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
Using NRICH Tasks Richly describes ways in which teachers and learners can work with NRICH tasks in the classroom.
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.
Then discuss the proof that $\sqrt 2$ is irrational.
The students can work with the interactivity Proof Sorter and perhaps some of them might read this article which was written by two undergraduates.
Key question
If the difference between$\sqrt 2$ and $\sqrt 3$ is an integer multiple of the common difference in an arithmetic series, and the difference between $\sqrt 3$ and $\sqrt 5$ is also an integer multiple of that common difference, can you use these two facts to write down two expressions, eliminate the unknown common difference and then find an impossible relationship?
Possible support
Proof Sorter and article
Possible extension
Can you prove that $\sqrt{1}$, $\sqrt{2}$ and $\sqrt{3}$ cannot be terms of ANY arithmetic progression?