Prove that all integers can be represented as powers of 2 multiplied by an odd number

Question

I would like to prove that $\forall n \in \mathbb{N},n\geq1$ $, n = 2^k\times m$ with $k \in \mathbb{Z}$ and $ m $ odd and $m \in \mathbb{N}$.

Answer 1

As a next move, why not consider the cases that $j$ is even or odd? For example, if $j$ is even, you can consider $j'=j/2$ which you know does have the property that $j'=2^k m$ for some $k,m\in \mathbb N$ and $m$ odd, because its smaller than $j$.

Answer 2

Let $a$ and $n$ be positive integers with $a > 2$. Clearly, the nonempty set $\{j \in \mathbb N \text{ s.t } a^j | n\}$ is finite and therefore contains a largest element; call it $k$. Now set $m = \frac{n}{a^k}$, a positive integer. It is clear that is clear that $n = a^km$ and $a$ doesn't divide $n$, by definition of $k$.

The case $a = 2$ is a particular case of the above argument.