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**Class :**10th**Chapter :**1**Subject :**Mathematics**Topic :**Real Number Exercise 1.2**Resource :**Notes, Important Question & Practice Paper

## Chapter 1 Real Number Exercise 1.2

**Q.1 Prove that √5 is irrational.**

**Solution :-** Let us assume, to the contrary, that √5 is rational.

That, we can find integrate a and b (≠ 0) such that √5 = a/b

Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.

So, b√5 = a.

Squaring on both sides and rearranging we get 5b² = a².

Therefore a² is divisible by 5, and by theorem 1.3 follows that a is also divisible by 5.

So, we can write a = 5c for some integer c.

Substituting for a, we get 5b² = 25c², that

**b² = 5c².**

This means that b² is divisible by 3, and so b also divisible by 5 (using Theorem 1.3 with p = 5).

Therefore a and b have at least 5 as a common factor.

But this contradiction has arisen because of our incorrect assumption that √5 is rational.

So, we conclude that √5 is irrational.

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

**Solution :-** We have to prove 3 + √5 is irrational

Let us assume the opposite,

i.e., 3 + √5 can be written in the form a/b

Where a and b (b ≠ 0) are coprime (no common factor other than 1)

**Hence, **

3 + 2√5 = a/b

2√5 = a/b – 3

2√5 = a – 3b/ b

2√5 = 1/2 × a – 3b/b

√5 = a- 3b/2b

Here, a – 3b is rational number But √5 is irrational

Since,

Rational ≠ Number

This is contradiction

Our assumption is incorrect

Therefore, 3 + 2√5 is irrational

**Hence Proved,**

**Q.3 Prove that the following are irrationals :**

**(i).** 1/√2**(ii).** 7√5**(iii).** 6+√2

**(i). 1/√2**

**Solution :-**

We have to prove 1/√2 is irrational

Let us assume the opposite,

i.e, 1/√2 is rational

Hence, 1/√2 can be written in the form a/b

Where a and b (b ≠ 0) are co-prime (no common factor other than 1)

Hence,

1/√2 = a/b

b/a = √2

Here, b/a is rational number

But √2 is irrational

Since,

Rational ≠ irrational

This is a contradiction

Our assumption is incorrect

Therefore, 1√2 is irrational

Hence Proved

**(ii). 7√5 **

**Solution :-**

We have to prove 7√5 is irrational

Let us assume the opposite,

i.e., 7√5 is rational

Hence, 7√5 can be written in the form a/b

Where a and b (b ≠ 0) are co-prime (no common factor other than 1)

**Hence, **

7√5 = a/b

√5 = 1/7 × a/b

Here, a/7b is a rational number

But √5 is irrational

Since,

Rational ≠ irrational

This is a contradiction

Our assumption is incorrect

Therefore, 7√5 is irrational

Hence Proved

**(iii). 6+√2**

**Solution :-**

We have to prove 6+√2 is irrational

Let us assume the opposite,

i.e., 6+√2 is rational

Hence, 6+√2 can be written in the form a/b

Where a and b (b ≠ 0) are co-prime (no common factor other than 1)

**Hence, **

6+√2 = a/b

√2 = a/b – 6

Here, a – 6b/b is a rational number

But √5 is irrational

**Since, **

Rational ≠ irrational

This is a contradiction

Our assumption is incorrect

Therefore, 6+√2 is irrational

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