 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.
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Correct up to this point.
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Thus if a is rational, √(1+a) is rational.
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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:
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The contrapositive is equivalent.
Therefore if a is irrational, √(1+a) is irrational.!
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