Show that any natural number $n$ can be written on the form $n = 2^k \cdot m$

Question

Show that any natural number $n$ can be written on the form $n = 2^k \cdot m$, where $m$ is an odd integer and $k \geq 0$.

I know that if $n$ is an integer, it can be written on the form $n = \frac{a}{b}$ for two integers $a$ and $b$. Also, $m$ can be written as $m = 2c +1$ for an integer c, since it's odd. But I don't know how to use this to come up with a proof. Also, I have to show that $n$ is a non-negative integer

Answer 1

Let we consider the following algorithm having $n\geq 1$ as input.

1. Initialization: we set $k=0$ and $m=n$;
2. If $m$ is odd, we return $(k,m)$;
3. If $m$ is even, we replace $m$ by $\frac{m}{2}$ and $k$ by $k+1$. Go to $2.$

If $n\leq 2^H$, we perform step $(3.)$ at most $H$ times. It follows that the above algorithm terminates for every input $n$, providing us the wanted representation $$ n=2^k \cdot m.$$ It is the same as considering the binary representation of $n$ and removing the trailing zeroes: that gives $m$. $k$ is just the number of the removed trailing zeroes.

Answer 2

Hint:

Use prime decomposition, and remember that $2$ is a prime number ( and $2^0=1$ if $2$ is not a factor of $n$).