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Proving that \sqrt{p} is irrational 
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#1
Nov2713, 10:05 PM

P: 615

I'm aware of the standard proof.
What I'm wondering is why we can't just do the following. Given, I haven't slept well and I'm currently out of caffeine, so this one might be trivial for you guys. Suppose, by way of contradiction, that ##\sqrt{p}=\frac{m}{n}##, for ##m,n\in\mathbb{Z}## coprime. Then, ##p=\frac{m^2}{n^2}##. Because ##p## is an integer, ##\frac{m^2}{n^2}## must be as well. However, because ##m## and ##n## are coprime, so are ##m^2## and ##n^2##. Thus, ##n^2=1## is necessary for ##\frac{m^2}{n^2}## to be an integer. But that means ##m^2=p##, and ##p## is prime. Thus, a contradiction is met and we see that ##\sqrt{p}## is irrational. Is this valid? I'll probably figure out my mistake (if I made one) by the time I get back with adequate caffeination, but until then, I'd like to make sure I figure it out. Thank you. Edit: Gee, I guess it might be important to mention that ##p## is prime. :facepalm: 


#2
Nov2713, 10:32 PM

Newcomer
P: 341

In my opinion, it's valid.



#4
Nov2713, 11:55 PM

HW Helper
P: 2,940

Proving that \sqrt{p} is irrational
Completely valid. The last step basically assumes that a prime cannot be a perfect square, which is true and fairly obvious. But if you really want to make the proof completely obvious, you can state: ##m^2 = p##. Hence ##mp## (m divides p), which is a contradiction.



#5
Nov2813, 02:47 AM

P: 354




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