Prove that (√7 +

We will prove this by contradiction. So, assuming the statement (√7 + √11) is irrational to be true.

Let k = √7 + √11

√11 = k - √7

Squaring both sides, we get,

11 = (k - √7)²

11 = k² + (√7)² – 2 × √7 × k

k² + 7 - 2√7k = 11

2√7k = 11 – 7 – k²

2√7k = 4 – k²

√7

As, k is a rational number, is also n rational number.

But √7 is an irrational number.

So, there is a contradiction.

Therefore, statement is wrong.

Hence, √7 + √11 is an irrational number.

OR

Any odd integer is of the form (2q + 1) for some integer q.

Let a = 2m + 1 and b = 2n + 1

Therefore,

a² + b² = (2m + 1)²+ (2n + 1)²

a² + b² = 4m² + 4m + 1 + 4n² + 4n + 1

a² + b² = 4{(m² + n²) + (m + n)} + 2

Let k = {(m² + n²) + (m + n)}

Then,

a² + b² = 4k + 2

a² + b² = 2(2k + 1)

Therefore, a² + b² = even number.

As the multiple is of 2 and not 4. The number is not divisible by 4.

Hence, Proved.