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Proving that \sqrt{p} is irrational

by Mandelbroth
Tags: irrational, proving, sqrtp
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Mandelbroth
#1
Nov27-13, 10:05 PM
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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:
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R136a1
#2
Nov27-13, 10:32 PM
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In my opinion, it's valid.
jbunniii
#3
Nov27-13, 11:14 PM
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It looks OK to me.

Curious3141
#4
Nov27-13, 11:55 PM
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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 ##m|p## (m divides p), which is a contradiction.
PeroK
#5
Nov28-13, 02:47 AM
P: 438
Quote Quote by Mandelbroth View Post
Edit: Gee, I guess it might be important to mention that ##p## is prime. :facepalm:
Actually, it doesn't matter that p is prime. It's valid for any +ve integer: the square root is either an integer or irrational; it cannot be a proper rational.
Mandelbroth
#6
Nov28-13, 09:33 AM
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Thank you, all.


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