Question

prove that a) 6-√2 are irrational no​

Answer 1

Proof: By contrdiction.

Assumed that 6-√2 is rational,

Therefore, 6-√2 = a/b [Where a & b are coprimes]

6-√2 = a/b

or, -√2 = a/b - 6

or, -√2 = (a-6b)/b

or, √2 = (6b-a)/b

since, a & b are rational hence (6b-a)/b is rational. This contradicts the fact that √2 is irrational.

Hence, (6b-a)/b is irrational.

This contradiction has been arise due to our wrong assumption that 6-√2 is rational.

Hence, 6-√2 is irrational.

Answer 2

Assumed that 6-√2 is rational,

Therefore, 6-√2 = a/b [Where a & b are coprimes]

6-√2 = a/b

or, -√2 = a/b - 6

or, -√2 = (a-6b)/b

or, √2 = (6b-a)/b

since, a & b are rational hence (6b-a)/b is rational. This contradicts the fact that √2 is irrational.

Hence, (6b-a)/b is irrational.

This contradiction has been arise due to our wrong assumption that 6-√2 is rational.

Hence, 6-√2 is irrational.