### Be Reasonable

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

### The Root Cause

Prove that if a is a natural number and the square root of a is rational, then it is a square number (an integer n^2 for some integer n.)

### Good Approximations

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

# Rational Roots

##### Stage: 5 Challenge Level:

David sent in this solution, using the hints we gave you.

$\sqrt{a}+\sqrt{b}$ rational
$\Rightarrow (\sqrt{a}+\sqrt{b})^2=a+b+2\sqrt{a b}$ rational
$\Rightarrow 2\sqrt{a b}$ rational
$\Rightarrow \sqrt{a b}$ rational
$\Rightarrow a+\sqrt{a b}$ rational
i.e., $\sqrt{a}(\sqrt{a}+\sqrt{b})$ rational
$\Rightarrow \sqrt{a}$ rational (and so $a$ is a square)
$\Rightarrow \sqrt{b}$ is also rational and hence $b$ is a square.

For the second part:

$\sqrt{a}+\sqrt{b}+\sqrt{c}$ rational
$\Rightarrow (\sqrt{a}+\sqrt{b}+\sqrt{c})^2$ rational
$\Rightarrow \sqrt{a b}+\sqrt{b c}+\sqrt{c a}$ rational
$(\sqrt{a b}+\sqrt{b c}+\sqrt{c a})^2=a b+b c+c a+2\sqrt{a b c}(\sqrt{a}+\sqrt{b}+ \sqrt{c})$
so $\sqrt{a b c}$ is also rational
$\Rightarrow \sqrt{a}(\sqrt{a b}+\sqrt{b c}+\sqrt{c a})-\sqrt{a b c}=a(\sqrt{b} +\sqrt{c})$ is rational
so $\sqrt{b}+\sqrt{c}$ is rational and $\sqrt{a}$ is rational, and use the above.