Suppose the quotient of two odd integers is an integer. Make and prove a conjecture about whether the quotient is either even or odd.
If you had an even (2n) divided by and odd (2m+1), it wont work.
so my odd integers would be a= 2n+1 and b=2m+1
a/b = c = (2n+1)/(2m+1) which is also odd so c = (2w + 1)
c|a, so a = [(2n+1)/(2m+1)] * k for some integer k
?????
or I have
Let a = 2n +1 and b = 2m + 1. From the definition a/b = c and c|a, then we get a/b = c . Thus a = b*c = (2m+1)(c) = 4mc + c = c(4m + 1). Then, we have an equation that is (c)*(odd) making the final result odd.
examples: 9/3 = 3
21/7 = 3
81/9 = 9
49/7 = 7
35/5 = 7
Best Answer
Let $a=2n+1$ and $b = 2m+1$ be odd integers. I see now you have a conjecture that $c=\frac{a}{b}$ is odd i.e $c=2k+1.$
Let's prove it:
Now assume the opposite, i.e assume $c$ is even, $c=2k.$ What happens in this case? Use the fact that if $\frac{x_1}{x_2}=\frac{y_1}{y_2}$ then $x_1y_2=x_2y_1$.
You will find that $a=cb=2kb=2(kb)=2l$ (for some integer $l=kb$), but this mean $a$ is even which contradicts our original hypothesis that $a$ is odd. Hence our assumption $c$ is even is wrong, this implies $c$ must be an odd integer. QED