1st: I took the contrapositive. If $a^b$ is irrational, then a and b are irrational.
2nd: I found an example to disprove my contrapositive. Let $a = 2$ Let $b = 1/2$ Both are rational.
$a^b = 2^{(1/2)} = \sqrt{2}$, which is irrational.
It's too difficult to do html formatting on my phone, so I apologize in advance. Thanks.
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$\begingroup$Technically, your proof is fine. You showed that for all a and b are rational numbers, a^b is not always rational, by counterexample. Unless you're going to prove the statement by contraposition, finding the contrapositive is unnecessary. Just for future reference, the contrapositive of "if a and b then d", is "if ~d then ~a or ~b" (this is due to DeMorgan's Law).
$\endgroup$ $\begingroup$Your counter example is completely valid way to prove that your statement is false.
Your statement is:
Forall rational $a,b$, $a^b$ is also rational
We will say that $\lnot=$ the negative, so the negative of the statement is:
$\lnot($Forall rational $a,b$, $a^b$ is also rational$)$
By DeMorgan's Law:
exists rational $a,b$, $\lnot(a^b$ is also rational$)$
The negative of rational is irrational so we left with
exists rational $a,b$, $a^b$ is irrational.
So a counter example proves the negative of the statement, i.e disprove the statement
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