Answers to Homework Set #1 - MATH 242 (Fall 2012) - WikiNotes

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12/21/2014 Answer s to homewor k set #1 - MATH 242 ( Fall 2012) - Wi ki Notes

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MATH242

Fall 2012

Coursesummary

Answers to homework set #1

PrintHistoryEdit

Answers to homework set #1

Professor: Ivo Klemes

Student-provided answers to homework set #1, due date unspecified (not to be handed in and thus not marked). The content on this page is solely intended to

functionasa study aid for students and should constitute fair dealing under Canadian copyright law.

Problems to complete: (2.1) #1-12, (2.2) #1-13,16,17, (2.3) #1-14, (2.4) #1-8, #13-19, (2.5) #1-13, (3.1) #1-18, (3.2) #1-23, (3.3) #1-12, (3.4) #1-16 (omi

limsup and liminf), (3.5) #1-11 (omit contractive sequences), (3.6) #1-10.

If you notice any errors, please registeror log inand edit this page, or contact @dellsystemabout it.

1 Section 2

1.1 Section 2.1

1.1.1 Question 1

1.1.2 Question 2

1.1.3 Question 3

1.1.4 Question 4

1.1.5 Question 5

1.1.6 Question 6

1.1.7 Question 7

1.1.8 Question 8

1.1.9 Question 9

1.1.10 Question 10

1.1.11 Question 11

1.1.12 Question 12

1.2 Section 2.2

1.2.1 Question 1

1.2.2 Question 2

1.2.3 Question 3

1.2.4 Question 4

1.2.5 Question 5

1.2.6 Question 6

1.2.7 Question 7

1.2.8 Question 8

1.2.9 Question 9
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1.2.10 Question 10

1.2.11 Question 11

1.2.12 Question 12

1.2.13 Question 13

1.2.14 Question 16

1.2.15 Question 17

1.3 Section 2.3

1.3.1 Question 1

1.3.2 Question 2

1.3.3 Question 3

1.3.4 Question 4

1.3.5 Question 5

1.3.6 Question 6

1.3.7 Question 7

1.3.8 Question 8

1.3.9 Question 9

1.3.10 Question 10

1.3.11 Question 11

1.3.12 Question 12

1.3.13 Question 13

1.3.14 Question 14

1.4 Section 2.4

1.4.1 Question 1

1.4.2 Question 2

1.4.3 Question 31.4.4 Question 4

1.4.5 Question 5

1.4.6 Question 6

1.4.7 Question 7

1.4.8 Question 8

1.4.9 Question 13

1.4.10 Question 14

1.4.11 Question 15

1.4.12 Question 16

1.4.13 Question 17

1.4.14 Question 18

1.4.15 Question 19

1.5 Section 2.5

1.5.1 Question 1

1.5.2 Question 2

1.5.3 Question 3

1.5.4 Question 4

1.5.5 Question 5

1.5.6 Question 6

1.5.7 Question 7

1.5.8 Question 8

1.5.9 Question 9

1.5.10 Question 10

1.5.11 Question 11

1.5.12 Question 12

1.5.13 Question 13

2 Section 32.1 Section 3.1

2.1.1 Question 1

2.1.2 Question 2

2.1.3 Question 3

2.1.4 Question 4

2.1.5 Question 5

2.1.6 Question 6

2.1.7 Question 7

2.1.8 Question 8

2.1.9 Question 9

2.1.10 Question 10

2.1.11 Question 11

2.1.12 Question 12


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2.1.13 Question 13

2.1.14 Question 14

2.1.15 Question 15

2.1.16 Question 16

2.1.17 Question 17

2.1.18 Question 18

2.2 Section 3.2

2.2.1 Question 1

2.2.2 Question 2

2.2.3 Question 3

2.2.4 Question 4

2.2.5 Question 5

2.2.6 Question 6

2.2.7 Question 7

2.2.8 Question 8

2.2.9 Question 9

2.2.10 Question 10

2.2.11 Question 11

2.2.12 Question 12

2.2.13 Question 13

2.2.14 Question 14

2.2.15 Question 15

2.2.16 Question 16

2.2.17 Question 17

2.2.18 Question 182.2.19 Question 19

2.2.20 Question 20

2.2.21 Question 21

2.2.22 Question 22

2.2.23 Question 23

2.3 Section 3.3

2.3.1 Question 1

2.3.2 Question 2

2.3.3 Question 3

2.3.4 Question 4

2.3.5 Question 5

2.4 Section 3.4

2.4.1 Question 1

2.4.2 Question 2

2.4.3 Question 3

2.4.4 Question 4

2.5 Section 3.5

2.6 Section 3.6

1Section 2edit

1.1Section 2.1edit

1.1.1Question 1edit

If , prove the following:

(a) If , then

(b)

(c)

(d) .

(a)

(b) From axiom A4, we know that . If we add to both sides, we have . By associativity

(A2) and A4 again, we can rewrite that .

(c)

a , b R

a + b= 0

b=

a

( a

) =a

( 1 ) a

= a

( 1 ) ( 1 ) = 1

b = b+ 0 =

b+ (

a a) = =

a= 0

a=

a( b + a)

a

a s s o c i a t i v i t y

( a + b )

c o m m u t a t i v i t y

( ( a ) ) + ( a ) = 0 a ( ( ( a ) ) + ( a ) ) + a=

a

( ( a

) ) + ( ( a

) +a

) = ( ( a

) ) + 0 = ( a

) =a

( 1 ) a

= ( 1 ) a

+ 0 = ( 1 ) a

+ ( ( a

) +a

) ( A 4 )
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(d) Using the statement proved in part (c), with , we have that . From part (b), we have that , and so

.

1.1.2Question 2edit

Prove that if , then

(a)

(b)

(c)

(d) if .

(a)

(b)

(c)

(d)

1.1.3Question 3edit

Solve the following equations, justifying each step:

(a)

(b)

(c)

(d) .

(a)

( 1 ) a

= ( 1 ) a

+ 0 = ( 1 ) a

+ ( ( a

) +a )

= a( 1 ) + ( (

a) + (

a) )

= (a

( 1 ) + ( a

) ) + ( a )

= (a

( 1 ) + a

1 ) + ( a )

= (a

( ( 1 ) + 1 ) ) + ( a )

= (a

( 0 ) + ( a )

= 0 + ( a )

= ( a )

( A 4 )

( A 1 )

( A 2 )

( M 3 )

( D )

( A 4 )

( M 3 )

( A 3 )

a= ( 1 ) ( 1 ) ( 1 ) = ( 1 ) ( 1 ) = 1

( 1 ) ( 1 ) = 1

a , b R

(a + b

) = ( a

) + ( b )

( a

) ( b

) =a b

1 / ( a) = ( 1 /

a )

(a / b

) = ( a

) / b b 0

(a + b

) = ( 1 ) ( a + b )

= ( 1 ) ( a

) + ( 1 ) b

= ( a

) + ( b )

( b y 1 ( c ) )

( D )

( b y 1 ( c ) a g a i n , u s e d t w i c e )

( a

) ( b

) = ( ( 1 ) a

) ( ( 1 ) b )

= ( ( 1 ) ( a

( 1 ) ) ) ( b

) = ( ( 1 ) ( ( 1 ) a

) ) ( b )

= ( ( ( 1 ) ( 1 ) ) a

) b

= ( 1 a

) b

= a b

( b y 1 ( c ) )

( b y M 2 , a p p l i e d t w i c e )

( b y M 1 )

( b y 1 ( d ) )

( b y M 3 )

1 / ( a )

= ( 1 / ( a

) ) 1

= ( 1 / ( a

) ) ( a ( 1 / a

) )

= ( 1 / ( a

) ) ( ( a

) ( ( 1 / a

) )

= ( ( 1 / ( a

) ) ( a

) ) ( ( 1 / a

) )

= 1 ( ( 1 / a

) )

= ( 1 / a )

( b y M 3 )

( b y M 4 )

( b y p a r t ( b ) )

( b y M 2 )

( b y M 4 )

( b y M 3 )

(a / b

) = ( 1 ) ( a / b )

= ( 1 ) ( a ( 1 / b

) )

= ( ( 1 ) a

) ( 1 / b )

= ( a

) ( 1 / b )

= ( a

) /b

( b y 1 ( c ) )

( b y t h e d e f i n i t i o n o f d i v i s i o n )

( b y M 2 )

( b y 1 ( c ) a g a i n )

( b y t h e d e f i n i t i o n o f d i v i s i o n )

2 + 5 = 8

= 2

2

1 = 3

2

( 1 ) (

+ 2 ) = 0

( s u b t r a c t i n g 5 f r o m b o t h s i d e s )
http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-3http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-2


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(b)

(c)

Not sure how to use this to conclude that . What are square roots???

(d) Since the product of and is zero, we know that at least one of the factors is zero. If , then adding 1 to both sides gives us

. If , then subtracting 2 from both sides gives us .

1.1.4Question 4edit

If satisfies , prove that either or .

First, we assume that . Then . So if . Since

satisfies , as by part (c) of theorem 2.1.2, then is a possible solution consequently, if , then either or

.

Not entirely sure if this is a valid proof.

1.1.5Question 5edit

If and , show that .

I can't seem to formulate this proof in a way that doesn't require ((((()))))) and 9000 uses of M1 and M2 so I'm just going to give the answer in words. The idea

that , so you can multiply by either, then when you multiply those together, you get (after some

manipulation) and then since you just get .

1.1.6Question 6edit

Use the argument in the proof of Theorem 2.1.4 to show that there does not exist a rational number such that .

(For reference, the proof of Theorem 2.1.4 is the famous proof, which proceeds by contradiction and uses even/odd properties.)

Proof by contradiction: we assume that there doesexist a rational number such that . So , where and . Let the ratio b

written in its lowest form, such that and have no common factors other than 1. Since , and so . must

be even, since 2 is a factor. It follows that is even as well, since if were odd, then and so which

would also be odd. So since is even, it follows that must be odd, because otherwise and would have 2 as a common factor.

Now, since is even, we can write in the form , where . tells us that .

Multiplying by on both sides, we have , which indicates that is even. Since 3 is odd, must be even in order for their product

to be even. But can only be even if is even, as if were odd then would be the product of two odd numbers and would thus be odd. So must be

2 + 5 5 = 8 5

2 + 0

2

( 1 / 2 ) ( 2 )

( ( 1 / 2 ) 2 )

1

= 3

= 3

= ( 1 / 2 ) 3

= ( 3 / 2 )

= ( 3 / 2 )

= ( 3 / 2 )

( s u b t r a c t i n g 5 f r o m b o t h s i d e s )

( f r o m a b o v e )

( b y A 3 )

( m u l t i p l y i n g b y 1 / 2 o n b o t h s i d e s )

( b y M 2 , a n d t h e d e f i n i t i o n o f d i v i s i o n )

( b y M 4 )

( b y M 3 )

2

( ) ( 1 /

)

( 1 ( 1 /

) )

1

= 2

= 2

= ( 2

) ( 1 / )

= 2 ( ( 1 /

) )

= 2 1

= 2

( b y t h e d e f i n i t i o n o f e x p o n e n t s )

( m u l t i p l y i n g b y 1 / x o n e a c h s i d e )

( b y M 2 )

( b y M 4 )

( b y M 3 )

( 1 ) + 1

2

+ ( ( 1 ) + 1 )

2

+ 0

2

2

= 3 + 1

= 4

= 4

= 4

( a d d i n g 1 t o e a c h s i d e )

( b y A 2 )

( b y A 4 )

( b y A 3 )

= 2

( 1 ) (

+ 2 )

1 = 0

= 1

+ 2 = 0

= 2

a

R a

a=

a a= 0 a = 1

a 0 a = 1 a = ( ( 1 / a ) a ) a = ( 1 / a ) ( a

a ) = ( 1 / a ) a = 1 a 0 , a = 1 a = 0

a a = a0 0 = 0

a= 0

a a =a a = 0

a= 1

a 0

b 0 1 / ( a b ) = ( 1 /

a ) ( 1 / b )

( 1 /a a

) = ( 1 / b b

) = 1 ( ( 1 / a

) ( 1 / b

) ) ( a b)

a b ( 1 / (

a b ) ) = 1 ( 1 / a

) ( 1 / b )

= 6

2

Q2

= 6

2

= /

, Z 0

= 6

2

/ = 6

2

2

= 6 = 3 2

2

2

2

2

= ( 2

+ 1 ) , Z

= 4 + 4

+ 1

2

2

= 2 Z

= 6

2

/ = ( 2

/ = 4 / = 6

2

2

)

2

2

2

2

( 1 / 2 )

2

2 = 3

2

2

3

2

2

2

2
http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-6http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-4http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-5


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even. But this is a contradiction to the conclusion that is odd, which was reached in the previous paragraph. Consequently, it must be that there does notexist

rational number such that .

1.1.7Question 7edit

Use the argument in the proof of Theorem 2.1.4 to show that there does not exist a rational number such that .

Proof by contradiction: we assume that there doesexist a rational number such that . So , where and . Let the ratio be

written in lowest terms, such that and have no common factors other than 1. Since , and so . Now, there are two

possibilities: either is even, or is odd. If is even, then must also be even (if it were odd, then the product of and 3 - both odd numbers - would

also be odd, contradicting the premise that is even). However, if both and are even, then and must both be even as well, by the reasoning in the

previous question. Consequently, and are both even, and so they share a factor of 2. But this contradicts the premise that and have no common factors

other than 1. So cannot be even.

Now let's look at the case where is odd. For this to happen, must also be odd, otherwise the product of and 3 would be even. So and are both

odd. We can write them as and , . So and . Substituting

these into the equation , we have:

By performing some algebraic manipulations we can simplify this to:

On the left side of the equation, we have an even integer, as 2 is a factor. On the right side of the equation, we have an odd integer, as is even due

to 2 being a factor of 6. A number cannot be both even and odd - this is a contradiction. Consequently, it must be that there does notexist a rational number

such that .

1.1.8Question 8edit

(a) Show that if are rational numbers, then and are rational numbers.

(b) Prove that if is a rational number and is an irrational number, then is an irrational number. If, in addition, , then show that

is an irrational number.

(a) Let and , where and . Then

(skipping some steps, and using some properties proved in previous exercises). Similarly,

. Then, (don't know if we're allowed to just do this?), where both the denominator and the

numerator are integers (and the denominator is non-zero). So it's rational. For , the argument is similar.

(b) I think this was mentioned in class. Later.

1.1.9Question 9edit

Let . Show that satisfies the following:

(a) If , , then and .

(b) If and , then .

(a) . . We can write their sum as follows:

. Also, we can write their product as

(b) Don't know

1.1.10Question 10edit

(a) If and , prove that .

(b) If and , prove that .

(a) This is similar to (though more complicated than) something we did in the first lecture. means that , and means that

. From property (i) of , we know that (this assumes that ). Now we just need to perform

some algebraic manipulations:

= 6

2

= 3

2

= 3

2

= /

, Z 0

= 3

2

/ = 3

2

2

= 3

2

2

2

2

2

2

2

2

2

2

2

2

2

= 2 + 1 = 2 + 1 ,

Z = 4 + 4 + 1

2

2

= 4 + 4 + 1

2

2

= 3

2

2

4 + 4

+ 1 = 3 ( 4 + 4

+ 1 )

2

2

2 ( +

) = 6 ( + 1 ) + 1

2

2

6 ( + 1 )

2

= 3

2

,

+

+

0

= /

1

1

= /

2

2

, , , Z

1

2

1

2

, 0

1

2

= ( / ) 1 = ( / ) ( / ) = ( ) / ( )

1

1

1

1

2

2

1

2

1

2

= ( ) / ( )

2

1

2

1

+ = ( + ) / ( )

2

1

1

2

1

2

K = { +

:

,

Q}2

K

1

K

2

+ K

1

2

K

1

2

0

K1 /

K

= +

1

1

1

2

= +

2

2

2

2

+ = ( + ) + ( + ) = ( + ) + ( + ) K

1

2

1

1

2

2

2

2

1

2

1

2

2

= ( + ) ( + ) = + + + = ( + 2 ) + ( + ) K

1

2

1

1

2

2

2

2

1

2

1

2

2

1

2

2

1

2

2

2

1

2

1

2

1

1

1

2

2

a


7/20



Consequently, and so .

In the case where , then we can't use property (i) directly, but since and

then so the above still applies. I'm not actually sure if this extra step needs to be made explicit, but, whatever.

(b) Later


(a) Show that if , then and .

(b) Show that if , then .

(a) means that . By some axiom1, we know that . By the Trichotomy Property (for ), there are three possibilities: either

or, or, .

Now, the third case is clearly false, because , by axiom M4. But if , then . So we have that which

is not true, by axiom M3.

Let's look at the second case. If this is true, then by the closure of under multiplication. This simplifies to . But it was proved inclass that , and so , which means that it cannotbe that . Thus, we've reached a contradiction, and this case is not valid.

So there's only one possibility left: that .

To show that :

(b) We know that . Adding to both sides gives us , i.e., . Dividing both sides by 2, we get .

Additionally, if we add to both sides of the original inequality, we get , and so . So then .


Let be numbers satisfying and . Give an example where , and one where .

If , , and , then and , hence (so ).

If , , and , then and , hence (so ).

1.2Section 2.2edit

1.2.1Question 1edit

If and , show that

(a)

(b) .

(a) From theorem 2.2.2 (b), we have that . Taking the square root of both sides, we have that and so .

(b) From theorem 2.2.2 (a), we have that for all . Since is equivalent to , then

. But since , then .

1.2.2Question 2edit

( b a) + (

c )

= ( (b a

) +

) c )

= ( (b +

) a

) c

= (b +

) + ( a c )

= (b +

) (a + c )

( b y A 2 )

( b y A 1 , A 2 )

( b y A 2 )

( b y s o m e t h i n g )

(b

+ ) ( a

+c ) P

(a

+c ) < ( b

+

)

c= 0

b a P ( b a) = (

b a) + 0 = (

b a) + (

c )

( b a) + (

c)

P

a> 0 1 /

a> 0 1 / ( 1 /

a) =

a

a 0 1 P

1 P

1 /a

P

1 / ( 1 / a) =

a

1 / ( 1 / a) = ( 1 / ( 1 /

a) ) 1

= ( 1 / ( 1 / a ) ) ( ( 1 / a ) a)

= ( ( 1 / ( 1 / a

) ) ( 1 / a

) ) a

= 1 a

= a

( b y M 3 )

( b y M 4 )

( b y M 2 )

( b y M 4 )

( b y M 3 )

a


8/20



If , show that if and only if .

First we prove the leftwards implication (that if the statement holds, then ). Assume that . There are 4 cases for the values of

and . Either and are both , or and , or and , or and both are . For the first case, if either is 0, then

otherwise, both are , and so their product is as well, which means that . For the second case,

for . Similarly for the third case. In the fourth case, since both are negative, their product is positive

(since , and so theirproduct, , and from exercise 2 (b) in section 2.1 we have that ), which means tha

(and consequently ). (It doesn't actually matter if the statement holds or not, just that it's false when the inequality is - material implication!)

Now we prove the rightward implication. If , then either or . In the former case, then either and or and

(from 2.1.10). In the latter case, either in which case the statement simplifies to on the left side and

on the right side, which is trivially true, or in which case it simplifies to which is again trivially true. If

, then as well (by 2.1.5 (i)), and so . Also, and , so we hvae on the left and

on the right, which proves that the statement is true. However, if , then the right side simplifies to . We magically realise2

that , which means that . So the left and right sides match up, which means that the statement holds, and so everything is

proved.

1.2.3Question 3edit

If and , show that if and only if . Interpret this geometrically.

Probably want to use triangle inequality for this. Skipping.

1.2.4Question 4edit

Show that if and only if .

means that . Adding to each term gives us . That was nice.

1.2.5Question 5edit

If and , show that . Interpret this geometrically.

Just need to show that . Well, . Also, . So

, i.e., . Adding the first and last inequalities together gives us

. Subtracting from each term gives us which is what

we wanted (since ).

1.2.6Question 6edit

Find all that satisfy the following inequalities.

(a)

(b)

(a) We need to solve . Well, if , then , and if , then . So

works.

(b) We can factor that as . So . Well, if , then we have and

, so that works for the right inequality. And if , then and , so . So

works.

1.2.7Question 7edit

Skipping

1.2.8Question 8edit

Skipping

1.2.9Question 9edit

Skipping

a , b R | a + b| = |

a| + |

b |a b 0

a b 0 |a + b

| = |a

| + |b | a

b a b 0a

< 0b

0a

0b

< 0 a b < 0

a b = 0 P

a b a b 0

| a| + |

b| =

a + b = b a |

a + b | a< 0

a,

b P(

a) (

b)

P(

a) (

b) = a b

a b > 0 a b 0

a b 0 a b > 0 a b = 0a

> 0b

> 0a

< 0

b< 0

a= 0 | 0 +

b| = |

b |

| 0 | + | b

| = 0 + | b

| = |b | b

= 0 |a

| = |a |

a , b> 0

a + b> 0 |

a + b| =

a + b | a| =

a | b| = b a

+b a

+ b

a , b< 0

a b= (

a + b )

a+

b < 0 | a+

b | = ( a+

b)

, , R

| | + |

| = |

|

a


9/20




Skipping


Skipping


Skipping


Skipping


Skipping


Skipping

1.3Section 2.3edit

1.3.1Question 1edit

Let . Show in detail that the set has lower bounds, but no upper bounds. Show that .

No upper bound by the Archimedean property. 0 is a lower bound, and if is any lower bound, we know that because 0 is in the set. Thus the infimu

is 0.

1.3.2Question 2edit

Skip

1.3.3Question 3edit

Let . Show that and .

Coming soon

1.3.4Question 4edit

Let . Find and .

First few terms: , , , . So the inf is , the sup is 2.

1.3.5Question 5edit

Skip

1.3.6Question 6edit

Skip

1.3.7Question 7edit

Skip

1.3.8Question 8edit

= { R :

0 }S

1

S

1

i n f = 0 S

1

0

= { 1 / :

N}

S

3

s u p = 1 S

3

i n f 0 S

3

= { 1 ( 1 / : N }

S

4

)

i n f

S

4

s u p

S

4

1 ( 1 ) / 1 = 2 1 ( 1 ) / 2 = 1 / 2 1 ( 1 ) / 3 = 4 / 3 1 ( 1 ) / 4 = 3 / 4 1 / 2
http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-17http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-3_2http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-8_2http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=section-23http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-1_2http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-6_2http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-4_2http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-11_1http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-5_2http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-7_2http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-12_1http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-2_2http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-10_1http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-16http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-13


10/20



Skip

1.3.9Question 9edit

Skip


Skip


Skip


Skip


Skip


Sigh

1.4Section 2.4edit

1.4.1Question 1edit

Show that .

First, we show that 1 is an upper bound: since for , then . So yeah, 1 is an upper bound. Now we need to show that if

then there exists some in such that . Let's write as for some . Well, by the Archimedean property, we know that there exists

such that . So then . So then there exists such that , proving that nothing less than 1 is an

upper bound. We conclude that 1 is the supremum.

This is probably not standard procedure but it seems to work. Also, !! 0

1

> 0 1 < 1

1

< 1

S

> 1 > 0

N 1

=

1

S

>

S= { 1 /

1 /

: , N} i n f

Ss u p

S
http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-11_2http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=section-24http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-2_3http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-9_2http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-4_3http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-6_3http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-12_2http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-13_1http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-14http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-7_3http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-1_3http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-10_2http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-5_3http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-3_3


11/20



Skipping

1.4.8Question 8edit

Skipping


Skipping


Skipping


Skipping


Skipping


Skipping


Skipping


Skipping

1.5Section 2.5edit

1.5.1Question 1edit

If and are closed intervals in , show that if and only if and .

First, we show that implies that and . Let . Then, from the set definition of an interval, we have that . Since we

also have that , then . Clearly, , and also . So and . From this we can conclude that

and . .

Now, we show that if and , then . Let . From the set definition of , we know that . Since and

, then by transitivity. Similarly we have that . Putting that together, we get that , and so .

1.5.2Question 2edit

If is nonempty, show that is bounded if and only if there exists a closed bounded interval such that .

First, we show that being bounded implies that there exists a closed bounded interval such that . Well, if is bounded, then it has an infimum and

supremum. We let and , and we create an interval . Then, if , by the definition of infima and suprema we have

that . So , by the set definition of the interval.

Now, we show that if there exists a closed bounded interval such that , then is bounded. Let be the upper bound of and let be the lower

bound of some interval satisfying this property. We can write as . Since , then for any , we have that

. is thus a lower bound of , and is a lower bound. Consequently, is bounded.

1.5.3Question 3edit

If is a nonempty bounded set, and , show that . Moreover, if is any closed bounded interval

containing , show that .

I= [

a , b] = [ , ]

I

a

b

R I I

aa

b b

I I

aa

b b

I a

b

I

a

b

a

I b

I

a

a

b

b

a

b

aa

b b

aa

b b

I I

I I a

b aa

a a

b

a

b

I

S RS I S

I

S I S

I S

a= i n f S b = s u p S I = [

a , b ] S

a b

I

I S

I S a I b

I I{ R : a b } S

I S

a b a S b S

S R= [ i n f

S, s u p

S ]I

S

S I

S

J

S JI

S
http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-19http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-1_4http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-16_1http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-8_3http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-17_1http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=section-25http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-18http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-14_1http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-15http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-3_4http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-13_2http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-2_4


12/20



Not very interesting (fairly trivial, and similar to previous questions)

1.5.4Question 4edit

In the proof of Case (ii) of Theorem 2.5.1, explain why exist in .

Wow, asking us to answer a "(Why?)". Okay. Well, since , then there always exists a such that (from the supremum definition). Also, since

is notbounded below, there is always an such that . Is that enough

1.5.5Question 5edit

Write out the details of the proof of case (iv) in Theorem 2.5.1.

Seriously? This is a question? That's one way to write a textbook, I guess.

In any case, if is not bounded above or below, I don't even care

1.5.6Question 6edit

If is a nested sequence of intervals and if , show that

and .

I can't even pretend to care

1.5.7Question 7edit

Let for . Prove that .

Let . Clearly for every , hence . Suppose . Then for all . So for all . Hence

by the Archimedean property (why? seriously, I'm not sure). Thus .

This is taken almost verbatim from the solutions to assignment 1, question 8 (b) (part two). See MyCourses for the solutions.

1.5.8Question 8edit


By the Archimedean property, the only such that for all is 0. So only 0 could be in the intersection. But 0 is not in the intersection as it's

an open set. So the intersection is empty.

1.5.9Question 9edit


Pretty boring


skipping


skipping


Give the binary representations of and .

, S


13/20



Only not skipping because it's so easy:


(a) Give the first four digits in the binary representation of .

(b) Give the complete binary representation of .

(a) (draw the diagram and you'll see why)(b)

2Section 3edit

2.1Section 3.1edit

2.1.1Question 1edit

The sequence is defined by the following formulas for the th term. Write the first five terms in each case:

(a) ,

(b) ,

(c) ,

(d) .

(a)

(b)

(c)

(d)

2.1.2Question 2edit

The first few terms of a sequence are given below. Assuming that the "natural pattern" indicated by these terms persists, give a formula for the

th term .

(a) ,

(b) ,

(c) ,

(d) .

(a)

(b)

(c)

(d)

2.1.3Question 3edit

List the first five terms of the following inductively defined sequences.

(a) 1, 4, 13, 40, 121

(b) 3/2, 17/12. 577/408, etc

(c) 1, 2, 3, 5, 4

(d) 8, 13, 21, 34, 55

2.1.4Question 4edit

For any , prove that .

See Example 3.1.6 (a), only multiply by , let , etc.

= =

3

8

0 . 0 1 1

2

7

1 6

0 . 0 1 1 1

2

1

3

1

3

0 . 0 1 0 1

0 . 0 1

( )

= 1 + ( 1

)

= ( 1 /

)

=

1

( + 1 )

=

1

+ 2

2

( 0 , 2 , 0 , 2 , 0 )

( 1 , 1 / 2 , 1 / 3 , 1 / 4 , 1 / 5 )

( 1 / 2 , 1 / 6 , 1 / 1 2 , 1 / 2 0 , 1 / 3 0 )

( 1 / 3 , 1 / 6 , 1 / 1 1 , 1 / 1 8 , 1 / 2 7 )

( )

5 , 7 , 9 , 1 1 ,

/ 1 / 2 , 1 / 4 , 1 / 8 , 1 / 1 6 ,

1 / 2 , 2 / 3 , 3 / 4 , 4 / 5 ,

1 , 4 , 9 , 1 6 ,

3 + 2

( 1 / )

+ 1

2

/ (

+ 1 )

2

b Rl i m (

/ ) = 0

1 / K b=

b
http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=section-3http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-2_5http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-13_3http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-4_5http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=section-31http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-1_5http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-3_5


14/20



2.1.5Question 5edit

Use the definition of the limit of a sequence to establish the following limits.

(a) ,

(b) ,

(c) ,

(d) ,

DON'T SKIP THIS. IT LOOKS USEFUL.

2.1.6Question 6edit

Show that

(a) ,

(b) ,

(c) ,

(d) ,

SAME AS ABOVE.

2.1.7Question 7edit

Skipping

2.1.8Question 8edit

Skipping

2.1.9Question 9edit

Skipping


Everything sucks


Idk


Idk


Idk


l i m

= 0

+ 1

2

l i m

= 2

2

+ 1

l i m =

3 + 1

2 + 5

3

2

l i m( ) =

1

2

2 + 3

2

1

2

l i m( )

= 0

1

+ 7

l i m = 2

2

+ 2

l i m( )

= 0

+ 1

l i m( )

= 0

( 1)

+ 1

2

l i m ( ( 2

) = 1

1 /
http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-12_4http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-9_4http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-13_4http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-7_5http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-8_5http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-5_5http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-14_2http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-10_4http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-11_4http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-6_5


15/20



Show that .

I give up.


Show that .

Might as well write the questions out since I'll have to do them one day


I think I have the wrong version of the textbook. My question numbers don't agree.


What is a limit? Seriously, how do you prove shit


sigh

2.2Section 3.2edit

2.2.1Question 1edit

Converge/diverge?

(a)

(b)

(c)

(d)

(a) Converges to 1. Proof: ?

(b) I think it diverges. Proof: ?

(c) Diverges. Proof: ?

(d) Converges to 2. Proof: ?

2.2.2Question 2edit

2.2.3Question 3edit

2.2.4Question 4edit

2.2.5Question 5edit

2.2.6Question 6edit

2.2.7Question 7edit

2.2.8Question 8edit

2.2.9Question 9edit





l i m ( ( 2

) = 1)

1 /

l i m ( /

! ) = 0

2

= / ( + 1 )

= ( 1

/ (

+ 1 )

)

= / (

+ 1 )

2

= ( 2 + 3 ) / ( + 1 )

2

2
http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-8_6http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-9_5http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-12_5http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-2_6http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-10_5http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-7_6http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-4_6http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-13_5http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-5_6http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-15_1http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-1_6http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-16_2http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-11_5http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-3_6http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=section-32http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-18_1http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-6_6http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-17_2


16/20













2.3Section 3.3edit

2.3.1Question 1edit

Let and for . Show that is bounded and monotone. Find the limit.

In any problem like this, with inductively-defined sequences, we need to three things: show the sequence has a bound (upper if the sequence is increasing, and

lower if it's decreasing), prove the sequence is increasing or decreasing, and find the limit (usually by invoking the property of sequences that if , the

-tail of also converges to .

Just by finding , we see that the sequence is decreasing. Let's prove this with induction:

Base case: we've already shown .

Induction step: . Our induction hypothesis is that for some . Thus by the rules we know about inequalities, this

implies . Thus we can invoke the I.H. in our induction step as follows:

Thus , thus is decreasing. Since it's decreasing, we now need to show that the sequence is bounded below. Looking at the

sequence definition, we can guess that the sequence is at least bounded below by . This may not be the greatest lower bound (or the infimum of the set consistin

of all ) but we just need to show any lower bound exists. So let's prove that 2 is a lower bound, once again with induction:

Base case:

Induction step: . Our Induction Hypothesis is that . Thus using the I.H.:

As the implication is true, we have proved the sequence is bounded below by 2. By the Monotone Convergence Theorem, we know the sequence converges as

we've proven it's monotone and bounded. Now, finally, we need to find the limit. By 3.1.9 in the book, we know the tail of a sequence converges to the same

thing as the sequence (as mentioned before). So if then . So we can write:

And now we're done.

2.3.2Question 2edit

Let and for . Show that is bounded and monotone. Find the limit.

= 8

1

= + 2

+ 1

1

2

N( )

= 6

2

N

+ 1

+ 2

( )

2

= 8 , 2 8

1

= + 2

+ 1

1

2

2 N

= + 2 2 + 2 = 3 3 2

+ 1

1

2

1

2

L

L

+ 1

L = L+ 2

L= 4

1

2

> 1

1

= 2

+ 1

1

N

( )
http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-23http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-2_7http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-20http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-17_3http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-1_7http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-14_3http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=section-33http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-21http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-15_2http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-22http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-18_2http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-16_3http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-19_1


17/20



This problem is similar to the one above, but there is some extra challenge. It's tougher to see if the sequence is increasing or decreasing, and because of the

inequality, the base case is hard to prove. Solving for we get , which doesn't tell us anything about the sequence's behavior. Let's do this a different

way, by solving . If the difference is positive, the sequence is (in the base case) increasing, and if it's negative, the sequence is decreasing. Our strategy

is to use the fact that to make some inference about the difference.

As and since the entire expression must be negative. Therefore the base case implies the sequence is

decreasing. Having done the base case, let's prove it for the rest of using induction, as we did in the previous question:

Thus . Note I wasn't as formal here about stating our use of the induction hypothesis, and I didn't explicitily show that

(it just follows from the rules of inequalities from earlier in the course). A more formal treatment can be found i

Question 1 above. When doing quizzes/tests, it's a good idea to completely rationalize each step.

Anyway, now we show there is a lower bound. As we know can't be less than 1, 1 seems like a good candidate for a lower bound. Proving by

induction:

Base case:

Induction step: .

Thus by the Monotone Convergence Theorem, converges. Using the property of -tails explained in Question 1:

And we're done.

2.3.3Question 3edit

Let and for . Show that is decreasing and bounded below by 2. Find the limit.

Because is defined with an inequality, this problem also requires the trick we used in Question 2. Let's see what we get when we take :

Since , we know . Thus the whole expression is negative, and for the base case, the sequence is

decreasing. Let's prove it for the rest of the natural numbers with induction. Having already done the base case, here's the induction step:

Bam, easy. Just remember, when doing quizzes or tests, explain each step, like I did in Question 1. Now let's show the sequence is bounded below by 2, as the

question asks.

Base case:

Induction step:

Now for the limit, using the same method as in Questions 1 and 2 (rationalization for it is given in Question 1):

We get two roots, . The limit must be 2 in this case, as the sequence is decreasing, so it will hit 2 before it hits 1.

2.3.4Question 4edit

Let and for . Show that converges and find the limit.

Same as Question 1, pretty much. There is no extra difficulty like in Questions 2 and 3. I know you can do the proofs for showing the sequence converges, but

the limit part might be a bit tricky. As always, if , we can say:

2

> 1

2

2

1

> 1

1

= 2 (

+ 2 1 )

( 1

2

1

1

1

1

1

1

2

1

1

1

1

1

)

2

( 1 > 0

1

)

2

> 1

( 1 / ) < 0

1

1

N

= 2 > 2 =

+ 1

1

1

+ 1

+ 2

>

N

+ 1

> 2 > 2

+ 1

1

1

+ 1

2

1

> 1

1

= 2 2

1 1

+ 1

1

1

1

( )

L = 2 = 2 L 1 L = 1

1

L

L

2

2 1

1

= 1 +

+ 1

1

N( )

1

2

1

= 1 + = ( 1 ) = ( 1

2

1

1

1

1

1

1

1

1

1

1

1

2

1

1

1

1

1 0 1

1

= 1 + > 1 + =

+ 1

1

1

+ 1

+ 2

2

1

= 1 + 1 + = 2

2 2

+ 1

1

2 1

L= 1 +

( L 1 =

L 1 3

L+ 2 = 0

L 1

)

2

L

2

L= 1 ,

L= 2

= 1

1

=

+ 1

2 +

N( )

L

L = L 2 = 0

( L 2 ) (

L+ 1 ) = 0

2
http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-3_7http://www.wikinotes.ca/MATH_242/summary/fall-2012/answers-to-homework-set-1/edit?section=question-4_7


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So we get two limits, and . Since is defined by a square root, all terms of must be positive. Therefore a negative limit makes no sense, so the limit

must be 2.

2.3.5Question 5edit

Let , where , and for . Show that converges and find the limit. [Hint: One upper bound

is .]

Now this one looks fun - we have a variable in here! We want to show it converges, so if we show that the sequence is both monotone and bounded then we'veaccomplished that. So let's first show that the sequence is increasing (the hint sort of gives away the fact that it increases).

Base case: .

Induction step:

Now we need to show the sequence is bounded above. The question hints that an upper bound is . The reason they give this particular bound is will

become apparent when we do the induction step for the proof.

Base case:

Induction step:

Now we know the sequence, let's find the limit:

Using our old middle school friend Mr. Quadratic Equation, we obtain solutions:

As the latter solution is negative, we can ignore it since the sequence is strictly positive. Thus the limit is:

2.4Section 3.4edit

2.4.1Question 1edit

Consider the sequence . This sequence is unbounded but the subsequence converges. It is easy to define piece-wise

sequences like this that satisfy the stated condition.

2.4.2Question 2edit

Use the method of Example 3.4.3(b) to show that if , then .

If we define , then we can show that is monotone and use the Monotone Convergence Theorem to prove the sequence converges to some

limit. First, we'll show that the sequence is increasing. Since we can say that:

If we can show that is positive, then we know is increasing. Thus:

As we can continue with the above equation as follows:

L =

L 2 = 0

( L 2 ) (

L+ 1 ) = 0 2 +

L

L

2

1 2

=

1

> 0 =

+ 1

+

N( )

1 + 2

= > = =

2

+

+ 0

1

= < =

+ 1

+

+

+ 1

+ 2

1 + 2

= 1 + 2

1

= < = = + 1 1 + 2

+ 1

+

+ 1 + 2

( + 1

)

2

L = L = 0

+ L

L

2

L= ,

L =

1 + 1 + 4

2

1 1 + 4

2

L =

1 + 1 + 4

2

= ( 1 , 1 / 2 , 3 , 1 / 4 )

+ 1

0


19/20



As the exponentiation of any positive number is positive, we know . By our definition of we know , so

Thus the expression between the brackets is also positive, hence the entire expression is positive. Therefore is increasing.

Now we need to show there is an upper bound. It makes sense to choose 1, and we can prove it using the same definition of . Since

and , thus .

So we now know that converges to some limit such that . By Theorem 3.4.2 it follows that . By the definition of we

see that:

By Theorem 3.2.10, . Thus which means is either 0 or 1. Since is increasin

and greater than for all we deduce the limit is 1.

2.4.3Question 3edit

Let be the Fibonacci Sequence (see Example 3.1.2(d)). Let . Given that exists, determine the value of .

Notice that . Then we have

Notice that . By Theorem , we get

.

This gives us two possible solutions. One is negative we do not need to consider it. This leaves us with

2.4.4Question 4edit

Show that the following sequences are divergent.

Consider the sequence . . Now consider the sequence . . Thus by theorem 3.4.5

is divergent.

We know that (at least you should probably know that). This is the subsequence , and it is

the constant sequence , the limit of which is of course . Also, is the constant sequence , with limit . This is the

subsequence . Since we have two subsequences with different limits, we can invoke Theorem 3.4.5 again to conclude that is divergent.

2.5Section 3.5edit

Later

2.6Section 3.6edit

= 1 = 1

a

b

1

( + 1 )

a

b

1

a

b

1

a

b

1

a

b

1

( + 1 )

a

b

1

b

a

1

( + 1 )

> 0

a

b

1

a ,b b

/ a> 1 (

b / a> 1

)

1 / (

+ 1 )

c = a / b =

c

1 /

a

1 /

b

1 /


20/20


Later

1. I don't really know which. Taken from the lecture notes for Tuesday, September 11, which I haven't typed up yet.

2. Kind of like how once in a while, the textbook will present some step without justifying it and then enclose "Why?" in parentheses. Except in this case I

don't actually know why, whereas I'm assuming the authors do. (Why don't they just tell us then???? (Why?)) In both cases, the "Why?" is left as an

exercise for the reader

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Answers to Homework Set #1 - MATH 242 (Fall 2012) - WikiNotes