# 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?

It helps to have seen a proof that $\sqrt 2$ is irrational and
to appreciate how the logic of arguments by contradiction
work.

See
Proof Sorter and, for some further reading on proofs by
contradiction, see
this article written by two undergraduates.

Then you only need to know the definition of an arithmetic
series to do this problem. 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?

## 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.

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?