Can an irrational number be expressed as a sum of other irrational numbers, at least one of which is not an integral multiple of the required number?

Question

For example, $\pi = Ae + B\sqrt 2+ \cdots$ ($A,B,\ldots\in\mathbb R$)

(Equations like "$\pi = 3\pi - 2\pi$" are not allowed.)

Answer 1

$$ \pi = \sqrt 2 + (\text{another irrational number}). $$

This "other irrational number" is of course $\pi-\sqrt 2$, and so we have the question of how we know that that is irrational. Just suppose it's rational, so then we have $$ \pi = \sqrt 2 + \frac m n $$ where $m,n$ are positive integers. Then it would follow that $$ \left( n\pi - m \right)^2 = 2n^2. $$ That would make $\pi$ an algebraic number. Just how to prove that $\pi$ is not an algebraic number takes some work, as does, for that matter, proving $\pi$ is irrational.

Answer 2

hi i have one example

$$\sqrt{3}=\frac{2}{\sqrt{3}}+\frac{1}{\sqrt{3}}$$