If $x^{ 1/3 } \in\mathbb{R}$ is an irrational number, then $x$ is also irrational.

I want to use the contrapositive to disprove this. The contrapositive would be:

If $x$ is rational, then $x^{ 1/3 }$ is rational.

I am thinking that if I use $10$ as a counter example, then it will disprove the claim. I am not sure that this is the right approach. I am also unsure of whether or not I'll have to prove that the cubed root of $10$ is irrational to make the proof complete.

  • $\begingroup$ If you can show that $\alpha=10^{\frac 13}$ is irrational then you will indeed have a counterexample. But you do need to show this. After all, both $8$ and $8^{\frac 13}$ are rational. $\endgroup$ – lulu Nov 2 '15 at 13:17
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    $\begingroup$ $x$ might not be the cube of a rational number. So 10 is a good counter-example. But this is not a contrapositive method. Rather it is counter example method used in the reverse way. What you have doe here could have been done straightaway taking $2^{\frac{1}{3}}$ as $x^{\frac{1}{3}}$. $\endgroup$ – SchrodingersCat Nov 2 '15 at 13:18
  • $\begingroup$ @Aniket don't I have to prove that $2^{ 1/3 }$ is irrational to make the proof complete? $\endgroup$ – nikolita Nov 2 '15 at 13:36

Yes, your approach is correct. An implication is true if and only the contrapositive is true (since the contra positive of the contra positive is the implication itself - the implication being true will imply the contrapositive to be true).

You just have to select a suitable integer and proove that $x^{1/3}$ is irrational (so while your approach is correct, you haven't completed the task). This is done in similar way that you proove that $\sqrt2$ is irrational. A suitable candidate is $x=2$ ($x=10$ would do to, but I think that complicates it a bit).


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