I'm not familiar with approximation, I have tried to get some of two irrational number which is equal to $\pi$ , I have looked to $\sqrt{2}+\sqrt{3}$ as simple example , but since i'm not familiar with approximation field , i'm not sure if $\sqrt{2}+\sqrt{3}$ could be equal to $\pi$ however $(\sqrt{2}+\sqrt{3})-pi \to 0$ , then my question here is :
Question: ls $\sqrt{2}+\sqrt{3}$ the only sum of two irrational which close to $\pi$?
There are many algebraic approximations of $\pi$, but they are just approximations, since $\pi$ is trascendental over $\mathbb{Q}$ as proved by Lindemann. For instance, by approximating a circumference with the union of a large number of arcs of parabola, then exploiting the nice formula (due to Archimedes) for the area of the parabolic segment, we have that
$$ \pi\approx 8\sqrt{2-\sqrt{3}}-1, $$ $$ \pi\approx 2\left(8\sqrt{2-\sqrt{2+\sqrt{3}}}-\sqrt{2-\sqrt{3}}\right) $$ and so on. If you are confident with Italian, you may have a look at it here (page 19), too.
Another famous approximation (besides the Archimedean $\pi\approx\frac{22}{7}$ and the Chinese $\pi\approx\frac{355}{113}$, giving two convergents of the continued fraction of $\pi$) is $\pi\approx\sqrt[3]{31}$, related to the fact that $$ \sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^3}=\frac{\pi^3}{32}$$ as shown here.