**NCERT Solutions for Class 10 Maths**

Chapter 1 Real Numbers Ex 1.3

Chapter 1 Real Numbers Ex 1.3

**Page 14**** **

**1. Prove that √5 is irrational.**

**Answer :**

Let’s assume, that **√**5 is rational number.

i.e. **√**5 = x/y (where, x and y are co-primes)

y**√**5= x

Squaring both the sides, we get,

(y**√**5)^{2} = x^{2}

⇒ 5y^{2} = x^{2}……………………………….. (1)

Thus, x^{2} is divisible by 5, so x is also divisible by 5.

Let us say, x = 5k, for some value of k and substituting the value of x in equation (1), we get,

5y^{2} = (5k)^{2}

⇒ y^{2} = 5k^{2}

is divisible by 5 it means y is divisible by 5.

Therefore, x and y are co-primes. Since, our assumption about is rational is incorrect.

Hence, **√**5 is irrational number.

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

**Answer :**

Let’s assume that 3 + 2√5 is a rational number.

So we can write this number as

3 + 2√5 = *a*/*b*

Here a and b are two co prime number and b is not equal to 0

Subtract 3 both sides we get

2√5 = *a*/*b* – 3

2√5 = (*a*-3*b*)/*b*

Now divide by 2, we get

√5 = (*a*-3*b*)/2*b*

Here *a* and *b* are integer so (*a*-3*b*)/2*b* is a rational number so √5 should be a rational number But √5 is a irrational number so it contradicts.

Hence, 3 + 2√5 is a irrational number.

**3. Prove that the following are irrationals : (i) 1/√2 (ii) 7√ 5 (iii) 6 +√2**

**Answer :**

**(i) 1/**√**2**

Let us assume 1/√2 is rational.

Then we can find co-prime x and y (y ≠ 0) such that 1/√2 = x/y

Rearranging, we get,

√2 = y/x

Since, x and y are integers, thus, √2 is a rational number, which contradicts the fact that √2 is irrational.

Hence, we can conclude that 1/√2 is irrational.

**(ii) 7**√**5**

Let us assume 7√5 is a rational number.

Then we can find co-prime a and b (b ≠ 0) such that 7√5 = x/y

Rearranging, we get,

√5 = x/7y

Since, x and y are integers, thus, √5 is a rational number, which contradicts the fact that √5 is irrational.

Hence, we can conclude that 7√5 is irrational.

**(iii) 6 +**√**2**

Let us assume 6 +√2 is a rational number.

Then we can find co-primes x and y (y ≠ 0) such that 6 +√2 = x/y⋅

Rearranging, we get,

√2 = (x/y) – 6

Since, x and y are integers, thus (x/y) – 6 is a rational number and therefore, √2 is rational. This contradicts the fact that √2 is an irrational number.

Hence, we can conclude that 6 +√2 is irrational.

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