Quote by mathhelpme
that was sort of my attempt at a solution. This is what i've got now.
Assume a irrational, but √(1+a) is rational.
Then √(1+a) = p/q for p,q integers q≠0.
1+a = p^2/q^2 → a = p^2/q^2  1 → (p^2 q^2)/q^2
Since p,q integers, p^2q^2 and q^2 must be integers.
Thus a must also be rational by definition, a contradiction.

Correct up to this point.
Thus if a is rational, √(1+a) is rational.

No, it's the other way around. If [itex]\sqrt{1+a}[/itex] is rational, then [itex]a[/itex] is rational. This is exactly what you just proved.
And therefore:
The contrapositive is equivalent.
Therefore if a is irrational, √(1+a) is irrational.!
