Real Number Presentation

7/28/2019 Real Number Presentation

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Topic Real Numbers By : _________________


Real numbers consist of all the rational and irrational numbers. The real number systemhas many subsets: Natural Numbers Whole Numbers Integers Real Numbers


Natural Numbers Natural numbers are the set of counting numbers which starts from 1 .

They are denoted by N Example : {1, 2, 3,}


Whole Numbers Whole numbers are the set of numbers that include 0 plus the set of

natural numbers. Example : {0, 1, 2, 3, 4, 5,}


An integer is a whole number (not a fractional number) that can be positive, negative, or

zero. It is denoted by Z . Example : Z = {..., -3, -2, -1, 0, 1, 2, 3, ...} Integers


Rational Numbers Rational numbers are any numbers that can be expressed in the form

of a/b , where a and b are integers, and b 0. They can always be expressed by usingterminating decimals or repeating decimals. Example : 2/3, 6/7,1


Terminating Decimals Terminating decimals are decimals that contain a finite number of

digits. Examples: 36.8 0.125 4.5 Repeating Decimals Repeating decimals are decimalsthat contain a infinite number of digits. Examples: 0.333 7.689689 Non Terminating

Decimals While expressing a fraction into a decimal by the division method, if

the division process continues indefinitely, and zero remainder is never obtained then

such a decimal is called Non-Terminating Decimal OR A non-terminating decimal is adecimal never repeats. Example : 0.076923...., 0.05882352.....


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Euclid's Division Lemma Euclid's division lemma states that " For any two positive

integers a and b, there exist integers q and r such that a= bq+r , 0 r< b Example : For a=

15,b=3 it is observed that 15=3(5)+0 where q=5 and r=0 Lemma : A lemma is a proven

statement used for proving another statement


Euclid Division Algorithm Algorithm : An algorithm is a series of well defined steps

which gives a procedure for solving a type of problem. Euclid division algorithm can be

used to find the HCF of two numbers. It can also be used to find some commonproperties of numbers . To obtain the HCF of two positive integers,say c and d, with c>d ,

we have to follow the steps below: STEP 1 : Apply euclid division lemma, to c and d. So,

we find whole numbers,q and r such that c= dq+r STEP 2 : If r=0,d is the HCF of c and d.

If r does not equal to 0 , apply the division lemma to d and r. STEP 3 : Continue theprocess till the remainder is zero. The divisior at this stage will be the required HCF.


Question : Use Euclid algorithm to find the HCF of 455 and 42 ? Solution : Since

455>42, we apply division lemma to 455 and 42 , to get 455= 42 10+35 Remainder isnot zero therefore we apply lemma to 42 and 35, 42=35 1+7 Again remainder is not

zero therefore we apply lemma to 35 and 7 35=7 5+0 The remainder has become zero ,

and we cannot proceed any further ,therefore the HCF of 455 and 42 is the divisor at this

stage , i,e 7


The Fundamental Theorem of Arithmetic Every composite number can be expressed as aproduct of primes. This representation is called prime factorisation of the number. This

factorisation is unique, apart from the order in which the prime factors occur.


The HCF of two numbers is equal to the product of the terms containing the least powersof common prime factors of the two numbers. Highest Common Factor(HCF) : The LCM

of two numbers is equal to the product of the terms containing the greatest powers of all

prime factors of the two numbers. Lowest Common Multiple(LCM) :



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Relationship between HCF and LCM For any two positive integers a and b , HCF ( a , b )

LCM ( a , b ) = a b Example : if a=3 and b=6 HCF(3,6) LCM(3,6) = 3 6 3 6 =

18 18 = 18 Hence verified..


Question : Given that HCF(306,657)=9, find LCM(306,657) S olution : We know that the

product of the HCF and the LCM of two numbers is equal to the product of the given

numbers . Therefore HCF(306,657) LCM(306,657) = 306 657 9 LCM(306,657) = 306657 LCM(306,657) = 306 657/9 LCM =22338


Irrational Numbers Irrational numbers are any numbers that cannot be expressed as a/b .

They are expressed as non-terminating, non-repeating decimals ; decimals that go onforever without repeating a pattern. Examples of irrational numbers:0.34334333433334 45.86745893 Pi


Things to Remember . Let p be a prime number. If p divides a 2, then p also divides a ,

where a is a positive integer. . When prime factorisation of q is of the form 2 m 5 n . Then

x has a decimal expansion which terminates and when q is not of the form 2 m 5 n , then

x has a decimal expansion which is non-terminating repeating or recurring.


. Is 17/8 has a terminating decimal expansion? Sol : We have 17/8 =17/2 3 5 2 So , the

denominator 8 of 17/8 is of the form 2 m 5 n therefore it has a terminating decimal

expansion. . Is 29/343 has a terminating decimal expansion? Sol : We have 29/343 =29/3

5 Clearly 343 is not of the form 2 m 5 n therefore it has a non terminating decimalexpansion.

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Real Number Presentation