Proving a set is unbounded 
Prove that the set $S = \{x\in \Bbb R: x^2-25x > 0\}$ is unbounded. 

I started off by assuming by way of contradiction that is has an upper bound, specifically $\beta$ = sup S. I'm getting stuck on where to go from here. 
 A: Note that:
$$x^2-25x=x(x-25)$$
Hence $S\supset (25,\infty)$, which is unbounded. 
A: Don't assume  it is bounded, and then try and show a contradiction. Just show that there are large enough (in absolute value) real numbers satisfying this equality.
For this, simply note that $x^2-25x = x(x-25)$. Hence, this is greater than zero  whenever both are greater than zero, which happens whenever $x>25$. Thus, for any $x > 25$, it follows that $x(x-25)>0$, so $x \in S$, so $S$ is unbounded.
Since you are proceeding by contradiction, we should probably discuss that way as well. So suppose that $\beta = \sup S$. Note that $26 \in S$ (which you verify on your own), so $\beta \geq 26$. Now, consider $\beta+1$. Note that
\begin{align}
(\beta+1)^2-25(\beta+1) &= \beta^2+2\beta+1-25\beta-25 \\
 &\geq \beta^2 -25\beta + 2(26)+1-25 = \beta^2-25\beta + 28\\
 &> \beta^2-25\beta\\
 &> 0
\end{align}
Hence, $\beta+1 \in S$, contradicting that $\beta = \sup S$. Hence, $S$ is unbounded.
A: If $\beta = \sup S$ then $x^2 - 25 x  \le \beta$
So $x^2 - 25x + (25/2)^2 \le \beta + (25/2)^2$
$0 \le (x - 25/2)^2 \le \beta + (25/2)^2$
$x - 25/2 \le \sqrt{\beta + (25/2)^2}$
$x \le \sqrt{\beta + (25/2)^2} + 25/2$.
So just take $x > \sqrt{\beta + (25/2)^2} + 25/2$ and we will have $x^2 - 25 > \beta$.   A contradiction that is unavoidable.
But this isnt a very good way to do it.
