# Real Numbers - Irrational Numbers Proof

Ex 1.3 Q3. CBSE 10th Maths NCERT exercise solution

#### Question: Prove that the following are irrationals.

i. $\frac{1}{\sqrt2}$
ii. 7√5
iii. 6 + √2)

##### i. $\frac{1}{\sqrt2}$

Let us assume the contrary that $\frac{1}{\sqrt2}$ is rational.
∴ we can write $\frac{1}{\sqrt2}$ = $\frac{a}{b}$
Where a and b are coprime integers such that a, b ≠ 0. In this case a ≠ 0. Else the value of $\frac{1}{\sqrt2}$ = 0, which it is not.
So, √2 = $\frac{b}{a}$
Because 'b' and 'a' are integers, $\frac{b}{a}$ is rational implying that √2 is rational.
But this contradicts with the fact that √2 is irrational.
So, the assumption that $\frac{1}{\sqrt2}$ is rational is incorrect.
∴ we can conclude $\frac{1}{\sqrt2}$ is irrational.

##### ii. 7√5

Let us assume the contrary that 7√5 is rational.
∴ we can write 7√5 = $\frac{a}{b}$ where a, b are co-prime integers such that b ≠ 0
√5 = $\frac{a}{7b}$
Because a, b and 7 are integers, $\frac{a}{7b}$ is rational implying that √5 is rational.
But this contradicts with the fact that √5 is irrational.
So, the assumption that 7√5 is rational is incorrect.
∴ 7√5 is irrational.

##### iii. 6 + √2

Let us assume the contrary that 6 + √2 is rational.
So, we can write 6 + √2 = $\frac{a}{b}$, where a, b are co-prime integers such that b ≠ 0
6 + √2 = $\frac{a}{b}$
√2 = $\frac{a}{b}$ – 6
Or √2 = $\frac{a - 6b}{b}$
Because a, b and 6 are integers $\frac{a - 6b}{b}$ is rational implying that √2 is rational.
But this contradicts with the fact that √2 is irrational.
So, the assumption that 6 + √2 is rational is incorrect.
∴ 6 + √2 is irrational.