Real numbers

REAL NUMBERS
Exercise 1.3
Rational Number :
A number which can be express in the form of p/q ,
where p and q are integers, q ≠ 0.
Irrational Number :
A number which can not be express in the form of
p/q.

Qn. Prove that √5 is irrational.
Proof:
Let us assume that √5 is rational ,then
√5 = a/b ( a & b are integers ,Coprime, b≠0)
(√5)2 = (a/b)2 (squaring on both the sides)
5 = a2 / b2 => 5b2 = a2  ①
therefore, a2 is divisible by ‘5’, then a also divisible by 5.
now, a = 5k ( where a is multiple of 5 , k some integer)
( a )2 = (5k)2 (squaring on both the sides )
=> a2 = 25 k2
=> 5b2 = 25 k2 ( from ① ) => b2 = 5k2  ②
therefore, b2 is divisible by 5, then b also divisible by 5.
from ① and② , a & b have 5 as a common factor
which is contradiction to our assumption.
So, we conclude that √5 is an irrational number.

Qn. Prove that √11 is irrational.
Proof:
Let us assume that √11 is rational ,then
√11 = a/b ( a & b are integers ,Coprime, b≠0)
(√11)2 = (a/b)2 (squaring on both the sides)
11 = a2 / b2 => 11b2 = a2  ①
therefore, a2 is divisible by ‘11’, then a also divisible by 11.
now, a = 11k ( where a is multiple of 11 , k some integer)
( a )2 = (11k)2 (squaring on both the sides )
=> a2 = 121 k2
=> 11b2 = 121 k2 ( from ① ) => b2 = 11k2  ②
therefore, b2 is divisible by 11 , then b also divisible by 11.
from ① and② , a & b have 11 as a common factor
which is contradiction to our assumption.
So, we conclude that √11 is an irrational number.

Qn. Show that 3 + √5 is irrational
Solution :
Let us assume, to the contrary, that 3 + √5 is rational.
=> 3 + √5 = a / b , ( a & b are intezers , b ≠ 0)
= > √5 = a / b - 3
=> √5 = ( a - 3b ) / b
Since ( a-3b ) / b is rational number, then 3+ √5 is also rational.
But this contradicts the fact that √5 is irrational.
So, our assumption 3 + √5 as rational is wrong.
Hence , we conclude that 3 + √5 is irrational.

Qn. Show that 11√7 is irrational.
Solution:
Let us assume, to the contrary, that 11√7 is rational.
=> 11√7 = a / b , ( a & b are intezers , b ≠ 0)
=> √7 = a / 11b
Since a/11b is rational number, then √7 is also rational
But this contradicts the fact that √7 is irrational.
So, our assumption 11√7 as rational is wrong.
Hence , we conclude that 11√7 is irrational.

General forms:
General form an even number is:
2k , where k is a +ve integer
General form an odd number is:
2k + 1 , where k is a +ve integer
Euclid Division lemma:
a = b * q + r, 0 < r < b

d) Show that any positive odd integer is of the form
6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Ans. Let a be any positive odd integer and b = 6.
Then, by Euclid divison algorithm ,
a = 6q + r for q ≥ 0, and r = 0, 1, 2, 3, 4, 5
hence , a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5
since a is odd, a cannot be 6q or 6q + 2 or 6q + 4
as these are divisible by 2.
therefore, any odd integer can be expressed in the form
6q + 1 or 6q + 3 or 6q + 5.

Qn . Use Euclid division lemma to show that the square of any
+ve integer is either of form 3m or 3m + 1 for some integer m.
Ans. Let a be any positive odd integer and b = 3.
By Euclid divison algorithm ,
a = 3q + r for q ≥ 0, and r = 0, 1, 2
then , a = 3q or 3q + 1 or 3q + 2
a2 = (3q)2 or (3q + 1)2 or (3q + 2)2
= 9q2 or 9q2 + 6q + 1 or 9q2 + 12q + 4 => 3x 3q2 + 3 x 4q + 3 + 1
= 3(3q2) or 3(3q2 + 2q) + 1 or 3(3q2 + 4q +1) + 1
= 3 k1 or 3 k2 + 1 or 3 k3 + 1
where k1 , k2 and k3 are some positive integers.
therefore, square of any +ve integer is either of the
form 3m or 3m + 1.

Qn.Use Euclid’s division lemma to show that the cube
of any +ve integer is of the form 9m, 9m + 1or9m + 8
Ans
Let a be any positive integer and b = 3
By Euclid divison algorithm
a = 3q + r, for q ≥ 0 and r = 0, 1, 2
then, a = 3q or 3q + 1 or 3q + 2

An algorithm is a series of well
defined steps which gives a
procedure for solving a type of
problem.

A lemma is a proven statement used
for proving another statement.

Euclid’s Division Lemma:
If a and b are two positive integers
then they must satisfy the condition
a = bq + r, where 0 ≤ r < b.
# it is stated for only positive
integers, it can be extended for
all integers except zero

Euclid’s division lemma and algorithm
are so closely inter linked to that of
division algorithm .
Dividend = Divisor * Quotient + Remainder

Find HCF(113,17) .... Division method
17 ) 113 ( 6
102
11) 17 ( 1
11
6 ) 11 ( 1
6
5 ) 6 ( 1
5
1) 5 (5
5
0
Therefore ,HCF (113,17) = 1

• By Euclid Division Lemma method
• Let a = 113, b = 17
• Since a > b , then by applying E D L ( a = b * q + r )
• 113 = 17 x 6 + 11
•
• 17 = 11 x 1 + 6
•
• 11 = 6 x 1 + 5
•
• 6 = 5 x 1 + 1
•
• 5 = 1 x 5 + 0
• Therefore, HCF(113,17) = the last divisor = 1

Finding HCF of given numbers by using
Euclid’s Division Algorithm:
a) 28 and 72
We have 72 >28, then
72 = 28 x 2 + 16
28 = 16 x 1 + 12
16 = 12 x 1 + 4
12 = 4 x 3 + 0
Therefore, HCF (28,72) is 4

b) 867 and 255
we have 867 > 255, then
867 = 255 x 3 + 102
255 = 102 x 2 + 51
102 = 51 x 2 + 0
Therefore, HCF (255,867) is 51

c)An army contingent of 616 members is to march behind an army band of 32
members in a parade. The two groups are to march in the same number of columns.
What is the maximum number of columns in which they can march?
Sol: We have to find the HCF (616,32) to find the
max. No. Of columns in which they can march
Since 616 > 32, then
616 = 32 x 19 + 8
32 = 8 x 4 + 0
therefore, HCF( 616,32) is 8
Hence they can march in 8 columns each.

Problem set:
1. Use Euclid’s division algorithm to find the HCF of :
(i) 135 and 225
(ii) 196 and 38220
2.A sweet seller has 420 kaju barfis and 130 badam barfis. She
wants tostack them in such a way that each stack has the
same number, and they take up theleast area of the tray.
What is the number of that can be placed in each stack for
this purpose?
3. What is the HCF of
a) two prime numbers b) two even numbers
c) two odd numbers

Real numbers

More Related Content

What's hot (20)

Similar to Real numbers (20)

Recently uploaded (20)

Real numbers