

-+-+

+

+-

÷÷

ø

ö

çç

è

æ

-

-¸-

-÷÷

ø

ö

çç

è

æ

-

-

-+--+-¸´+---+-

-÷÷

ø

ö

çç

è

æ

--

´-

--

j

i

5. Use the order of operations algorithm (see 16.4) to evaluate the following expressions:

222)22()2)222(2)222)

222)222))222(22)

22222)22222)22222)

22222)22222)22222)

22222

2222222

222

222

´+¸--+´¸-+´

¸-´-¸´++

´+¸+-+´¸--+¸-´

¸´+-´-+¸¸´-+

¸-

´++¸

lkj

ihg

fed

cba



27. Divisors

31


1. Find the logical value (true or false) of each statement:

© La Citadelle

27.1 (Divisors) An integer a is a divisor of the integer b if dividing b by a leaves no remainder. In this case there is an integer c so:

cabca

b´=Û=

27.2 (Definitions) Both a and c are called divisors or factors of a.Both a and c divide a.b is called a multiplier of both a and c.b is divisible by both a and c. For the example above:Both -3 and -4 are divisors of 12Both -3 and -4 divide 1212 is divisible by both -3 and -4

27.3 (Improper Divisors) Given any integer a (except 0 and 1) there are at least 4 divisors of this number that are improper divisors: 1, -1, a, -a. Indeed:

1;1;1

;1

-=-

=-=-

=a

a

a

aa

aa

a

27.4 (Proper Divisors) Any divisor of an integer that is not an improper divisor is called a proper divisor.Example: The proper divisors of -6 are:+2, -2, +3, and -3

27.5 (Prime Numbers) A prime number is a positive integer (a natural number) with no proper divisors. The only divisors of a prime number are 1 and the number itself (if you ignore -1 and the opposite of that number).

Example: 7 is a prime number. The only natural numbers that are divisors of 7 are the improper divisors 1 and 7.

27.6 (Composite Numbers) A natural number that is not a prime number is called composite number. A composite number has proper divisors (different from 1 and the number itself).Example: 12 is composite number. The improper divisors are 1 and 12. The proper divisors are 2, 3, 4, and 6. For this example the negative divisors were ignored.

Example: -3 is a divisor of 12 because:

Example: The improper divisors of -5 are:1, -1, -5, +5.

27.7 (Number 1) Conventionally, number 1 is neither a prime nor a composite number.

)4()3(1243

12-´-=Û-=

-

112)614)

205)83)

224)416)

103)205)

-+-

-++-

+--

-+--

divideshofmultipleaisg

dividesfofdivisoraise

bydivisibleisdofmultipleaisc

dividesbofdivisoraisa

2. Find the improper divisors of each integer:

3)11)20)5)8)10) ---+- fedcba

3. Find the proper divisors of each integer:

16)6)10)15)6)12) ----+- fedcba

4. Find all divisors of each integer:

21)30)20)24)18)9) ----+- fedcba

5. Find all prime numbers between 1 and 30.

6. Find all composite numbers between 31 and 50.

7. Classify each of the following numbers as either prime, composite or neither.

1)57)53)51)41)31)21)11) hgfedcba

8. Use the Sieve algorithm (below) to find the prime numbers between 1 and 100:

Eratosthenes

1. Create the list of all whole numbers between 2 and 100: list A (see the list below).2. Create an empty list and name it: list B (this is the list of prime numbers).3. Move the first number from list A to list B, then remove from list A all that number’s multiples.4. Repeat step 3 until no more numbers are left in list A.

The list of prime numbers is the list B.

100999897969594939291

90898887868584838281

80797877767574737271

70696867666564636261

60595857565554535251

50494847464544434241

40393837363534333231

30292827262524232221

20191817161514131211

1098765432List A:

List B:



28. Divisibility Rules

32


1. For each of the following numbers, find whether or not they are divisible by 2:

© La Citadelle

28.1 (Divisibility by 2) An integer is divisible by 2 if the last digit is 2, 4, 6, 8, or 0. A number divisible by 2 is called an even number. Examples: 24, -12, -30.

28.2 (Divisibility by 3) An integer is divisible by 3 if the sum of all digits is divisible by 3. Examples: -111, 102, 750.

28.3 (Divisibility by 4) An integer is divisible by 4 if the number defined by the last two digits is divisible by 4. Examples: 1020, -132, -144..

28.4 (Divisibility by 5) An integer is divisible by 5 if the last digit is 5 or 0.

28.5 (Divisibility by 6) An integer is divisible by 6 if the number is divisible by 2 and 3. Examples: 102, -222, 1014.

28.6 (Divisibility by 7) Use the following algorithm:a) Drop the last digit and double it b) Subtract the result from the remaining number. c) If the final result is divisible by 7 then the original number is also divisible by 7. Example:28.7 (Divisibility by 8) An integer is divisible by 8 if the number defined by the last three digits is divisible by 8. Examples: 816, -8192, -2000.

28.8 (Divisibility by 9) An integer is divisible by 9 if the sum of all digits is divisible by 9. Examples: 369, -252, 126.

28.9 (Divisibility by 10) An integer is divisible by 10 if the last digit is 0.

28.10 (Divisibility by 11) Alternately add and subtract the digits from left to right. If the number you get is divisible by 11 then the original number also is. Example: 1353.

28.11 (Divisibility by 12) An integer is divisible by 12 if the number is divisible by 3 and 4. Examples: 22344, -132, -8652.

28.12 (Divisibility by 13) Use the following algorithm:a) Drop the last digit and multiply it by 9 b) Subtract the result from the remaining number. c) If the final result is divisible by 13 then the original number also is. Example: 3003.

771279;791 =´-

011

0;03531 ==-+-

013

0;03927

27339300

==´-

=´-












13. Use a calculator and the algorithms in 28.1 to 28.12 to find if:

36)31)112)21)0)14)55) gfedcba +--

12345)130)765)201)1111)222)1234) --+-- gfedcba

12386)148)966)701)510)312)120) --+-- gfedcba

12777)444)465)700)1225)115)220) --+-- gfedcba

9654)888)465)900)1158)135)246) --+-- gfedcba

23345)885)465)861)1152)635)231) --+-- gfedcba

9870)896)890)556)1152)620)248) --+-- gfedcba

5553)666)981)777)108)6588)1234) --+-- gfedcba

999)7890)1000)5555)1230) -+-- edcba

23432)642)22)1111)308)6171)1232) --+-- gfedcba

66)12345)111)222)324)600)240) --+-- gfedcba

74191)100)11)1313)182)3900)130) --+-- gfedcba

012345678998765432109)

12655555666661223334444)

1234513579864206)

11665554449998887776)

765432112345678983)

divided

ofmultipleaisd

ofdivisoraisc

ofmultipleaisb

ofdivisoraisa



29. Prime Factorization

33


1. Use the following factor trees and write down for each case the prime factorization of the number placed at the root of the tree:

© La Citadelle

29.1 (Fundamental Theorem of Arithmetic) Any natural number except 1 can be written as a product of prime numbers in a unique way called the prime factorization of that number. Use exponents to make the expression simpler. Example:

312 5325553221500 ´´=´´´´´=

29.2 (Factor Tree Method) The factor tree method is a method to find the prime factorization of a natural number. The root of the tree is the number itself. For example:

24

Using divisibility rules try to find a divisor of the given number. In our case 2 is a divisor. By dividing 24 by 2 you get 12. So:

2 is a prime factor and so is a leaf of the tree. Let’s continue the process with 12 until we get only prime factors (leaves):

24

2 12

24

2 12

2 6

2 3

All leaves of the tree generate the prime factorization of the given number:

13 32322224 ´=´´´=

29.3 (Uniqueness) Although for a natural number more than one factor tree is possible, the prime factorization of that number is unique. Example:

24

4 6

2 32 2

24

3 8

2 4

2 2

29.4 (Integers) Prime factorization can be extended to integers (except 0, 1, and -1) if only positive factors are considered. Example: -36

36

4 9

3 32 2

22 3236 ´-=-

So:

210

7 30

2 15

3 5

225

25 9

3 35 5

150

5 30

3 10

5 2

330

10 33

11 35 2

A) B) C) D)

3. Build a factor tree for each number and write down its prime factorization:

83)350)1024)250)100)80)64)40) hgfedcba

2. Complete the following factor trees and write down the prime factorization:

2

3

5 7

2 311 5

7

11

5 2

3 33 3

A) B) C) D)

4. Build at least 4 different factor trees for the number 600.

5. Build a factor tree for each of the following negative integers and write down its prime factorization:

12345)17)4096)300)88)12) ------ fedcba

6. Find the value of each number given by its prime factorization and then build a factor tree:

111112558

2222412123

117532)52)3)2)

532)52)753)32)

´´´´´

´´´´´-´

hgfe

dcba

7. Complete the following factor trees and write down the prime factorization:

24

A) B) C)

200 360

8. Three numbers a, b, and c are given by their prime factorization:

13122341223 75327532532 ´´´=´´´=´´= cba

Use the exponents rules to find the prime factorization for:

2

22 ))))/))

c

bafcbae

c

badacabcbbaa

´´´

´´



30. Set of Divisors

34


1. Use the set notation to express the following sets:

A) the set of prime numbers less than 10B) the set of divisors of 10C) the set of first 5 natural numbers starting with 0D) the set of first positive 5 multiples of 3E) the set of odd numbers between 10 and 20

© La Citadelle

30.1 (Set) A set is a collection of elements. One way to define a set is to enumerate all the elements that belong to the set.. Example: }5,2,1{

Note that the elements are separated by commas and written inside braces {}. This is called the set notation.

30.2 (Multiplying Sets) The product of two sets is a set of products of each element of the first set by each element of the second set. Example:

}21,15,14,10,7,5{

}73,53,72,52,71,51{

}7,5{}3,2,1(

=

=´´´´´´=

=´

30.3 (The Set of Divisors) To find the set of all divisors of a natural number:a) find the prime factorizationb) for each prime factor, build a set having elements the prime factor raised to all possible powers starting 0 and ending at the maximum power as found in the prime factorizationc) multiply the sets you get in step b) Example:

}90,18,30,6,10,2,45,9,15,3,5,1{

}518,118,56

,16,52,12,59,19,53,13,51,11{

}5,1{},18,6,2,9,3,1{

}5,1{}92,32,12,91,31,11{

}5,1{}9,3,1{}2,1{

}5,1{5

}9,3,1{3

}2,1{2

53290

1

2

1

121

=

=´´´

´´´´´´´´´=

=´=

=´´´´´´´=

=´´

Þ

Þ

Þ

´´=

30.4 (Integral Divisors) Given an integer you can use the algorithm presented above to find all integral divisors of that number. Example:

}24,8,12,4,6,2,3,1{

}38,18,34,14,32,12,31,11{

}3,1{}8,4,2,1{

}3,1{3

}8,4,2,1{2

3224

1

3

13

=

=´´´´´´´´=

=´

Þ

Þ

´-=-

So, the set of all integral divisors is:

}24,8,12,4,6,2,3,1{ ±±±±±±±±

2. Use the rule of “multiplying sets” presented in 30.2 and find the “product” of the following sets:

}3,2,1{}2,1{}1{)}2,1{}2,1{)}4,3{}2,1{)}5{}1{)

}2,1{}2,1{}2,1{)}4,3{}5,2{)}3,2{}3,2,1{)}2,1{}2,1{)

´´-´-´´

´´´´´

hgfe

dcba

3. For each of the following powers, build the set of all its divisors (see 30.3):

31032

4321

5)2)3)3)

2)2)2)2)

hgfe

dcba

Example:

}16,8,4,2,1{}2,2,2,2,2{2) 432104 =Þd

4. For each of the following natural numbers, find the set of positive divisors (see 30.3):

2048)625)100)36)20)16)10)6) hgfedcba

5. For each of the following integers, find the set of all divisors (positive and negative) (see 30.3):

512)1000)360)200)160)40)8) ----- gfedcba

6. For each natural number, find the number of divisors (see the example below):

5120)2500)6000)640)200)64)40)12) hgfedcba

Example:

.

40425)13()11()14(:

5326000:

6000)

314

divisorspositiveallofnumbertheisThis

themmultiplyandexponentsallto1Add

ionfactorizatprimethePerform

f

=´´=+´+´+

´´=

7. Compute the following operations with sets of numbers (see 30.2 for multiplication). One case is solved for you as an example:

}2,1{}6,4,2{)}2,1{)}5,2{)}3,2{}3,2,1{)}2,1{}2,1{) 32 ¸-+ edcba

Example:

}25,10,4{}25,10,10,4{}55,25,52,22{}5,2{}5,2{}5,2{) 2 º=´´´´=´=c

Note: In a set any element is unique (appears only once).



31. Highest Common Factor (HCF)

35


1. For each case: find the set of the positive divisors of all the given numbers then choose the highest common divisor (factor):

© La Citadelle

31.1 (HCF) The Highest Common Factor (HCF) of two or more natural numbers is the highest common natural number into which you can divide these numbers evenly (without a remainder). Example:

54

20;3

4

12;2

4

8

4

20,12,8:

===

=HCF

andGiven

31.2 (The Algorithm for finding HCF) The following algorithm allows you to find the HCF for any set of natural numbers:a) perform the prime factorization of each numberb) complete the missing prime factors with prime factors raised to power 0c) HCF is the number with all the prime factors raised to the lowest exponents.Example:

12327532)

7532180;7532120

;753284)

532180

;532120;73284)

180,120,84:

120012

10120113

1012

122

113112

=´=´´´=

´´´=´´´=

´´´=

´´=

´´=´´=

HCFc

b

a

andGiven

31.3 (HCF for integers) To calculate the HCF for a set of integers, consider only the positive divisors of these numbers. In this way, the HCF will always be a natural number. Example:

20532)

532180;532200)

532180;52200)

180200:

102

122203

12223

=´´=

´´=´´-=-

´´=´-=-

+-

HCFc

b

a

andGiven

31.4 (Factoring) Given an expression with two or more terms you can always factor the HCF of these terms. Example:

)752(6423012 -+-´=-+-

31.5 (HCF Function) The function HCF has as input a list of natural or integral numbers separated by commas, and as output the HCF of these numbers. Examples:

6)60,30,12(

5)15,10(

=-

=

HCF

HCF

60;18)100;250)90;36;81)24;64)48;36;8)20;12) fedcba

Example:

6

}60,30,20,15,12,10,6,5,4,3,2,1{}60,12,20,4,30,6,10,2,15,3,5,1{

}5,1{}12,4,6,2,3,1{}5,1{}3,1{}4,2,1{53260

}18,9,6,3,2,1{}18,6,2,9,3,1{}9,3,1{}2,1{3218

60;18)

112

21

=

==

=´=´´Þ´´=

==´Þ´=

HCF

f

3. For each case use the algorithm presented in 31.2 to find the HCF:

300;225)100;40;64)81;128)72;60;18)30;24) edcba

4. For each case use the algorithm presented in 31.2 to find the HCF (see also 31.3):

625;225;125)40;21)60;36;30)250;400) ------- dcba

5. Rewrite the following expressions by factoring out the HCF:

907260)25664)500250200)183012)

350100240)225125)8064)3624)

--+-+----

+-----

hgfe

dcba

7. Simplify the following expressions by canceling out the HCF.

52

37)

13065

5226)

150100

1257550)

9648

806432)

115150

906045)

8472

603624)

5664

4024)

4221

3514)

+-

-

+-

-

+-

+-

+-

+-+-

+-

+-

+-

--

--

+-

-

hgfe

dcba

Example: 22

4

64

532

)64(25

)532(25

150100

1257550) ==

+-

+-=

+-´

+-´=

+-

+-f

2. For each case, find the HCF by factoring out the common factors one by one (one case is solved for you as an example):

105;70)48;40)20;45;35)32;24)36;20;6)12;8) fedcba

Example:

3575

}3;2{75}21;14{5}105;70{)

=´

´´=´=

isHCFThe

f

6. Simplify the following expressions by canceling out one by one the common factors (one case is solved for you as an example):

10234

5117)

150125

5075)

10836

1207260)

9040

806030)

9680

642440)

5442

361830)

2520

3015)

1215

69)

+-

-

+-

-

+-

+-

+-

+-+-

+-

+-

+-

--

--

+-

-

hgfe

dcba

Example:

42

8

97

635

)97(3

)635(3

2721

18915

)2721(2

)18915(2

5442

361830) ==

+-

+-=

+-´

+-´=

+-

+-=

+-´

+-´=

+-

+-c



32. Least Common Multiple (LCM)

36


1. For each set of numbers, list the multiples until you can identify the LCM (see 32.2):

© La Citadelle

32.1 (LCM) The Least Common Multiple (LCM) of two or more natural numbers is the least common natural number which can be divided evenly (without a remainder) by each of these numbers. Example:

32.3 (The Algorithm for finding LCM. Solution 2) The following algorithm allows you to find the LCM for any set of natural numbers:a) perform the prime factorization of each numberb) complete the missing prime factors with prime factors raised to power 0c) LCM is the number having all the prime factors raised to the greatest exponents.Example:

32.5 (LCM for integers) To calculate the LCM for a set of integers, consider only the positive multiples of these numbers. In this way, the LCM will always be a natural number. Example:

32.6 (LCM Function) The function LCM has as input a list of natural or integral numbers separated by commas, and as output the LCM of these numbers. Examples:

120)15,8(;30)15,10( =+-= LCMLCM

26

12;3

4

12

12;64

==

=LCMandGiven

32.2 (The Algorithm for finding LCM. Solution 1) List all the multiples of each given number until you find a first common multiple. That is the Least Common Multiple. Example:

24

,...30,24,18,12,66

,...28,24,20,16,12,8,44

,...30,27,24,21,18,15,12,9,6,33

6,4,3

=

Þ

Þ

Þ

LCM

andGiven

600532

53225;53212

5328;25,12,8

213

200012

003

=´´=

´´=´´=

´´=

LCM

andGiven

30532;53215

5326;15,6

111110

011

=´´=´´=

´´-=-+-

LCM

andGiven

32.4 (LCM & HCF) The product of any two natural numbers (except 1) is equal to the product of LCM and HCF of these numbers. Example:

HCFLCM

HCFLCM

andGiven

´==´

==

1801512

3;60

1512

20;12)35;25)15;6;4)12;5)10;8;3)15;6) fedcba

2. For each set of numbers, use the algorithm presented in 32.3 to find the LCM:

32;24)25;30)8;12;5)20;16)18;15;10)10;4) fedcba

3. For each pair of numbers, prove that the product of the numbers is equal to the product between the HCF and the LCM of these numbers:

16;14)25;10)8;6)30;12)15;10)12;10) fedcba

4. For each set of integers find the LCM:

25;20)5;12;18)16;12)30;25;20)8;3) -------- edcba

5. Find the value of the following expressions containing HCF and LCM functions (see 31.5 and 32.6):

)25,15,10()25,15,10(

251510)

)5,10,16(

)50,100,160()

)80,100,120())35,25()35,25(

3525)

)20,15(

)32,48())8,6()24,20()

)16,12())8,6()

HCFLCMh

LCM

HCFg

HCFfHCFLCM

e

LCM

HCFdLCMHCFc

LCMbHCFa

´

´´´

´

-

Example:

2052)80,100,120(

5280;52100;532120

)80,100,120()

12

1422113

=´=

´=´=´´=

HCF

HCFf

6. LCM is used to add (or subtract) two or more fractions with unlike denominators. The Least Common Denominator (LCD) is the LCM of all denominators. To add the following fractions, convert them into equivalent fractions having a denominator equal to LCD and then add (or subtract) numerators. One case is solved for you as an example:

10

1

8

3)

16

1

12

1)

15

1

10

1)

6

5

4

3)

6

1

3

2)

3

1

2

1) ---+-+ fedcba

Example:12

19

12

109

12

10

12

9

6

5

2

2

4

3

3

3

6

5

4

3) =

+=+=´+´=+c

7. LCM is used to add (or subtract) two or more rational numbers with unlike denominators. The Least Common Denominator (LCD) is the LCM of all denominators. To add the following rational numbers, convert them into equivalent numbers having a denominator equal to LCD and then add (or subtract) numerators. One case is solved for you as an example:

16

1

12

5)

15

1

10

1)

6

5

4

1)

6

5

3

2)

3

1

2

1)

-+

-

--

--+

-

-+

-

--

-edcba

Example:

30

1

30

23

30

2

30

3

15

1

2

2

10

1

3

3

15

1

10

1

15

1

10

1)

-=

+-=+

-=´+

-´=+

-=

--

-d



33. Square Roots

37


1. Find the squares of the following numbers:

© La Citadelle

33.1 (Square Roots) The square root operation is the inverse operation to raising a number to the power 2:

axsoxthatisa =2

33.2 (Radicand) Because any real number (including an integer) squared is positive or 0, the radicand (the expressions inside the radical sign) is positive or 0:

Example: 42;24 2 ==

0³aifrealisa

Example:

01616 <-- becauserealnotis

33.3 (Principal Root) If more numbers squared are equal to the radicand, then by convention the square root is represented by the positive number called the principal root:

0³a

Example: 39;93;9)3( 22 ===-

33.6 (Square root of a square) The square root of a negative number squared gives you the opposite of the original number:

33.5 (Square Root of a Square) The square root of a positive number squared gives you back the original number:

33.7 (Algorithm for finding the square root) To find the square root of a number:a) do the prime factorizationb) the square root has the same factors as the original number, but all exponents are divided by 2Example:

33.8 (Real Numbers) If the radicand of a square root is not a perfect square, then the square root of that number is a real number. You can use a calculator to find a decimal approximation of the real number. Example:

3;39)3( 2 -==- not

847327056

7327056?;7056

112

224

=´´=

´´==

3932 ==

141.12 @

33.4 (Basic Square Roots) Here is a list of some basic roots (square roots):

;525255;416164

3993;2442

1111;0000

22

22

22

=Þ==Þ=

=Þ==Þ=

=Þ==Þ=

200)100)50)20)15)12)11)10)9)

8)7)6)5)4)3)2)1)0)

rqponmlkj

ihgfedcba

--

--

2. Find the following square roots (use the results you have got at Exercise 1):

16)144)16)49)40000)2500)

10000)100)400)121)64)9)

36)81)25)0)4)1)

rqponm

lkjihg

fedcba

3. Find the value of the following expressions containing square roots of squares (see 33.5):

222222

222222

99999)123)100)20)13)10)

9)5)7)0)3)1)

lkjihg

fedcba

4. Find the value of the following expressions containing square roots of squares (see 33.6):

22222

22222

)125())100())30())10())11()

)3())8())7())1())5()

-----

-----

jihgf

edcba

5. Find the value of the following expressions (see 33.7):

64422442

624222422

)5()10()3)20())3()5()5)7()

310)332)53)32)

-´-´--´-´-

´´´´´

hgfe

dcba

6. Use the prime factorization algorithm to find the following square roots (see 33.7):

Example:

12500125100510510510)5()10() 322/62/46464 =´=´=´=´=-´-h

202500)7056)15625)2500)2916)1089) fedcba

7. Use a calculator to find the following roots. Round the answer to the nearest hundredth (see 33.8):

99999)1234)50)10)7)3)2) gfedcba

8. Use a calculator to find the following roots. Round the answer to the nearest thousandth (see 33.8):

12345)777)200)30)10)3)2) gfedcba

Example:

23.31699999) @f



34. Cubic Roots

38


© La Citadelle

34.2 (Useful Cubic Roots) Here is a list of some basic cubic roots:

51251255;464644

327273;2882

1111;0000

3333

3333

3333

=Þ==Þ=

=Þ==Þ=

=Þ==Þ=

34.4 (Sign) The cubic root of a positive number is positive (because a positive number cubed is positive). The cubic root of a negative number is negative (because a negative number cubed is negative). Examples:

34.3 (The Cubic Root of a Cube) The cubic root of a number cubed is the original number:

aa =3 3

34.5 (The Algorithm for finding the cubic root) To find the cubic root of a number:a) perform the prime factorizationb) the cubic root has the same factors as the original number, but all exponents are divided by 3Example:

12321728

321728

?1728

123

36

3

=´=

´=

=

34.6 (Real Numbers) If the radicand of a cubic root is not a perfect cube, the cubic root of that number is a real number. You can use a calculator to find a decimal approximation of that real number. Example:

154.2103 @

32727)3(

288)2(

111)1(

33

33

33

-=-Þ-=-

-=-Þ-=-

-=-Þ-=-

Examples:

28)2(;3273 33 333 3 -=-=-==

5125;5125 33 -=-=

34.1 (Cubic Roots) The cubic root operation is the inverse operation to raising a number to the power of 3:

axsoxthatisa =33

Example:

82;28 33 ==

1. Find the cubes of the following numbers:

12)11)200)100)50)20)10)

6)5)4)3)2)1)0)

nmlkjih

gfedcba

2. Find the following cubic roots (use the results you have found at Exercise 1):

333333

333333

1331)125000)8000)27)1000)125)

64)8)216)1)1728)0)

lkjihg

fedcba

3. Find the cubes of the following numbers:

12)300)100)40)20)10)

6)5)4)3)2)1)

------

------

lkjihg

fedcba

4. Find the following cubic roots (use the results you found in Exercise 3):

33333

33333

1728)125)8000)64)1000)

27)216)8)216)1)

-----

----

jihgf

edcba

5. Find the value of the following expressions containing cubic roots of cubes (see 34.3):

3 33 33 33 33 33 3 12345)12)10)5)3)1) fedcba

6. Find the value of the following expressions containing cubic roots of cubes (see 34.3):

3 33 33 33 33 3 )123())10())4())2())1() ----- edcba

7. Find the value of the following expressions:

3 363 633 363 63

3 3693 363 633 33

)10()2()3)10())4()5()5)4()

532)310)52)32)

-´-´--´-´-

´´´´´

hgfe

dcba

8. Use the prime factorization algorithm to find the following cubic roots (see 34.5):

Example:

40)10(4)10()2()10()2()10()2() 123/33/63 36 -=-´=-´-=-´-=-´-h

33333 1728000)32768)13824)91125)5832) edcba

9. Use a calculator to find the following roots. Round the answer to the nearest thousandth (see 34.6):

333333 99999)1234)100)80)20)7) fedcba --

Example:

416.4699999) 3 @f



35. Roots of superior order

39


© La Citadelle

35.1 (n-th Order Roots) The n-th order root operation is the inverse operation to raising a number to the power n:

35.4 (Basic n-th order roots) Here is a list of some basic n-th order roots:

212818)2(

381813;216162

1111;0000

77

4444

5544

-=-Þ-=-

=Þ==Þ=

=Þ==Þ=

35.5 (The n-th order root of a positive number raised to power n) The n-th order root of a positive number raised to power n is the original number:

0; >= aaan n

35.7 (The Algorithm for finding the n-th root) To find the n-th root of a number:a) perform the prime factorizationb) the n-th root has the same factors as the original number, but all exponents are divided by nExample:

12323220736 124 484 =´=´=

Examples:

1010;536 65 5 ==

35.6 (The n-th order root of a negative number raised to power n) The n-th order root of a negative number raised to power n is the original number if n is odd and the opposite of the original number if n is even. Examples:

22256)2(;2)2(8 888 85 5 ===--=-

axsoxthatisa nn =Example:

32)2(;232

162;216

55

44

-=--=-

==

35.2 (Radicand) Because any real number (including an integer) raised to an even power is positive or 0 the radicand must be positive or 0:

evenisnifaan 0; ³

35.3 (The Principal Root) If n is even, more numbers raised to the power of n are equal to the radicand. By convention the n-th root is represented by the positive number called the principal root:

Example:

081814 <-- becauserealnotis

evenisnifan 0³

Example:

381;81)3(;813 444 ==-=

1. Find the values of the following powers:

5445465

4645445

)10())10())3())2())2())1())1()

100)10)5)4)3)2)0)

------- nmlkjih

gfedcba

2. Find the value of the following roots (use the results you found in Exercise 1):

954557

464455

512)243)1)100000)32)1)

81)1000000)625)16)1)0)

------

-

lkjihg

fedcba

3. Find the value of the following expressions containing roots of powers (see 35.5):

11 117 79 97 74 45 5 54321)12)10)2)4)1) fedcba

10 105 57 75 56 6 )123())10())4())1())1() ----- edcba

4. Find the value of the following expressions containing roots of powers (see 35.6):

5. Find the value of the following expressions (see 35.7):

5 10157 1475 1054 48

5 510104 484 1245 105

)10()2()3)10())4()5()5)4()

332)310)52)32)

-´-´--´-´-

´´´´´

hgfe

dcba

6. Use the prime factorization algorithm to find the following roots (see 35.7):

76654 78125)4096)46656)59049)20736) edcba

7. Use a calculator to find the following roots. Round the answer to the nearest hundredth:

745654 12345)100)100)125)100)10) -- fedcba

Example:

100000)10(10100000) 55 -=-Û-=-i

Example:

5432154321) 11 11 =f

Example:

123123)123() 10 1010 10 ==-e

Example:

8001008)10()2()10()2()10()2() 235/105/155 1015 -=´-=--=-´-=-´-h

Example:

84.312345) 7 -@-f



36. Roots Rules

40


1. Use the product rule (see 36.1) to simplify the following expressions:

© La Citadelle

36.1 (Product Rule) The product of two or more like roots is equal to the root of the product of the radicands:

nnn baba ...... ´´=´´

Examples:

101010101010

284242

2

2333

==´=´

==´=´

36.2 (Simplifying Radicals) The product rule allows you to simplify radicals. Example:

252522522550 =´=´=´=

36.3 (Quotient Rule) The ratio of two like roots is equal to the root of the ratio of the radicands:

nn

n

b

a

b

a=

Example:

4642

128

2

128 333

3

===

36.4 (Simplifying Radicals) The quotient rule allows you to simplify radicals. Example:

2

5

4

25

4

25==

36.5 (Power Rule) A root raised to a power is equal to the root of the radicand raised to the same power:

n mmn aa =)(

Example:33 223 255)5( ==

36.6 (Canceling Out) The n-th order root raised to the power of n is equal to the radicand:

aa nn =)(Examples:

32)32(;3)3( 552 -=-=

36.7 (Simplifying Expressions) You may use all the above rules to simplify expressions containing radicals. Example:

65

152

)5(

)15()2(

5

152

:;6)6(5

30

5

30

5

152

5

152

2

222

2

2

222

=´

=´

=÷÷

ø

ö

çç

è

æ ´

==÷÷

ø

ö

çç

è

æ=

=÷÷

ø

ö

çç

è

æ=

÷÷

ø

ö

çç

è

æ ´=

÷÷

ø

ö

çç

è

æ ´

or

82)1010)555)44)

162)42)444)22)

17 47 333344

5533333

´´--´-´-´

´-´´´´

-hgfe

dcba

2. Use the product rule to simplify the following radicals (see 36.2):

335 63 4

4 6335

500)32)128)10)10)200)

10)32)54)128)18)8)

lkjihg

fedcba

--

-

3. Use the quotient rule (see 36.3) to simplify the following expressions:

3

3

4

4

5

5

5

5

3

3

3

3 4

3

3

3

3

500

4)

120

15)

64

4)

3

96)

4

128)

24

3)

75

3)

10

10)

4

500)

2

250)

2

16)

2

50)

-

-

-

--

lkjihg

fedcba

4. Use the quotient rule to simplify the following expressions (see 36.4):

43353

81

625)

27

1000)

64

125)

243

32)

27

8)

25

9) fedcba

-

-

-

-

5. Use the power rule (see 36.5) to rewrite (simplify) the following expressions:

2324435233 )5())10())5())3())5())2() -fedcba

6. Use the canceling out rule (see 36.6) to simplify the following expressions:

5544332332 )5())10())10())20())5())2() -- fedcba

Example:

333 33 33 443 5555555)5() ´=´=´==d

7. Simplify the following expressions with radicals:

2

3

2

3)

54

32)

5

32)

5

2)

2

3)

3232

162)

52

108)

5

202)

1030

3)

2

63)

3

33

3323

3

32

55

55

33

33

3

33

´÷÷

ø

ö

çç

è

æ

´

´÷÷

ø

ö

çç

è

æ ´÷÷

ø

ö

çç

è

æ

-

-÷÷

ø

ö

çç

è

æ

´

´

´

´

-

-´-

´

´

jihgf

edcba

2

3

32

333

33

516

23)

654

321)

20

5

10

3

60

2)

62540

32)

62150

105)

86

322)

÷÷

ø

ö

çç

è

æ

´

´÷÷

ø

ö

çç

è

æ

´´

´´´´

´´

-´

´´

´

´

´´

pon

mlk

Example:

2

1

4

1

4

1

48

12

48

12

86

322

86

322) =====

´

´´=

´

´´k



37. Equations with powers and radicals

41


1. Check by substitution if the given x is a solution of the given equation:

© La Citadelle

37.1 (Roots) Consider the equation:

axn =

Any number x satisfying the equation is called a root of the equation. Example:

42 =x

4)2(2 2 =±±= becausex

The two roots of this equations are:

37.2 (n is even) In the case that n is even (e.g. 2, 4, 6, ...) the equation:

0; ³= aaxn

has two real roots given by:

n ax ±=

There are no real roots if a<0. Examples:

rootsrealnox

xx

Þ-=

±=±=Þ=

1

38181

2

44

37.3 (n is odd) In the case that n is odd (e.g. 3, 5, 7, ...) the equation:

has always a real solution given by:

axn =

n ax =

Examples:

23232

32727

55

33

==Þ=

-=-=Þ-=

xx

xx

37.4 (Unknown Radicand) To find the solution of the equation:

axn =

raise left and right sides to power n:

nnnn axax =Þ=)(

Examples:

8)2(2

2555

33

2

-=-=Þ-=

==Þ=

xx

xx

37.5 (More Steps Solution) You may need to do more steps in order to solve an equation with powers and radicals. Example:

28

8816

1648

4138

32718

;27)18(

3

3

23

3

33

33

==

=-=

==+

=+=+

==-+

=-+

x

x

x

x

x

x

1024;2)10000;10)125;5)

216;6)125;5)1;1)

64;4)16;4)9;3)

1045

337

322

-=-==-=-=-=

=====-=

=-==-==-=

xxixxhxxg

xxfxxexxd

xxcxxbxxa

2. Solve the following equations (see 37.2):

1024)625)64)81)256)

10000)64)625)4)25)

104648

42422

=====

===-==

xjxixhxgxf

xexdxcxbxa

3. Solve the following equations (see 37.3):

100000)216)128)64)32)

1000)128)125)32)64)

53735

37353

=-==-==

-=-==-==

xjxixhxgxf

xexdxcxbxa

4. Solve for x (see 37.4):

10)5)1)2)3)

1)2)3)2)2)

43333

43

=-=-===

===-==

xjxixhxgxf

xexdxcxbxa


327)464)21024)5125)

264)11)28)525)

=-=-=-=-

=-=-==

xxxx

xxxx

hgfe

dcba

Example:

3)4()4(64)4(464) 3 =Û-=-Û-=-Û-=- xg xxx


43

343

33

3

335 34 2

223333

3 3

3

27

3

3

103)027)14()1()

5

1

8

5)

8

64

425

4)83)1(5)222)

)5()3()813

)16)5(2)

3)2(3)242)21)

=´

´=-+-´+=

--

=-

--=++´=-´

-=+-=+=+´

=-´-=-´=-

xlxk

xj

xixhxg

xfx

exd

xcxbxa

6. Find x that satisfies the following equations:

xexdxcxbxa xxxxx ===== 46656)3125)256)27)4)

Example:

64665664665646656) 6 =Û=Û=Û= xxxe xx



38. Link between Radicals and Powers

42


1. Convert these radical notations into exponential notations:

© La Citadelle

38.1 (The Link between Radicals and Powers) If the radicand is positive, a radical expression can be converted to a power according to the following formula:

Example:0;

1

>= aaa nn

2/133 =

38.4 (Multiplying Unlike Radicals) Using the link between radicals and powers, we can now multiply unlike radicals:

nm nmnm

nm

nm

nmnmnm

aaa

aaaaaa

´ +

´

++

=´

Þ==´=´

1111

38.2 (Generalization) The precedent relation can be generalized to:

0; >= aaa n

mn m

Example:3/23 2 55 =

Example:

66 532 323 322222 ===´´ +

38.5 (Dividing Unlike Radicals) Using the link between radicals and powers, we can now divide unlike radicals:

mn mnmn

mn

nm

n

m

n

m

aaa

a

a

a

a ´ -´

--

====

11

1

1

Example:

55 215 6

15 253 35

5

3

5

3

5

3

422

888

8

8

8

8

8

===

====-

-=

-

- ´ -

38.6 (Radical of Radicals) If the radicand of a radical is another radical, you can use the following formula to simplify the expression:

nmm n aa ´=

Example:

2646464 6233 === ´

Indeed:

nmnmmnmnm n aaaaa ´´ ====

1111

)()(

38.3 (Simplifying Radicals) The following formula allows you to simplify radicals:

n kn

k

nm

kmnm km aaaa === ´

´´ ´

Example:

555 26 3 ==

5104543 2)1024)64)2)5)27)2) -gfedcba

2. Convert these exponential notations into radical notations:

4/12/13/12/13/12/12/1 625)100)27)10)8)3)2) gfedcba

3. Convert these radical notations into exponential notations:

3 23 67 36 55 23 23 2)3)7)10)5)7)2) gfedcba

4. Convert these exponential notations into radical notations:

3/55/43/43/24/52/35/3 5)20)7)9)10)5)3) gfedcba

5. Simplify the following radicals (see 38.3):

6 310 55 103 94 23 94 27)2)2)2)25)2)3) gfedcba

6. Write the following products of radicals as unique radicals (see 38.4):

543644354

453433

3333)222)101010)22)

22)55)1010)33)

´´´´´´´´

´´´´

hgfe

dcba

7. Write the following ratios of radicals as unique radicals (see 38.5):

43

6

6

4

4

53

4

3

3 3

3)

10

10)

5

5)

2

2)

5

5)

10

10)

2

2) gfedcba

8. Write the following expressions with radicals as unique radicals (see 38.6):

65 33 34 333 64)4)10)5)10)2)2) gfedcba

9. Write the following expressions with radicals as unique radicals (one case is solved for you as an example):

4

33

33

3 24

53

34

5

4

34

3

5

5)55)

5

5)10)2)

3

3)

2

2

2

2)

33

33)

10

1010)2

2

2)

jihgf

edcba

´

´´

´´´

Example:

1212

11

12

13

12

3

12

4

12

6

4

1

3

1

2

1

4

1

3

1

2

143

101010

101010101010101010)

´==

====´´=´´

+

++++f

Example:

12 512

5

12

3

12

2

4

1

6

1

4

1

6

1

4

1

3

1

2

1

4

1

3

1

2

1

4

3222222222

2

22

2

2) ====´=´=´=´

++-a



39. Order of Operations (V)

43



© La Citadelle

39.1 (Radical Symbol) The radical symbol is also considered a grouping symbol:

)( )(nn aa Û

So, to evaluate an expression containing a radical symbol, follow these steps:a) evaluate the order of the radical (n). This must be a positive integer.b) evaluate the radicand (a).c) evaluate the value of the radical.Example:

416652652 2432 ==+´=+´-´

39.2 (Nested Radicals) If the radicand of a radical contains other radical symbols, start the evaluation of the expression with the innermost radical. Example:

39)4(5645 3 ==--=--

39.3 (Order of Operations) If an expression contains all kind of operations including radicals, then consider radicals as independent units. Evaluate the radicals and substitute their values in the original expression. All previously explained rules for the order of operations still apply. Examples:

5)3(2274.1 3 =--=--

72314))2(5(2

2

)2()2()3225(4

8.2 5

3

-=´--=´-+--

=

=-´-´-+--

1394))2(1()24(

)81()216(.3

22

232

=+=--+-=

=--+-

1323662

332322

22)32()32(.4

-=-=--+=

=´-´-´+

+´=+´-

1616)1()59(43

45

)125)3((6427

1625.5

2

3 634

33

-=´-=-´-

-=

=-+-´-+

-

8

27)27(

8

1)27(

2

1

)3(2)27()4(.6

3

339383

-=-´=-´=

=-´=-´ --

42

31)32( 33)4()8(

)2(2

5

452 2)3()1( 3

)1(

)2(1)6])2()3[(2))2()8()

)2()7)3()54)2()

25

22

-

----+-´--¸-

+-+´-

--¸-

-´-´-

fed

cba

2. Find the value of each expression (see 39.2):

33 278 3933

3 34 223 44

6464))2527()64)2()

)84(2)7)41()41()1231)

-----´-

--´++´----

fed

cba

Example:

39811

81

)1(

)2(1) 24

2

3

==+=-

-=-

--f

Example:

022426446464) 32278 3 33

=-=-=-=-f

3. Use the order of operations to find the value of each expression:

úú

û

ù

êê

ë

é

--+-

-----´

-¸

-´-¸-´-

---++-

+--÷øö

çèæ

--÷÷

ø

ö

çç

è

æ-´¸´-

-+-+---+

--

´÷÷

ø

ö

çç

è

æ+´+´-

--+-´-¸+

5

3

43

323255

27433

5

3333

22

2227

3

3 2

2

3

3

234325

53

3

3

2

33 2333 233

3

)64()32(

)4()3(64)81

3216

8898)

)321()321()16

256

1

1)

)21()21()100)

)3(38

16))54516()

)27()8()32()3264

2725)

5

4

125

64))4455()45()

8226)25162184)

nm

lk

ji

hg

fe

dc

ba

4. Use a calculator to evaluate the following expressions. Round the answer to the nearest thousandth.

Example:

242212211122122111

)21()21()21()21()21()21() 22

´-=´-´-´-´-´+´-´-´=

=+´+--´-=+--j

53 32

1

3

)21(5233

2

31))25())32(10)211)

-+

÷÷

ø

ö

çç

è

æ+---+- dcba



40. Substitution

44


1. For each case find whether or not the expression is well-formed:

© La Citadelle

40.1 (Algebraic Expressions) An algebraic expression contains a well-formed sequence of operations between numbers or/and variables. Examples:

formedwellisyx

formedwellnotisyx

-´+-

-´+-

2

/2 -

+

+--+-´--

-++-´-

+-- xj

zy

xixhxgxf

yxexdyxcxbxa

y 2)

2))5)2)3()

/)))5)3)

4(2

2

40.2 (Substitution) Substitution is the process of replacing the variables of an expressions with numbers in order to evaluate the expression. Example: The value of the following expression:

2-- yxif:

5

3

+=

-=

y

x

is: 10282)5()3( -=--=-+--

2. If x = -2, find the value of each expression:

)2()1())1

2)

2

4))1()

2)32))4())2()3)

3 5

3

+´-+

´

+-+

´-+´¸-´-´+

xxjxix

xh

xgxf

xxexdxxcbxxa

x

x

3. If x = -2 and y = +3, find the value of each expression:

)2()1())()3

2))

1

1)

))2(2)))2)

2222

233

-´++¸´´-

´÷ø

öçè

æ

+

-

+´-¸-¸´

yxjyxyxiy

xhyg

x

yf

yxeyxdxcxbyxa

x

yy

40.3 (Named Expressions or Formulas) You can give a name to an expression to make it available for further reference. Example:

ïî

ïíì

´=

=

2/

/

2tas

mFa

If F = 10; m = 5; t = 2 then:

ïî

ïíì

=´=´=

===

42/222/

25/10/

22tas

mFa

4. For each case, use substitution to find the value of the named expression:

3

2

3

:

3

4

:)

uv

zu

yxz

Find

y

x

Givena

=

-=

+-=

-=

-=

2

3

:

3

4

2

:)

edf

cbe

cbd

Find

c

b

a

Givenb

a

aa

´=

-´=

+=

-=

-=

=

222

)(2

:

3

5

2

:)

HWLD

HWLV

HWHLWLA

Find

H

W

L

Givenc

++=

´´=

´+´+´´=

=

=

=

40.4 (Function Notation) You can use an expression to define a function. A function has a name followed (in parentheses) by a list of parameters separated by commas. Example:

22

2

),(

3)(

yxyxg

xxxf

+=

´-=

You can use substitution to find the value of a function for the given values of parameters. Example:

5916)3()4()3,4(

431)1(3)1()1(

22

2

=+=++-=+-

=+=-´--=-

g

f

5. The functions f, g and h are defined below. Find the values of these functions for each case:

)2)1,3(())1),1(),0,0(()))1(),0(()))1(()

)0,1())2,4,4())2,1())1()

/

2),,(;),(;)(

22223

+

------

+

´+=--+=-+=

gfhfghgffgfffe

gdhcgbfa

zxy

yxzyxhyxyxyxgxxxxf

40.5 (Order of Operations) An expression can also contain functions. To find the value of the expression, replace the functions with their values. Example:

3201

222)2(;011)1(

?)]2([)1(1;)(

2

22

22

-=-+=

=-==-=

=-+=-=

E

ff

ffExxxf

6. The functions f(x) and g(x,y) are defined below. Find the value of each expression. One case is solved for you as an example:

)0,0()0(1))]1,2([)))1(1),0(1()))0(1()

)1,0()4(2))0,1(

)2())0,0()1,1())0()1()

)1(

1),(;)(

)2(

2

2

gfhggffgfffe

gfdg

fcggbffa

y

xyxgxxxf

f ´++-+

´---

--

+

-=-=

164)4(4

1

24222)2(;4

1

)11(

12)1,2(;)]1,2([)

2212

2

2

)2(

===÷ø

öçè

æ

-=-=-==+

-=

---

fggg f

Example:



Answers

45


1. Understanding integers (Page 5)

nokyesjnoiyeshnog

yesfyesenodyescnobyesa

)))))

)))))).1

101,)22,)3,)4,)10,)

11,)2,)0,)2,)3,).2

-+--+

-+-+

jihgf

edundefinedcba

3)0)1)2)5).3 edcba

55)41)300)20)1).4 +++++ edcba

negativejpositivei

negativehnegativegpositivefneithere

negativedpositivecnegativebpositivea

))

))))

)))).5

,...}4,3,2,1{.6 ----=-Z

}9,7,5,3,1{.7

falsef

falseetruedfalsecfalsebfalsea

)

))))).8

integerannotedorcba )0)55)11)7).9 -+-+

definednotjihgf

definednotedcba

)0))))

))))).10

¥¥¥

¥¥¥¥

2. Absolute value, sign, and opposite (Page 6)

7)7)12)

100)10)0)3)2)2)3).1

jih

gfedcba

123)5)10)3)2)1)0).2 ±±±±±± gfedcba

++-

+-+--+

)))

))))))).3

jih

gfedefinednotdcba

77)7)2)

11)0)20)3)3)20)12).4

+-+

+--++-

jih

gfedcba

5)2)2)

3)3)20)4)3)5)2).5

+--

------+

jih

gfedcba

4)15)4)

3)4)7)3)4)3)5).6

jih

gfedcba --

3)08)5)2)3)4)2)6)

5)8)3)10)6)2)6).7

--

-

nmlkjih

gfedcba

3. Number line (Page 7)

8)4)5)1)2).1 -+-+- EDCBA

+2+2+1 +3 +4 +5 +6 +7-5 -4 -3 -2 -1 0

A BC DE

.2

0)10)20)30)40).3 EDCBA ----

7)3)2)5)8).4 ++--- EDCBA

)6(')3(')1(')4(')6('.5 --+++ EDCBA

33)48)

82)2)7)5)5).6

+---

+++-++

orgorf

oredcba

4. Comparing integers (Page 8)

00)

01)33)50)35)21).1

=

>=<><

f

edcba

320123)321123)

32012)1023)

201)21).2

<<<-<-<-+<+<+<-<-<-

+<<<-<-<<-<-

<<-+<+

fe

dc

ba

52113)

542245)20234)

3210)201)21).3

->->->>+

->->->+>+>->>+>+>

->->->->>->-

f

ed

cba

13)

30)45)13)22)20).4

-<-

->+<-><-<

f

edcba

57)

55)42)24)73)

00)50)11)01)21).5

+<-

+=+<+-<-->-

=->-><-<

j

ihgf

edcba

falsejtrueitruehfalsegfalsef

falseefalsedfalsecfalsebtruea

)))))

))))).6

falsejtrueifalsehtruegtruef

trueetruedtruectruebtruea

)))))

))))).7

truejtrueitruehfalsegfalsef

falseefalsedfalsectruebtruea

)))))

))))).8

}2,1,0,1{)

}1,2{)}4,3,2,1,0{)}3,2,1,0{)}2{)

}5,4,3,2{)}1,0{)}0,1,2{)}0,1{)}2{).9

-

---

---

j

ihgf

edcba

7. Addition of integers using the number line (Page 11)



Answers

46


5. Applications of integers (Page 9)

2)1)0)4)5).1 -+-+ edcba

6. Addition of integers. Axioms (Page 10)

sumaddendsfsumaddendse

sumaddendsdsumaddendsc

sumaddendsbsumaddendsa

6;4,2)10;6,4)

4;3,7)4;2,2)

1;1,2)3;2,1).1

+

-+++

+-+

)4()5()

)3(4))1(0)50)23)01).2

-+-

-+-++++

f

edcba

5)4)1)5)1)0).3 -+- fedcba 8. Addition of integers using rules (Page 12)

5)3)0)2)2).2 -+-+ edcba

273)100)0)15)10).3 -+-+ edcba

10)3)0)5)10).4 -+-+ edcba

7)0)20)2)1000).5 +--+ edcba

0)3)7)5)2).6 edcba -+-+

125)31)0)150)500).7 +--+ edcba

0)45)40)550)50).8 edcba --++

casesallfor0.4

5)3)

3)37)33)7)25)5)15).5

ih

gfedcba -

0)30)27)8)20)

9)5)0)3)0)5)2).6

lkjih

gfedcba -

2)2)4)2)3)

3)5)1)1)5)3)2).1

--+--

--+++-

lkjih

gfedcba

)2(0))3()4()

)4()4())2()1())5()2().2

++++-

-++-+--++

ed

cba

2)1)

4)4)2)2)6)5)0).3

--

-----

ih

gfedcba

)5()2()3()5()1(.4 ++-+++-++

5421)

135)322)52)23)21).5

-++-

--+----+--

f

edcba

)2()1()2()3()4()5()

)3()3()2()2())4()3()2()1()

)5()3())3()2())2()1().6

-+-+-+++-++

++++-+-++-+++-

-+-++--++

f

ed

cba

3)6)5)4)

2)2)3)1)6)2)6)3).7

++-

----

lkji

hgfedcba

3)3)1)3).8 --- dcba

356)3210)65)

6)66666)4075)1575)1135)400)

235)444)125)54)22)10)4).1

pon

mlkjih

gfedcba

1905)90)9)1334)5555)4500)

800)250)225)6)5)3).2

------

------

lkjihg

fedcba

100)2222)

50)1250)425)200)100)50)

40)35)5)5)2)3)2)2).3

po

nmlkji

hgfedcba

225)2200)5100)1850)

100)450)25)150)25)5)

15)7)5)3)4)1).4

----

------

------

ponm

lkjihg

fedcba

3)20)200)300)10)222)

1000)111)80)6)2)10).5

-----

----

lkjihg

fedcba

15)90)1)6)0)10).6 -- fedcba

Page 15 - continue



Answers

47


9. Subtraction of integers (Page 13)

10. Order of operations (I) (Page 14)

12. Equivalent equalities and inequalities (Page 16)

falsef

falseetruedfalsecfalsebfalsea

)

))))).1

falseltruektruejtrueifalsehtrueg

trueftrueetruedfalsecfalsebtruea

))))))

)))))).2

trueftrueetrued

truectruebfalsea

;43);00);37)

;42);42);04).1

£³-<-

->-¹=

2)7)3)8)2)

3)1)6)0)1)4)3).1

---

--

lkjih

gfedcba

2)4()2()3)2()1()

5)3()2()4)4(0)2)4()2().2

-=+-++=--+

+=--+-=+-+=---

ed

cba

12)5)1)8)4)

5)2)1)7)2)4)4).3

--

---

lkjih

gfedcba

20)

5)6)3)10)5)4)4).4

-

-------

h

gfedcba

5)5)6)4)2)3)3)5).5 ---- hgfedcba

0)2)

3)8)10)10)2)5)2).6

ih

gfedcba --

2)2)7)3).7 dcba --

5)4)

0)2)4)2)3)2)0).1

--

--

ih

gfedcba

50)6)

3)3)1)4)5)5)0).2

-

-----

ih

gfedcba

1)5)

8)6)4)8)7)4)3).3

--

-

ih

gfedcba

5)4)8)4)5)3)5)2).4 ---- hgfedcba

9)

4)1)1)9)6)2)5).5

-

-----

h

gfedcba

0)0)6)1)1)4).6 fedcba --

falseitruehfalseg

falseftrueetruedfalsecfalsebtruea

)))

)))))).3

trueftrueetruedfalsecfalsebtruea )))))).4

truehfalseg

trueftrueefalsedtruectruebtruea

))

)))))).5

trueftrueetruedtruectruebfalsea )))))).6

falsehfalseg

trueffalseetruedfalsectruebtruea

))

)))))).7

1510101550

;155101510;101551510)

23312;312230;023312)

3215;35120;01235)

614;6410;0164)

342;3240;0432)

3210;0132;312)

..2

+--+-£

+-£+--£+-

³++--+-+-³³-++-

-¹+--+-¹¹+-+-

-->--+->>++-

+-<-++-<<+--

+--==+-+-=

f

e

d

c

b

a

varymayanswerThe

0151051510)

03221)0223)

021)0413)0422).3

£-+--

³++--¹--

>-<++-=+--

f

ed

cba

573230)

10515100)1130)

21210)3120)7430).4

-+--£

++--³--¹

--->++-<+--=

f

ed

cba

432321)22541)

34)32)21)211).5

++£+++³++

¹><=+

fe

dcba

11. Equalities and inequalities (Page 15)

531)

12)10)23)10)112).6

+-£

³¹->-<+=

f

edcba

10)

01)03)10)10)00).7

£

³¹-><=

f

edcba

truehtruegtruef

falseefalsedtruectruebtruea

)))

))))).8

15. Multiplication of integers (I) (Page 19)



Answers

48


13. Equations (Page 17)

14. Inequations (Page 18)

yesfnoeyesdnocnobyesa )))))).1

2)1)3)0)1)2)6)2)1).2 ihgfedcba -

10)0)

5)5)7)0)10)5)1).3

ih

gfedcba --

25)1)2)6)1)0).4 fedcba ---

3)4)6)3)0)4).5 --- fedcba

2)1)

1)5)5)5)2)0)1).6

±±

±±±±±±

ih

gfedcba

yesfnoeyesdyescyesbyesa )))))).1

15)

5)1)1)1)2).3

-³

-£-³<->¹

xf

xexdxcxbxa

)7(7)

)1(6))2(3))3(5)53)13).1

-´

-´-´-´´´

f

edcba

).2 a

)b

)c

)d

)e

)f

)g

)h

)i

)j

3)0)5)

7)0)5)1)1).4

-££³

-£>->-<-¹

xixgxf

xexdxcxbxa

5)

1)2)2)2)4).5

£

³-£><-¹

xf

xexdxcxbxa

0)

6)6)3)8)3).6

³

³³-><-¹

xf

xexdxcxbxa

15)

4)8)5)7)10).7

->

-£-³<->¹

xf

xexdxcxbxa

2)

3)6)2)7)8).8

<

³-£>->¹

xf

xexdxcxbxa

181818)1)1010101010)

444)33333)000)

55555)11)5555)222).2

++------

--------++

++++------++

jih

gfe

dcba

00)11)

333333)4443333)

2222255)33333555)

22222266)222244)

22233)111115).3

+=

++=++++=+++

++++=+++++=++

+++++=++++=+

++=+++++=

ji

hg

fe

dc

ba

30)50)100)6)30)15)10)1).4 hgfedcba

30)20)12)9)1)

12)4)3)5)6).5

-----

-----

jihgf

edcba

25)20)9)12)20)

8)10)3)6)1).6

jihgf

edcba

10)0)10)1)1)0)0)0).7 --- hgfedcba

111)0)

531)123)300)300)120)36)

125)200)60)100)100)100)

100)100)20)20)20)20).8

-

---

----

---

ts

rqponm

lkjihg

fedcba

0)0)9)2)333)0)8)6).9 hgfedcba

0 +1 +2 +3-1-2-3-4 +4

0 +1 +2 +3-1-2-3-4 +4

0 +1 +2 +3-1-2-3-4 +4

0 +1 +2 +3-1-2-3-4 +4

0 +1 +2 +3-1-2-3-4 +4

0 +1 +2 +3-1-2-3-4 +4

0 +1 +2 +3-1-2-3-4 +4

0 +1 +2 +3-1-2-3-4 +4

0 +1 +2 +3-1-2-3-4 +4

0 +1 +2 +3-1-2-3-4 +416. Multiplication of integers (II) (Page 20)

)3()12(36)

)4()15(60)12)10()10(100)

0)0(1))1(11)12)4()3()

)2()5(10))4()4(16))5(420).1

´-=-

-´-=+-=+´-=-

=´--´=--=+´-

-´-=+-´-=-´=-

i

hg

fed

cba

9000)3000)3600)1500)

1100)240)90)140)30).2

---

---

ihgf

edcba

24)4)

7)25)21)14)16)9)14).3

ih

gfedcba ----

50)4321(25)

75)32(15)30)432(10)

55)3251(11)4)431(2)

4)231(2)14)251(7)

20)53(10)20)51(5)

32)53(4)4)53(2)5)32(1).4

=+-+-´

-=--´=+-´

-=-+-´-=-+-´

-=+--´-=+-´

-=-´=+-´

-=--´-=-´=+´

l

kj

ih

gf

ed

cba

19. Division of Integers (I) (Page 23)



Answers

49


18. Division of integers (I) (Page 22)

20. Order of operations (III) (Page 24)

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50,40,30,20,10)40,32,24,16,8)

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5,4,3,2,1)30,24,18,12,6)

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RhRg

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3)5)

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mlkjih

gfedcba

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17. Order of operations (II) (Page 21)

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0)6)

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23. Inequalities and inequations (Page 27)

Page 27 - continue



Answers

50


21. Equalities and equations (Page 25)

22. Proportions and equations (Page 26)

24. Powers (Page 28)

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27. Divisors (Page 31)



Answers

51


25. Exponents rules (Page 29)

26. Order of operations (Page 30)

28. Divisibility rules (Page 32)

56475 0)5))10())3()2).1 edcba --

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10

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2

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yesgnofyeseyesdnocyesbnoa ))))))).2

nogyesfnoenodnocyesbyesa ))))))).3

nognofyeseyesdyescyesbyesa ))))))).4

yesgyesfnoeyesdyescnobyesa ))))))).5

yesgnofnoeyesdnocnobyesa ))))))).6

nogyesfnoenodyescnobyesa ))))))).7

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noeyesdyescnobyesa ))))).9

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30. Set of divisors (Page 34)



Answers

52


29. Prime factorization (Page 33)

31. Highest Common Factor (HCF) (Page 35)

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75)4)1)6)6).3 edcba

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3

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34. Cubic roots (Page 38)



Answers

53


32. Least Common Multiple (LCM) (Page 36)

35. Roots of superior order (Page 39)

60)175)60)60)120)30).1 fedcba

800)90)

80)80)108)300)250)18).5

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1)20)1)

15

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40

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1)

6

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48

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30

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13)

2

3)

6

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33. Square roots (Page 37)

40000)10000)2500)

400)225)144)121)100)81)64)

49)36)25)16)9)4)1)0).1

rqp

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hgfedcba

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125)100)

30)10)11)3)8)7)1)5).4

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180)75)175)270)54)45)6).5

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13.35)07.7)16.3)65.2)73.1)41.1).7

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142.14)477.5)162.3)732.1)414.1).8

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8000000)1000000)125000)8000)1000)

216)125)64)27)8)1)0).1

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100)100)360)300)50)6).7

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416.46)

726.10)642.4)309.4)714.2)913.1).9

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81)32)16)1)1)100000000)

1000000)625)1024)81)16)0).1

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gfedcba

54321)12)10)2)4)1).3 fedcba

123)10)4)1)1).4 edcba ---

5)4)6)9)12).6 edcba

84.3)16.3)51.2)24.2)51.2)78.1).7 -- fedcba

38. The link between radicals and powers(Page 42)



Answers

54


36. Roots rules (Page 40)

39. Order of operations (V) (Page 43)

4

3)3))32(4)

2)24)1000)2)1000)

3)2)5

4)1)2)15).3

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fedcba

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1)

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lkjihg

fedcba

3

5)

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3).4 fedcba

-

-

37. Equations with powers and radicals (Page 41)

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g

fedcba

33)2)4)8)5)8)9).5 gfedcba

3)4)2)4)2)3).1 fedcba -

0)8)2)2)2)1).2 fedcba --

148.3)070.2)310.2)503.0).4 dcba --

33

534

335

45)42)28)

1010)1010)210)10010)

42)23)42)23)22).2

lkj

ihgf

edcba

-

-

-

3433 25)100)55)39)25)32).5 fedcba

5)10)10)20)5)2).6 -- fedcba

20

3)

20

1)

20

1)

10

1)

6

1)

2

1)

2

3)

10

3)

5

6)

5

2)

2

3)

2

1)2)2)

10

1)3).7

ponmlkji

hgfedcba

-

-

3)3)10)3)

6))3)2).5

hgfe

dnumbernaturaloddanycba

definednotg

fedcba

)

1024)64)2)5)27)2).1 10

1

4

1

5

1

4

1

3

1

2

1

3

227

3

6

5

5

2

3

2

2

3

2)3)7)10)5)7)2).3 gfedcba

35 4

33 245 3

255)20)

77)9)1010)55)3).4

´

´´´

gf

edcba

60 1712 1112

20 94 315 812 76 5

33)2)1010)

2)2)5)10)3).6

´´ hgf

edcba

4

6

12

206

126

3)

10

1)5)

2

1)

5

1)10)2).7

g

fedcba

12

15941266

64)

4)10)5)10)2)2).8

g

fedcba

243 2

3

12

8

660 13

60 712 712 5

5)5)5

1)10)

2)3

1)

2

1)3)10)2).9

jihg

fedcba



Answers

55


40. Substitution (Page 44)

nojyesinohyesgnof

noeyesdyescyesbnoa

)))))

))))).1

0)2

1)2)1)1)

4)2)4)4

1)8).2

jihgf

edcba

-

---

38,30,62)

225,3,25)2,8,11).4

===

======

DVAc

fedbvuza

54)2)0)1)2)6)8)3).5 -- hgfedcba

1)36)3

2)

9

1)4)

1)2

1)2)8)3).3

-

---

jihgf

edcba

1)16)0)0)3)1)1)0).6 hgfedcba -

Documents

The Book of Integers - La Citadellela-citadelle.com/mathematics/the_book_of_integers.pdf · The Book of Integers 1. Understanding Integers 5 Iulia & Teodoru Gugoiu 1.1 (Definition)