Show that all real number can be approximated by the difference between square root of two different integers. Let $x \in \mathbb R$, show that for any $\epsilon > 0$, there exists $a,b \in \mathbb Z$ such that:
$$
\left|x - \left(\sqrt{a} - \sqrt{b}\right)\right| < \epsilon
$$
Not sure how to approach this, how do you show the difference between square roots of integers can be arbitrary.
any hints would be helpful!

There are several ways to approach this. Here's one of them:
Hint: For any integers $n, k$, show that there exists integers $a, b$ such that
$$ \frac{k}{2^n } < \sqrt{a} - \sqrt{b} < \frac{k+1}{2^n}. $$
Corollary: Given any $x$, we can first approximate it by $ \frac{k}{2^n}$, then by $ \sqrt{a} - \sqrt{b}$.
Fill in the gaps as needed. If you're stuck, write out your working and thought process to demonstrate where you're at.

Here's another approach.
Hint: For $ x \geq 0$, let $ b_n = n^2$ and $ a_n = \lceil  (n+x)^2 \rceil $. Show that
$$ |x - ( \sqrt{a_n} - \sqrt{b_n} )| < \frac{1}{ n}. $$
(We could get a tighter bound, but this is all that we need.)
