**NCERT Solutions for Class 10 Maths Ex 1.2**

**Q1. Prove that √5 is irrational.**

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**Solution: **Let us assume that √5 is a rational number.

Let *a* and *b* are two co-prime numbers and *b* is not equal to 0 such that √5 = *a*/*b*

Multiply by *b* on both sides, we get*b*√5 = *a*

To remove root, squaring on both sides, we get

5*b*^{2} = *a*^{2 } … (i)

Therefore, 5 divides *a*^{2} and according to theorem of rational number (Theorem 1.3), for any prime number *p* which is divides *a*^{2}, it will divide *a* also.

That means 5 divides *a*. So we can write*a* = *5*c*a*^{2} = (*5*c)^{2}

Putting value of *a*^{2} in equation (i), we get

5*b*^{2} = (5*c*)^{2}

5*b*^{2} = 25*c*^{2}

Dividing by 5, we get

b^{2} = 5c^{2}

Therefore, 5 divides *b*^{2} and according to theorem of rational number (Theorem 1.3), 5 divides *a*.

We have already find that *a* is divide by 5.

This means *a* and *b* both are divisible by 5.

So it contradicts our assumption that *a* and *b* are co-prime numbers.

Hence, √5 is not a rational number, it is irrational.

**Q2. Prove that 3 + 2√5 is irrational.**

**Solution:** Let us assume that 3 + 2√5 is a rational number.

So, we can write this number as

3 + 2√5 = *a*/*b *. . . (i)

Where *a* and *b* are two co-prime numbers and *b* is not equal to 0.

From equation (i), we get

2√5 = *a*/*b* – 3

2√5 = (*a *– 3*b*)/*b*

√5 = (*a *– 3*b*)/2*b *. . . (ii)

In eq (ii), *a* and *b* are integers, so (*a *– 3*b*)/2*b* is a rational number.

Therefor, √5 should be a rational number.

But √5 is an irrational number.

So, this contradicts our assumption.

Hence, 3 + 2√5 is irrational.

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

(i) 1/√2 (ii) 7√5 (iii) 6 + √2

**Solution:** **(i)** Let us assume that 1/√2 is a rational number.

So, we can write this number as

1/√2 = *a*/*b*

Where *a* and *b* are two co-prime numbers and *b* is not equal to 0.

In above equation, by cross-multiplication, we get*b* = (*a*√2)

Dividing by *a* on both sides, we get*b*/*a* = √2

Here, *a* and *b* are integers, so *b*/*a* is a rational number.

Therefore, √2 should be a rational number.

But √2 is an irrational number.

So this contradicts our assumption.

Hence, 1/√2 is an irrational number.

**(ii)** Let us assume that 7√5 is a rational number.

So, we can write this number as

7√5 = *a*/*b *. . . (i)

Where *a* and *b* are two co-prime numbers and *b* is not equal to 0.

From eq (i), we get

√5 = *a*/(7*b*)

Here, *a* and *b* are integers, so *a*/7*b* is a rational number.

Therefore, √5 should be a rational number.

But √5 is an irrational number.

So this contradicts our assumption.

Hence, 7√5 is an irrational number.

**(iii)** Let us assume that 6 + √2 is a rational number.

So, we can write this number as

6 + √2 = *a*/*b *. . . (i)

Where *a* and *b* are two co-prime numbers and *b* is not equal to 0.

From eq (i), we get

√2 = *a*/*b* – 6

√2 = (*a* – 6*b*)/*b*

Here, *a* and *b* are integers, so (*a* – 6*b*)/*b* is a rational number.

Therefore, √2 should be a rational number.

But √2 is an irrational number.

So, this contradicts our assumption.

Hence, 6 + √2 is an irrational number.