# Alternative prrof of $\sqrt 2$ (and $\sqrt {\text {of any non square integer}}$ ) not being rational

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?

• you are back to the start ! you haven't made any progress with this method. – DeepSea Oct 2 '15 at 3:55
• 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 $b$. – ChocolateAndCheese Oct 2 '15 at 3:58
• @A1DHTH not sure what you mean, it worked 20 years ago when I did the assignment. – Arjang Oct 2 '15 at 3:58
• @Arjang, if it worked 20 years ago, why are you asking then? – Hubble Oct 2 '15 at 4:00
• Sorry, I gave you the wrong link. I meant this page. Maybe you find something here. – mickep Oct 2 '15 at 4:11