
Direct Proof
Hi people,
I think its true that for every positive natural number, lets call it n, can be put in form of 2^k * m (where m is an odd number).
I think thats correct? Please confirm...
How would you use direct proof to proof this?? I cant get grips of it. Its abit advanced for me (Doh)
Thanks.

If we consider $\displaystyle 1$ an odd number your are right...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$

if n is odd choose k = 0 and m = n.
if n is even then write $\displaystyle n$ in its prime decomposition.
$\displaystyle n=2^rp_1p_2p_3\dots$
Choose k=r and choose $\displaystyle m$ to be the product of primes, $\displaystyle p_i$. The product of primes is guaranteed to be odd because an odd times an odd is an odd and no primes other than 2 are even. If there are no $\displaystyle p_i$s (ie. n is a power of 2), then take m=1.