Be Reasonable
Prove that sqrt2, sqrt3 and sqrt5 cannot be terms of ANY arithmetic progression.
Problem
Prove that there is no arithmetic progression containing all three of $\sqrt{2}$, $\sqrt{3}$ and $\sqrt{5}$.
Getting Started
Try a proof by contradiction. Suppose that the three irrational numbers do occur in some arithmetic series. Can you then go on to reach a contradiction?
Student Solutions
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?
Teachers' Resources
Why do this problem?
It is an exercise in proof by contradiction.
Possible approach
First discuss proof by contradiction so that students appreciate how the logic of arguments by contradiction work. You can draw on this article on Proof by Contradiction.