The division algorithm states

$\displaystyle b=aq+r$where $\displaystyle 0<r<a$ and $\displaystyle b,a,q,r\in{Z}$

Prove $\displaystyle r<\frac{b}{2}$

I shall attempt a proof by contradiction

Assuming $\displaystyle r>\frac{b}{2}$

Let $\displaystyle r=\frac{mb}{n}$ where $\displaystyle m<n<2m$

So

$\displaystyle b=aq+\frac{mb}{n}$

$\displaystyle b-\frac{mb}{n}=b(1-\frac{mb}{n})=b(\frac{n-m}{n})=aq$

$\displaystyle b=\frac{aqn}{n-m}$

Placing it into the original equation

$\displaystyle \frac{aqn}{n-m}=aq+r$

$\displaystyle aqn=aqn-aqm+(n-m)r$

$\displaystyle aq\frac{m}{n-m}=r$

And as $\displaystyle \frac{m}{n}>\frac{1}{2}$ and $\displaystyle m<n$, then $\displaystyle m>n-m$

So $\displaystyle r>aq$, but this is a contradiction to the division algorithm

So $\displaystyle r<\frac{b}{2}$

Is this alright?

N.B.

The case where $\displaystyle r=\frac{b}{2}$ works similarly. You end up with $\displaystyle r=aq$, which is also a contradiction