# Thread: Is there anything wrong with this proof?

1. ## Is there anything wrong with this proof?

The claim is that the square root of a prime number is irrational. This is a proof with flavors all over the Internet, but mine seems a teensy bit simpler than most, so I assume something is wrong with it

Assume sqrt (p) is in Q. so sqrt (p) = m/n for some relatively prime m,n in Z. squaring both sides, p= m^2 / n^2 or n*n*p = m*m. m divides n*n*p, but m and n have no common factors, so n must be 1 and m divides p. But p was prime, so contradiction and sqrt p is irrational.

2. Originally Posted by Eric Reed
The claim is that the square root of a prime number is irrational. This is a proof with flavors all over the Internet, but mine seems a teensy bit simpler than most, so I assume something is wrong with it

Assume sqrt (p) is in Q. so sqrt (p) = m/n for some relatively prime m,n in Z. squaring both sides, p= m^2 / n^2 or n*n*p = m*m. m divides n*n*p, but m and n have no common factors, so n must be 1 and m divides p. ... But p was prime, so contradiction and sqrt p is irrational.
Add in the red part that m is also different from 1, otherwise, the implication "m divides p & p prime = contradiction" is false, because 1 can divide a prime number.

Plus, what if $m=p$ ? It divides p but p is still prime !
If $m=p$, then we have : $n^2p=m^2 \implies n^2p=p^2$
$p=n^2$, which is a contradiction... I'll let your prove it because I've managed to confuse myself

3. Oooh, right...I guess if m=p then n^2 m = m^2 so n^2 = m and m divides n, contradiction.

Thanks!! I guess if I'm gonna hang around here I should learn how to make my formulas look pretty, eh?