Why is $\;n^2-\frac{n^2}{2} =\frac{n^2}{2}\;$? Could someone please expand on how to get from $\;\displaystyle\left( n^2-\frac{n^2}{2}\right)\;$ to $\;\left(\dfrac{n^2}{2}\right)\;?\;$ 
I can't seem to wrap my head around that. 

Actually, if you consider an integer division (so called Euclidian division) the proposition becomes: 
$$
n^2 - \frac{n^2}{2} = \frac{n^2}{2} + \text{mod}{(n^2,2)}
$$
Indeed, by definition of the Euclidian division of the integer $n^2$ by the integer $2$, one has: 
$$
n^2 = 2\times\frac{n^2}{2} + \text{mod}{(n^2,2)}
$$
where $\text{mod}{(a,b)}$ denotes the remainder of the euclidian division of an integer $a$ by an integer $b$.
Now, by (cleverly enough) remarking that the term $2\times \frac{n^2}{2}$ equals to $\frac{n^2}{2} + \frac{n^2}{2}$ one can write: 
$$
\begin{aligned}
n^2 &= 2\times\frac{n^2}{2} + \text{mod}{(n^2,2)} \\
n^2 &= \frac{n^2}{2} + \frac{n^2}{2} + \text{mod}{(n^2,2)} \\
\Leftrightarrow n^2 - \frac{n^2}{2}&= \frac{n^2}{2} + \text{mod}{(n^2,2)} \\
\end{aligned}
$$
that ends the proof ;)
