Be reasonable

Thank you to M. Grender-Jones for this solution.

To show that $\sqrt{2}$, $\sqrt{3}$ and $\sqrt{5}$ cannot form part of any arithmetic progression we give a proof by contradiction. Suppose they can, then $$\sqrt{3}-\sqrt{2}=px$$ $$\sqrt{5}-\sqrt{2}=qx$$ where $x$ is the common difference of the progression, and $p$ and $q$ are integers.

Eliminating $x$ from these two equations we get $$p(\sqrt{5}-\sqrt{2})=q(\sqrt{3}-\sqrt{2})$$ so $$q\sqrt{3}=p\sqrt{5}+(q-p)\sqrt{2}.$$ We know $p$ and $q$ are integers so to simplify this expression write $p-q = s$ where $s$ is an integer: $$q\sqrt{3}=p\sqrt{5}+s\sqrt{2}.$$ Squaring this: $$3q^2=5p^2+2s^2+2ps\sqrt{1}0.$$ Rearranging this expression gives: $$\sqrt{1}0=\frac{3q^2-5p^2-2s^2}{2ps}.$$ As $\sqrt 10$ is irrational and all the other terms in this expression are integers this is impossible and we have reached a contradiction. Therefore our assumption was false and $\sqrt 2$, $\sqrt 3$ and $\sqrt 5$ cannot be terms of an AP.

By the same method can you prove that $\sqrt{1}$, $\sqrt{2}$ and $\sqrt{3}$ cannot be terms of ANY arithmetic progression?