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## 'Be Reasonable' printed from http://nrich.maths.org/

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: $$3q^2 = 5p^2 + 2s^2 +
2ps\sqrt 10.$$ Rearranging this expression gives: $$\sqrt 10 =
{3q^2 - 5p^2 - 2s^2 \over 2ps}.$$ 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?