Prove That The no. Is Irrational

Let $2 + √6$

$So\; let\; 6$+√2 = a/b, where a and b are integers

so √2 = a/b  - 6 = (a - 6b)/b ...this should be a rational number since all integers are there

implies √2 is a rational number which is a contradiction.

So our assumption that  6+√2 be a rational number is wrong and actually

 $2 + √6$