### Be Reasonable

Prove that sqrt2, sqrt3 and sqrt5 cannot be terms of ANY arithmetic progression.

### Good Approximations

Solve quadratic equations and use continued fractions to find rational approximations to irrational numbers.

### Rational Roots

Given that a, b and c are natural numbers show that if sqrt a+sqrt b is rational then it is a natural number. Extend this to 3 variables.

# The Root Cause

##### Stage: 5 Challenge Level:

Good solutions were submitted by Hyeyoun from St Paul's Girls' School, London, Michal from Daramalan College, and Ling Xiang Ning from Raffles Institution, Singapore.

This is Hyeyoun's solution.

Suppose that $\sqrt{a}$ is rational. Therefore, where $x$ and $y$ are coprime integers and $y \ne 0$, we have $a = \frac{x^2}{y^2}$

We can always write the rational number so that x and y are coprime, that is they have no common factors except 1 so, as $\sqrt a$ is rational, it follows that $x$ and $y$ are coprime.

If $a$ is an integer, $y^2$ must be a factor of $x^2$ and so $y=1$.Therefore $a$ is a square number.

It follows (as the contrapositive) that if $a$ is an integer and is not a square number then $\sqrt{a}$ is irrational.