1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 2
    $\begingroup$ $x$ 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
3
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.