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Calculus theory proof- Suppose a is irrational, prove√(1+a) is irrational.

 Quote by math-help-me 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^2-q^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 $\sqrt{1+a}$ is rational, then $a$ is rational. This is exactly what you just proved.

And therefore:
 The contrapositive is equivalent. Therefore if a is irrational, √(1+a) is irrational.!