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.
This reduces to proving that $1-\frac 12 = \frac 12$. In order to do this, expand $1$ into the geometric series $$1=\sum_{i=1}^{\infty}\frac 1{2^i},$$ and divide both sides by $2$ to get $$\frac 12=\sum_{i=1}^{\infty}\frac 1{2^{i+1}}=\sum_{i=2}^{\infty}\frac 1{2^i}.$$ It follows that $$1-\frac 12 = \sum_{i=1}^{\infty}\frac 1{2^i}-\frac 1{2^1}=\sum_{i=2}^{\infty}\frac 1{2^{i}}=\frac 12.$$
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: 
$$
\begin{aligned}
n^2 &amp;= 2\times\frac{n^2}{2} + \text{mod}{(n^2,2)} \\
n^2 &amp;= \frac{n^2}{2} + \frac{n^2}{2} + \text{mod}{(n^2,2)} \\
\Leftrightarrow n^2 - \frac{n^2}{2}&amp;= \frac{n^2}{2} + \text{mod}{(n^2,2)} \\
\end{aligned}
$$
that ends the proof ;)