2
$\begingroup$

enter image description here

Am I completely mistake or is it suppose to say $n^2$ is a multiple of 2 and therefore $n$ must be a multiple of 4?

This is from MIT's Mathematics for Computer Science

$\endgroup$
  • $\begingroup$ Generally: If $m,n$ are positive integers and $m^{1/n}$ is not an integer then $m^{1/n}$ is irrational. Proved by a similar method. $\endgroup$ – DanielWainfleet Apr 26 at 1:48
  • 1
    $\begingroup$ What makes you think this is a mistake? $\endgroup$ – Servaes Apr 26 at 7:31
3
$\begingroup$

No, it is correct. The point is that $2d^2=n^2$ implies $n^2$ is even, and only even numbers square to give an even number, so $n$ much be even, so $n^2$ is then actually a multiple of $4$.

$\endgroup$
  • $\begingroup$ This makes sense thanks $\endgroup$ – doctopus May 2 at 1:47
3
$\begingroup$

No. $6^2$ is a multiple of $2$ but $6$ is not a multiple of $4$. If $n=2k$ then for sure $n^2=4k^2$. So, MIT is right.

$\endgroup$

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.