Question

.Prove that 6 + 3√2 is irrational, given that √2 is irrational.

Answer 1

Step-by-step explanation:

let us assume that 6+3√2 is a rational number and it can be expressed in the form of a/b

so ,

6+3√2=a/b

6+3√2=a/b 3√2=a/b -6

6+3√2=a/b 3√2=a/b -6 3√2=a-6b/b

6+3√2=a/b 3√2=a/b -6 3√2=a-6b/b √2=a-6b/3b

a-6b/3b is a rational number so

√2 should also be a rational number

but √2 is an irrational number

hence our assumption was wrong

and

hence prooved 6+3√2 is an irrational number