Question

prove that 6+sqrt2 is irrational and 7sqrt5 is irrational?

Answer 1

1. Let us assume to the contrary that 6 + √2 is a rational number. Then we can find integers a and b such that 6+√2 = a/b

i.e., √2 = a/b - 6.

It means that a/b - 6 is rational number i.e, √2 is rational number.

But √2 is a irrational number.

This contradiction has arisen because of our incorrect assumption that 6+ √2 is a rational number.

So we conclude that 6+ √2 is a irrational number.

2. Let us assume to the contrary that 7√5 is a rational number. Then we can find integers such as r and s such that 7√5 = r/s i.e, √5 = r/7s.

Now r/7s is a rational number. It means that √5 is a rational number.

But √5 is a irrational number.

This contradiction has arisen because of our incorrect assumption that 7√5 is a rational number.

Thus, we conclude that 7√5 is a irrational number.