let n in Z with n>0. prove that there exist k,m in Z with m odd such that n=2^km.
$\displaystyle n = 2^{k}m$
(1) If n is odd, then let k = 0 and m = n ($\displaystyle 2^{0} \times n = n$)
(2) If n is even, then factor out 2's one-by-one until you reach an odd number, in which case you'll set it to m. To illustrate: $\displaystyle n = 2n_{1} = 2^{2}n_{2} = 2^{3}n_{3} = ... \quad n_{1}, n_{2}, ... \in \mathbb{Z}^{+}$ Eventually, some $\displaystyle m_{i}$ will be odd and this will be your m - giving you $\displaystyle n = 2^{i}m_{i}$