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Prove that every positive integer $n$ has a unique expression of the form: $2^{r}m$ where $r\ge 0$ and $m$ is an odd positive integer

if $n$ is odd then $n=2^{0}n$, but I dont know what to do when $n$ is even

and to prove that this expression is unique is it a good choice to use the fundamental theorem of arithmetic?

I would really appreciate your help

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This is definitely where you use the Fundamental Theorem of Arithmetic.

Note that $2^rm$ actually is the unique prime decomposition of the number guaranteed by the theorem in disguise. The only difference is that $2$ and its powers have been separated from the rest of the primes (which have been combined into a single product, $m$).

In other words, consider a number $x$ and its unique prime decomposition. We can simply apply the associative property like so to get it into your form:

$$x = 2^{k_1}p_2^{k_2}p_3^{k_3} \cdot \cdot \cdot p_n^{k_n} = (2^{k_1}) \cdot (p_2^{k_2}p_3^{k_3} \cdot \cdot \cdot p_n^{k_n})$$

And of course if $x$ is odd, the power on $2$ would simply be $0$.

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  • $\begingroup$ thanks a lot kaj_H I knew that I needed to use the fundamental theorem but someone told me that I was wrong, thanks again!! $\endgroup$ – user140027 Apr 23 '14 at 5:08
  • $\begingroup$ No problem; I'm glad I could help! Don't forget to accept an answer by clicking the check mark if you're good to go. That way, the question doesn't show up as unanswered for future users. $\endgroup$ – Kaj Hansen Apr 23 '14 at 5:12
  • $\begingroup$ Typo: if $x$ is odd, the power on 2 would simply be $\color{#c0d}{0}$. $\endgroup$ – Anant Apr 24 '14 at 13:01
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If $n$ is odd, then $n = 2^{0}\cdot n$ which satisfies the condition. If $n$ is even, then let $r$ be the largest power of the factor $2^k$, then $n = 2^{r}\cdot m$. Claim: $m$ must be odd. If not then write $m = 2s$. So: $n = 2^{r+1}\cdot s$. So $r$ is no longer the max power of factor $2^k$, contradiction. Thus the conclusion follows.

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  • $\begingroup$ thanks for the help LAcarguy :D $\endgroup$ – user140027 Apr 23 '14 at 5:07

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