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

Question

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

Answer 1

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 $0$.

Answer 2

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.