2
$\begingroup$

Long time ago for an assignment, I submitted an alternative proof of $\sqrt 2$ not being rational along the following lines:

suppose $\frac{a^2}{b^2}=2$ then we must also have $a=b+n$ for $a,b,n$ positive integers.

$(b+n)^2=2b^2$

$b^2+n^2+2bn=2b^2$

$n^2+2bn-b^2=0$ At this point I managed to somehow show that no integer $n$ can satisfy this ( not sure if I used quadratic equation or not), in place of 2, substituting $k$ in the first equation lead to $n$ not being an integer whenever $k$ was not a square integer. without having to change anything in the proof.

I can not reconstruct the missing steps, what possible continuation in reasoning can work for this case?

$\endgroup$
  • $\begingroup$ you are back to the start ! you haven't made any progress with this method. $\endgroup$ – DeepSea Oct 2 '15 at 3:55
  • 4
    $\begingroup$ The first line of your proof is wrong. If $\frac{a^2}{b^2} > 1$ then $a>b$ so $a + n$ is always strictly greater than $b$ for any positive $n$. This doesn't necessarily change anything about your proof, however, other than switching $a$ and $b$. $\endgroup$ – ChocolateAndCheese Oct 2 '15 at 3:58
  • $\begingroup$ @A1DHTH not sure what you mean, it worked 20 years ago when I did the assignment. $\endgroup$ – Arjang Oct 2 '15 at 3:58
  • $\begingroup$ @Arjang, if it worked 20 years ago, why are you asking then? $\endgroup$ – Hubble Oct 2 '15 at 4:00
  • 1
    $\begingroup$ Sorry, I gave you the wrong link. I meant this page. Maybe you find something here. $\endgroup$ – mickep Oct 2 '15 at 4:11

Your Answer

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

Browse other questions tagged or ask your own question.