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Be Reasonable

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

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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.)

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Good Approximations

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

Rational Roots

Stage: 5 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

In this problem you are given that $a$, $b$ and $c$ are natural numbers. You have to show that if $\sqrt{a}+\sqrt{b}$ is rational then it is a natural number.

You could use the fact that if $\sqrt{a}+\sqrt{b}$ is rational then so is its square which means that $\sqrt ab $ is also rational. Knowing this the next step is to use $$\sqrt{a}(\sqrt{a}+\sqrt{b}) = a+\sqrt{ab}$$ to show that $\sqrt a$ is rational and to do likewise for $b$.

This is all you need because it has been proved that if $\sqrt a$ is rational then $a$ must be a square number.

See the problem The Root Cause .

Try to apply this method and then to extend it to three variables for the last part.