Prove that all naturals $n = 2^k j$. If we have a natural $n$, then $n = 2^k j$, where $k$ and $j$ are naturals, and this factorization is unique.  We add the restriction that $n \ne 0$ and $k \ne 0$ and $j$ is odd in order to make this statement valid.
Can someone prove this?
 A: Start with a number, say $n=792$.
If it's divisible by $2$, divide it by $2$, thus: $792/2 = 396$.
If that's divisible by $2$, divide it by $2$, thus: $392/2=198$.
If that's divisible by $2$, divide it by $2$, thus: $198/2=99$.
You divided by $2$ three times, so $k=3$ and $j=\text{what you got}=99$.
If you'd started with $n=793$, you'd have divided by $2$ zero times, so you'd have $k=0$ and $j=793$.
In other words, this is just mathematical induction.
A: Hint $\ $ Generally, in any ring, if $\rm\:c\:$ is a cancellable and $\rm\: c^j b = c^kd\,$ then $\rm\,c\nmid b,d\:\Rightarrow\:j = k.\:$ Proof: wlog $\rm\: j\le k\:$ so cancelling $\rm\:c^j\:$ yields $\rm\:b = c^{k-j}d,\:$ so $\rm\:k = j\:$ (else $\rm\:k>j\:\Rightarrow\: c\mid b,\:$ contra hypothesis).  
Thus such representations are unique if they exist; e.g. if $\rm\, 0\ne n\in \Bbb Z\,$ and $\rm\:c \ne 0,\pm1\:$ then we can let $\rm\:c^j\:$ be the highest power of $\rm\:c\:$ that divides $\rm\:n\:$ (which need not exist in general rings). 
Note that the proof does not use unique factorization or any related strong special properties. Rather, it depends only on cancellation (and that $\rm\,n\ne 0$ is not divisible by all powers of $\rm\:c).$
A: If in $n = 2^i j$ you take $n \ne 0$ and $j$ odd, then this is nothing more than the fundamental theorem of arithmetic: Each $n$ can be factored into primers (2 is prime) in essentially one way only.
A: In this answer I will always use the convention $0\in\mathbb{N}$.
Assume that for every $n<N$, $n=2^k(2j+1)$ for some $k,j\in\mathbb{N}$.
Then either $N$ is odd and hence $N=2^0(2j+1)$ for some $j\in\mathbb{N}$ or $N$ is even and hence $N=2n$ for some $n<N$. Since $n=2^k(2j+1)$ we then have $N = 2n = 2^{k+1}(2j+1)$.
Thus, since $1=2^0\cdot 1$ we may conclude by induction that $n=2^k\cdot(2j+1)$ for any $n\in\mathbb{N}$.
For uniqueness consider $2^k(2j+1) = 2^a(2b+1)$ and assume without loss of generality that $k\leq a$.
Then $2j+1 = 2^{a-k}(2b+1)$ and since the left hand side is odd we have $k=a$ and thus j=b follows as well.
A: Note: $k$ should also include $0$, if it doesn't all the odd naturals, can't be expressed in your form.
$2k+1=2^0 \cdot (2k+1) $, if you don't include $0$, you can only express only ($\mathbb{N}$-odd).
A: I will assume $0\in\mathbb{N}$. 
Define $ord_2 : \mathbb{N}\rightarrow \mathbb{N}$ as $ord_2(a)$ is the highest power of $2$ that divides $a$. Since $a$ is a number $ord_2(a)$ will be finite and well-defined. Let $ord_2(a)=k$. You have that $2^k|a$, but by definition $2^{k+1}\nmid a$. Therefore, you can write $a=2^k\cdot n$ from the first conclusion. 
