Divisibility Question. Show that if $a$ and $b$ are odd positive integers, and $b$ does not divide $a$, then there are integers $s$ and $t$ such that $ a = bs + t$ where $t$ is odd and $|t| <b$.
Let a = 2k +1 and b = 2j + 1 where k,j are positive integers.
since b does not divide a then there is a non-zero remainder on division by b.
we can write;
$a = bs + t$
now to show that $t$ is odd;
$(2k +1) = (2j + 1)s + t$
rearranging;
$2(k-2js) - ( s-1) = t$
$2(k-2js) + (1-s) = t$
$2[(k-2js) -s/2] + 1  = t$
Hence t is odd. but i have not shown that $|t| <b$.
In fact, i am questioning whether i have shown anything. 

The most straightforward method is by using the euclidian algorithm:
So by the euclidian algorithm, we find $q$ and $r$ such that
$$a = bq+r$$
and $0<|r|<|b| = b$. We consider two cases:


*

*$r$ is odd. We are done.

*$r$ is even, but in this case, $r-b$ is odd, so $$a = b(q+1)+(r-b)$$ and $-b < r-b < 0$ so $|r-b| < b$.


I think that almost every proof will require a (small) case-study.
