Question

Show that \[3 + \sqrt{2}\] is an irrational number.

Answer 1

Let us assume that\[3 + \sqrt{2}\] is rational .Then, there exist positive co primes a and b such that

\[3 + \sqrt{2}\] `=a/b`

`sqrt2=a/b-3`

`sqrt2=(a-3b)/b`

`sqrt2` is arational number which is a contradication.

Hence `3+sqrt3` is irrational