PSEB Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.3

 Q 1.Prove that √5 is irrational.

Solution:
Let us suppose that √5 is rational so we can find integers r and s where s ≠ 0
such that √5 = rs
Suppose r and s have some common factor other than 1, then divide r and s by the common factor to get :
√5 = ab where a and b are coprime and b ≠ 0
b√5 = a
Squaring both sides,
 (b√5)2 = a2
 b2 (√5)2 = a2
 5b2 = a2 

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5 divides a2.
By the theorem, if a prime number ‘p’ divides a2 then ‘p’ divides a where a is positive integer

 5 divides a 
So a = 5c for some integer c.
Put the value of a in (1),
5b2 = (5c)2
5b2 = 25c2
b2 = 5c2
or 5c2 = b2
 5 divides b2
 if a prime number ‘p’ divides b2, then p divides b ; where b is positive integer.
 5 divides b ………… (3)
From (2) and (3), a and b have at least 5 as common factor.
But this contradicts the fact that a and b are coprime i.e. no common factor other than 1.
 our supposition that √5 is rational wrong.
Hence √5 is irrational.

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Q 2. Prove that 3 + 2 √5 is irrational.

Solution:
Let us suppose that 3 + 2√5 is rational.
 we can find Co-Prime a and b, where a and b are integers and b ≠ 0

such that 3 + 2√5 = ab

Since,2 3, a and b  are integers, therefore a-3b/2b is retional, and so √5 is retional

                                                                                

But this contradicts the fact that √5 is irrational.

 our supposition is wrong.
Hence 3 + 2√5 is irrational.

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Q 3. Prove that the following are irrationals :

(i) 112√√2
(ii) 7√5
(iii) 6 + √2

Solution:


(i) Given that Let us suppose that is rational
we can find co-prime integers a, b and b ≠ 0.

 2√2=ab

 1. 1√2  = a/b

by rearranging,we get

        √2=b/a

since, a and b are integers, therefore b/a is rational, and so √2 is rational.

But this contradicts the fact that √2 is irrational.

so, we conclude that 1/√2 is irrational.

2. let us assume, to the contrary, thai 7√5 is rational.

that is, we can find coprime a and b (b≠0) such that

                7√5 = a/b

By rearranging, we get

                    √5 = a/7b

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since, a and b are intergers, therefore a/7b is rational, and so √5  is rational.

But this contradicts the fact that √5 is irrational.

so, we conclude that 7√5 is irrational.

3. Let us assume, to the contrary, that 6+√2 is rational.

That is, we can find integers a and b (b≠0) such that 

                            6+√2 = a/b

By rearranging, we get

                                √2 =a/b -6

Since, a, b and 6 are integers, therefore a/b -6 is rational, and so √2 is rational.

But this contradicts the fact that √2 id irrational.

So, we conclude that 6+√2 is irrational.

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more reed :-

Real Numbers Ex 1.1

Real Numbers Ex 1.2