I was thinking about proofs of the fact that square roots exist and ended up wondering if the set of numbers of the form $\sqrt{\dfrac{a^2}{b^2}+ \dfrac{c^2}{d^2}}$ with $a, b, c, d \in\Bbb N$ is a dense subset of $\left[0, \infty\right)$.
I have no idea how to go about determining this.
For any $x\in [0,\infty)$ and any $\epsilon>0$, we have a rational $\frac ab$ within $\frac\epsilon2$ of $x$. Also we may let $c=1$ and pick $d$ large enough that $\frac ab$ and $\sqrt{\frac{a^2}{b^2}+ \frac{c^2}{d^2}}$ are within $\frac\epsilon2$ of each other. Then we have $\sqrt{\frac{a^2}{b^2}+ \frac{c^2}{d^2}}$ within $\epsilon$ of $x$.
To see that for for sufficiently large $d$, $\sqrt{\frac{a^2}{b^2}+ \frac{1}{d^2}}$ can be arbitrarily close to $\frac ab$, note that:$$\left( \sqrt{\frac{a^2}{b^2}+ \frac{1}{d^2}}-\frac ab\right) \left(\sqrt{\frac{a^2}{b^2}+ \frac{1}{d^2}}+\frac ab\right)=\frac 1{d^2} $$
As $$\left(\sqrt{\frac{a^2}{b^2}+ \frac{1}{d^2}}+\frac ab\right)>\frac{2a}b,$$ we have $$0<\left( \sqrt{\frac{a^2}{b^2}+ \frac{1}{d^2}}-\frac ab\right)<\frac b{2ad^2}$$