HardMathematicsHigh School (9–10)GeneralNumber Systems › Rational and Irrational NumbersConceptualBoard Exam

Prove that the cube root of 2, the number whose cube is 2, is irrational…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Prove that the cube root of $2$ , the number whose cube is $2$ , is irrational. You may use this fact: if a prime divides $a^3$ , then it divides $a$ . It holds for the same reason as for $a^2$ .

Need a reminder? The key ideaShow

To prove that $\sqrt{p}$ is irrational, for a prime $p$ : suppose $\sqrt{p} = \dfrac{a}{b}$ in lowest terms. Then $a^2 = pb^2$ , so $p$ divides $a$ . Write $a = pc$ . Then $b^2 = pc^2$ , so $p$ divides $b$ . A common factor $p$ contradicts lowest terms.

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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