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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

Suppose that a and b are natural numbers. If $\sqrt{a} + \sqrt{b}$ is rational then show that it is a natural number. Show that, indeed both $\sqrt{a}$ and $\sqrt{b}$ are then integers.

Suppose that a, b and c are natural numbers. If $\sqrt{a} + \sqrt{b} + \sqrt{c}$ is rational then show that it is a natural number. Moreover show that $\sqrt{a}$ , $\sqrt{b}$ and $\sqrt{c}$ are then integers.