Show that every open ball in $R^2$ contains a point $x = (x_1, x_2)$ both of whose coordinates are rational Show that  every open ball in $R^2$ contains a point $x = (x_1, x_2)$ both of whose coordinates are rational.
What I have so far:
Let $B(y,r)$ be an arbitrary open ball centered at $y$ with radius $r$. My intuition tells me to proceed by some how using the denseness of the rational numbers, but i'm having trouble getting started with this.
 A: Your intuition is right.  And here's how to go from intuition to formal.  
Let $r > 0$.  Then there is rational $q$ so that $x_1 < q < x_1+r$. And a rational $p$ so that  $x_2 < p < x_2 + r$ and $d((x_1,x_2),(q,p)) = \sqrt{(q-x_1)^2 + (p-x_2)^2} < \sqrt{r^2 + r^2} = \sqrt{2}r$ and .... oh, buggernuts!...It's too big....
So.... make it smaller ... 
Okay, that's just fine...  if $r > r/500million > 0$ then there is a rational $q$ so that $x_1 < q < x_1 + r/500million$ and.....  $d((x_1,x_2),(q,p)) = \sqrt{(q-x_1)^2 + (p-x_2)^2} $$< \sqrt{(r/500million)^2 + (r/500million)^2} $$= \sqrt{2r^2/500million^2} $$= r*\frac {\sqrt{2}}{500million} < r$.
Okay, that was overkill.  .... but why NOT do overkill?  As long as we can get it small enough, it doesn't matter if we go way overboard.
Anywhooo.... if $r > r/\sqrt{2} > 0$ is the better choice.  Then "Then there is rational $q$ so that $x_1 < q < x_1+r/\sqrt{2}$. And a rational $p$ so that  $x_2 < p < x_2 + r/\sqrt{2}$ and $d((x_1,x_2),(q,p)) = \sqrt{(q-x_1)^2 + (p-x_2)^2} < \sqrt{(r/\sqrt{2})^2 + (r/\sqrt{2})^2} = r$" works perfectly.
Really.... all this stuff about getting the numbers to add up just right is just ...distracting, when the important thing is the analysis.  If $\mathbb Q$ is dense in $R$ we can find rationals arbitrarily close and we extend that to $\mathbb Q^m$ and $\mathbb R^m$.  The worrying about the arithmetic of square roots is .... not the relevant issue.
A: Given that this thread is several years old by now, this will most likely won't be of much use to you, but for those who are stuck:
It's indeed possible and the following is one way of proving it.
Let $d_E:\mathbb{R}^n\times\mathbb{R}^n\rightarrow \mathbb{R}$ given by $(x,y)\mapsto \sqrt{\sum_{i=1}^n (x_i-y_i)^2}$ be the Euclidean distance on $\mathbb{R}^n$.
For every $x,y\in\mathbb{R}^n$ we have $d(x,y) = \sqrt{\sum_{i=1}^n (x_i-y_i)^2} \leq \sum_{i=1}^n|x_i-y_i|$, where $|\cdot|$ is the usual norm on $\mathbb{R}$.
The trick is to choose $p=(p_1,\dots,p_n)\in \mathbb{Q^n}$ such that the $|x_i-p_i|<r\,/\,n$ for every $i \in \{1,\dots,n\}$.
If you're wondering where the inequality above comes from.
This follows from the fact that:
(1) The Euclidean and manhattan norm on $\mathbb{R}^n$, $\|\,\|_E$ and $\|\,\|_M$ respectively, are equivalent. Meaning that there exist positive real numbers $C,D$ such that for all $x\in\mathbb{R}^n$ $C\|x\|_E\leq\|x\|_M\leq D\|x\|_E$.
(2) The Euclidean distance can be defined as follows:
\begin{equation}
d_E:\mathbb{R}^n\times\mathbb{R}^n  \rightarrow \mathbb{R},\,\,
(x,y)\mapsto \|x-y\|_E.
\end{equation}
Further reading: https://en.wikipedia.org/wiki/Norm_(mathematics)
A: Let $x=(x_1,x_2)$. An open ball of $R^2$ $B(x,r)=\{(y_1,y_2):|y_1-x_1|^2+|y_2-x_2|^2<r^2\}$ contains $I_1\times I_2, I_i=(x_i-r/4,x_i-r/4), i=1,2$. Since the rational are dense in $R$, there exists $y_i\in I_i\cap Q, and (y_1,y_2)\in B(x,r)$.
