Prove that the following are irrational (i) 1/√2 (ii) 7√5 (iii) 6+√2

Question – Prove that the following are irrational (i) 1/√2 (ii) 7√5 (iii) 6+√2

(i) 1/√2

Let us assume the contrary that 1/√2 is not irrational.

This means that 1/√2 is a rational number.

Then, we can find two co prime integers a and b, where (b ≠ 0) such that

1/√2 = a/b

On rearranging, we get

⇒ b/a  = √2.

Since a and b are integers then b/a  is rational

Therefore, √2 is rational, which is a contradiction to the fact that √2 is irrational.

This contradiction has arisen because of our incorrect assumption that 1/√2 is rational.

Therefore, we concluded that 1/√2 is irrational.

(ii) 7√5

Let us assume, the contrary that 7√5 is not irrational.

This means that 7/√5 is a rational number.

Then, we can find two co prime integers a and b, where (b ≠ 0) such that

7√5 = a/b

On rearranging, we get

⇒ a/7b = √5.

Since, a and b are integers, then a/7b is rational

Therefore, √5 is rational, which is a contradiction to the fact that √5 is irrational.

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

Therefore, we concluded that 7/√5 is irrational.

(iii) 6+√2

Let us suppose, the opposite that 6+√2 is not irrational, i.e., rational number.

Then, we can find two co prime integers a and b, where (b ≠ 0) such that

6+√2 = a/b

⇒ a/b  – 6 = √2

⇒ (a – 6b)/b = √2

Since a, b and 6 are integers, then (a – 6b)/b is rational

Therefore, √2 is rational, which is a contradiction to the fact that √2 is irrational.

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

Therefore, we concluded that 6+√2 is irrational.

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