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Getting caught up on this problem. what i got so far.

Contrapositive: For $x \in \mathbb{R}$, if $\sqrt[3]{x}$ is rational, then $x$ is rational.

If $\sqrt[3]{x}$ is rational then there must exist an $a,b \in Z, b \neq 0$ such that

$$\sqrt[3]{x}=\frac{a}{b}$$ $$x=\frac{a^3}{b^3}$$ $$b^3\cdot x=a^3$$

As $b^3$ is a multiple of $a^3$, $x$ by defintion can not be rational as they share a common factor.

I think im taking the wrong approach to this problem and should try maybe going by contradiction? or both. Or is the original statement false?

Darío A. Gutiérrez
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Ringomain
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1 Answers1

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you were already finished at $x=\frac{a^3}{b^3}$, any number of this form is immediately rational.

Asinomás
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  • is this all i need to finish the problem or do i need to say more about x.......As ( a,b \in \Z ), the cubed root of (x) must be rational as by definition of a rational number and it follows that if the cubed root of (x) is rational then (x) must be rational. – Ringomain May 03 '18 at 21:21
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    well, you wanted to prove that if $\sqrt[3]{x}$ is rational then so is $x$, and that is exactly what you did. – Asinomás May 03 '18 at 21:23
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    If $a$ is an integer then $a^3$ certainly is... .likewise with $b$ and $b^3$. – Mason May 03 '18 at 21:25