Proving that $\mathbb{Q}$ is neither open or closed in $\mathbb{R}$ I need to figure out if $\mathbb{Q}$ is either open or closed in $\mathbb{R}$ and prove my assertions. I don't think it's either. I would like a review/critique of my proofs to make sure that there's not an easier way to do it.

To see that it's not open, take some $\frac{m}{n}$ with both $m$ and $n$ perfect squares so that $\left(\frac{m}{n}\right)^{\frac{1}{2}}$ is rational. We have that $x \mapsto x^{\frac{1}{2}}$ is continuous. Hence for any $\epsilon$-neighborhood of $\left(\frac{m}{n}\right)^{\frac{1}{2}}$, there exists a $\delta$-neighborhood of $\frac{m}{n}$ that maps into it. but

$$
\begin{eqnarray*}
\frac{m}{n} - \frac{cm}{cn + 1} &lt; \delta &amp;\Leftrightarrow&amp; \frac{m}{n(cn + 1)} &lt; \delta n(cn + 1) \\
&amp;\Leftrightarrow&amp; \frac{m}{\delta n^{2}} &lt; \left(nc^{2} + 1\right)^{2} \\
&amp;\Leftrightarrow&amp; \frac{m}{\delta n^{4}} &lt; c^2 + \frac{2}{n}c + \frac{1}{n^{2}} \\
&amp;\Leftrightarrow&amp; \frac{m}{\delta n^{4}} &lt; \left(c + \frac{1}{n}\right)^{2} \\
&amp;\Leftrightarrow&amp; \left(\frac{m}{\delta n^{4}}\right)^{\frac{1}{2}} - \frac{1}{n} &lt; c
\end{eqnarray*}
$$

So we can make $\frac{cm}{cm + 1}$ as close to $\frac{m}{n}$ as we like by taking $c$ sufficiently large. If $cm$ is a perfect square, then $c$ must be a perfect square because $m$ is so we can just choose a $c$ that's not a perfect square and hence $\left(\frac{cm}{cn +1}\right)^{\frac{1}{2}}$ is irrational. Because we can place an irrational arbitrarily close to at least one positive rational, $\mathbb{Q}$ is not open.

To show that $\mathbb{Q}$ is not closed, simply pick some irrational $\alpha$ and let $\epsilon_n = \frac{1}{n}$. Let $x_n$ be chosen from between $\alpha - \epsilon_n$ and $\alpha$. Then the sequence $x_n$ converges to $\alpha$ which shows that the complement of $\mathbb{q}$ in $\mathbb{R}$ is not open as we can place a rational in any $\epsilon$-neighborhood of any irrational.

Proving that the rationals aren't open was so much harder than proving that they're not closed because my book hadn't proved that the irrationals are dense in $\mathbb{R}$ and so I couldn't use that. Is there some other way that I could do this?
You are certainly right that all you need to show that $\mathbb{Q}$ is not open is to show that given a rational $q$, you can find irrationals arbitrarily closed to $q$. A very simple way of doing this is to take your favorite irrational, say $\sqrt{2}$, and then consider $\sqrt{2}/n$ with $n$ an integer. These numbers are never rational, and get arbitrarily small; so if $q$ is a given rational, then $q+\frac{\sqrt{2}}{n}$ will be as close as you want to $q$ by taking $n$ large enough.
To show Q is not open you can use the fact that Q is not open if and only if the complement of Q is not closed.....take a sequence Un in Q complement (R\Q)such that Un = (n+√2)/(n+1) .......
.lim Un = p....(p∈R) then p ∈Q so p∉Q complemet ....then Q complement is not closed therefore Q is not open