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Transcript
REAL
NUMBERS
INTRODUCTION
 We all know four fundamental operations of addition, subtraction,
multiplication and division, which we have applied on natural
numbers, integers and real numbers as well.
 When we think of division , then divisibility and all the four
components of division clicks our mind .But actually what is
divisibility? And on what basis do we say a no.’a’ is divisible by
another no. ‘b’? Also how do we check whether a number is odd or
even?
 This is the place where Euclid's division lemma works. It tells us
about the divisibility of integers.
DIVISIBILITY: A non-zero integer ‘b’ is said to divide an
integer ‘a’ if there exists an integer ‘c’ such that a=bc.
Here , ’b’ is called the divisor,
‘a’ is called the dividend,
‘c’ is called the quotient.
Example: 2 is a divisor of 6 as there exist an integer 3 such that
6=2*3.
Note : ‘b’ is a divisor of ‘a’ is written as a/b
For a/b we say that ‘a’ is divisible by ‘b’ or ‘b’ is a factor of ‘a’ or ‘a’
is a multiple of ‘b’ or ‘b’ is a divisor of ‘a’.
EUCLID’S DIVISION LEMMA
 Let ;a; and ‘b’ be any two positive integer .Then there exist unique
integers ‘q’ and ‘r’ such that
a=bq +r , 0 < = r < b.
let us consider the division of one positive integer with another ,say
45 by 8.
46=8*5+6, where 0<6<8.
the result of this division is that we get two integers - 5 which is called
the quotient and 6 which is the remainder.
 Any positive even integer is of the form 2m and any positive even
integer is of the form 2m+1.
 Any positive integer is of the form 3m,3m+1 and 3m+2.
EXAMPLES
1) Using Euclid’s Division Lemma show that every even integer is of the
form 2q and every odd integer is of the form 2q+1.
Sol: Let ‘a’ be any positive integer and b=2, then ,r=0,1.
case(1) when r=0
 a=2q+0
 a=2(q)
 a=2m ….even




Case(2) when r=1
a=2q+1
a=2(q)+1
a=2m+1 ……odd.
 a=2(q)+1
 a=2m +1 …odd


2)Show that an odd integer is of the 
form 4q+1 or 4q+3 ,where q is some 
positive integer.
Sol: Let a be the positive odd integer 
and b =4, then r=0,1,2,3.

 case(1)when r=0

 a=4q+0

 a=2(2q)

 a=2m ….even
 Case(2)when r= 1
Case(3)when r=2
a=4q+2
a=2(q+1)
a=2m ….even
Case(4)when r=3
a=4q+3
A=4q+2+1
A=2(2q+1)+1
A=2m +1 …odd
EUCLID’S DIVISION ALGORITHM
 Euclid’s Division Algorithm is an application of Euclid’s Division Lemma.
It is a technique to compute the HCF of two positive integers.
 To obtain the HCF of two positive integers ,say a and b with a>b, we
follow the steps given below.
STEP 1:Apply Euclid’s Division Lemma ta a and b.
So we find whole numbers q and r such that
a=bq+r , 0=<r<b.
STEP2:If r=0,b is the HCF of a and b .But if r is not equal to zero ,apply
the lemma to b and r.
STEP3:Continue the process till the remainder is zero. The divisor at this
stage will be the required HCF.
Example: Suppose we have to determine the HCF of 1756 and 144.
We start with the greater integer , i.e.,1756.By the lemma , we have
1756=144*12+28
Now again applying the lemma on the divisor 144 and the remainder 28 ,we
have
144= 28*5+4
Now again applying the lemma on the divisor 28 and remainder 4, we have
28=4*7+0
Here the last divisor is 4
Thus , 4= HCF(28,4)= HCF(144,28) =HCF(1756,144).
NOTE: HCF of (a,b) * LCM of (a,b) =a*b.
QUESTIONS
Q1.Using Euclid’s Division Algorithm, find the HCF of the following
a)65 and 170
b)1264 and 82
c)2165 and 272
Q2. If the HCF of 45 and 210 is expressible in the form 210x + 45 * 5, find x.
Q3.Find the HCF d of 117 and65. Also find integers x and y such that
d= 117x = 65y.
Q4.Find the largest positive integer that will divide 398, 436 and 542 leaving
remainder 7, 11 and 15 respectively.
Q5.Prove that n^2 –n is divisible by 2 for every positive integer n.
Q6Prove that if a positive integer is of the form 6q +5, then it is of the form
3q+2 for some integer q , but not conversely.
Q7.Prove that the product of two consecutive positive integer is divisible by 2.
Q8.Prove that the square of any positive integer of the form 5q+1 is of the
same form.
F U N DA M E N TA L T H E O R E M O F
ARITHMETIC
 Every composite number can be expressed as a product of
primes and this factorisation is unique, apart from the order
in which the prime factors occur.
For example (1) 6860 = 2^2*5*7^3.
(2) 1200 = 2^4*3*5^2
There is no other possible factorisations of 6860 and 1200 into prime
factors.
So the fundamental theorem of arithmetic tells us that prime numbers
are the building blocks for all the positive integers.
 Let p be a prime no. and a be a positive integer. If p divides
a^2, then p divides a.
R E V I S I T I N G I R R AT I O N A L
NUMBERS
RATIONAL NUMBERS: A no. Is said to be a rational number if it can be
written in the form of p/q where p and q are integers , p is not equal to 0
and the HCF of p and q is 1.
IRRATIONAL NUMBER:A no. is said to be an irrational no. if it’s Nth
root is not a whole no.
NOTE: A no. which is not a rational no. is an irrational no.
QUESTIONS
Q1.Find the HCF and LCM of the following integers
a) 84 b) 120 c) 256 d) 8008 e) 760
Q2.The HCF of two no.s is 54 and their LCM is 3024. If one no. is 336
find the other.
Q3. Fid the least no. that is divisible by all the no.s between 2 to 10 (both
inclusive).
Q4.. .Find the largest positive integer that will divide 398, 436 and 542
leaving remainder 7, 11 and 15 respectively.
Q5.Check whether 4^n can end with digit 0?
Q6.Prove that root 2,3,4,5,6 are irrational numbers.
Q7.Prove that there is no rational no. whose square is 6.
Q8.For any positive real no. x prove that there always exists an irrational no.
y satisfying 0<y<x.
DECIMAL REPRESENTATION
RATIONAL NUMBERS: They have either a terminating or a nonterminating repeating decimal expansion.
IRRATIONAL NUMBERS : They have non- terminating and non repeating
decimal expansion.
Let us take the following terminating decimal expansions:
a)0.35 = 35/100 =7/20=7/2^2*5
b)0.58 =58/100=29/50=7/2*5^2
In each case the terminating decimal expansion reduces to a rational no.of the
form p/q, where p and q are co-prime and the prime factorisation of q is of
the form 2^m*5^n, where m and n are no- negative integers.
 Let x be a rational no. whose decimal expansion terminates. Then x
can be expressed in the form p/q , where p and q are co –primes ,and the
prime factorisation of q is of the form 2^m*5^n, where m and n are nonnegative integers.
 Let x = p/q be a rational no. such that the prime factorisation of q is
of the form 2^m*5^n, where m and n are non- negative integers.
Then x has a decimal expansion which terminates.
 Let x = p/q be a rational no. such that the prime factorisation of q is
not of the form 2^m*5^n, where m and n are non- negative integers.
Then x has a decimal expansion which is non terminating repeating.
QUESTIONS
Q1.State whether the following will have a terminating or a nonterminating repeating decimal expansion.
a)11/125
b)137/260
c)23/6250
d)217/4000
e)215/4500
Q2.What can you say about the prime factorisation of the
denominators of the following rational no.
a)37.3507
b)572.1234569
c)272.25123
THANK YOU