# Is this a valid proof of the contrapositive?

The question is the following: if $a$ and $b$ are distinct group elements, then either $a^2 \neq b^2$ or $a^3 \neq b^3$. I find this difficult to prove directly, so I formulated the contrapositive to be: if both $a^2 = b^2$ and $a^3 = b^3$, then $a = b$.

The proof is simple: Suppose that $a^2 = b^2$ and $a^3 = b^3$. Then $a^3 = b^3 \Rightarrow a^2 = a^{-1}b^3 \Rightarrow a^{-1}b^3 = b^2$ which then implies that $\Rightarrow a^{-1}b^2 = b \Rightarrow a^{-1}b = e \Rightarrow b = a$.

Is this correct? I feel like there's a hole somewhere.

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Yes, it's fine. – user61527 Sep 25 '13 at 2:22
Nice proof. I like it. – Anonymous Sep 25 '13 at 2:24
I hate to put down an answer that is so trivial as to say "yes, you have answered the question with a good answer", but I don't know how to satisfy the need for the question to have a posted answer otherwise... – abiessu Sep 25 '13 at 2:25
@abiessu: There are at least two reasonable alternatives. One is the one that Adriano chose: find something to add, even if it’s just showing how to polish the proof or organize it more efficiently. The other is to make the answer community wiki. – Brian M. Scott Sep 25 '13 at 2:30
@BrianM.Scott: thank you, I like those alternatives. – abiessu Sep 25 '13 at 2:30

As the comments have pointed out, you've done your proof correctly. Good job! Here's how you can organize the steps into a single chain of equalities: $$a = a^3a^{-2} = (a^3)(a^2)^{-1} = (b^3)(b^2)^{-1} = b^3b^{-2}=b$$