Can anyone show me how to prove this statement? I'm very confused. I have tried substitute numbers into each x, k and m, it seems this statement is true but it needs to be proved by mathematical methods.
Thank you!
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If $x$ is even, then $x=2^k$ for some $k$. Thus $x=(2n+1)2^k$ for $n=0$ and $m=2\times 0+1=1$ is odd. – Mauro ALLEGRANZA Apr 06 '17 at 11:24
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If $x$ is odd, then $x=2n+1$ for some $n$, and thus $x=(2n+1) \times 2^0$. – Mauro ALLEGRANZA Apr 06 '17 at 11:25
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Thus, cooking them together: $\exists k \ \exists n \ [x=(2n+1)2^k]$. – Mauro ALLEGRANZA Apr 06 '17 at 11:48
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Please include a full statement of the problem you want help with in the body of the Question. Putting the problem statement only in the title often short-changes your Readers by being cryptic. – hardmath Apr 07 '17 at 16:40