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A proof that all square roots of primes are irrational
By Tom Hardcastle (P2477) on Wednesday, July 12, 2000 - 02:49 am:

Could someone check that I haven't gone mad; looks good to me but it's the holidays...

If x is an integer then
x = p1p2p3p4... where p1 etc. are primes
if and only if x2 = p1p1p2p2p3p 3p4p4...

as from the fundamental theorem of arithmetic this is the only way in which x2 can be created from primes.

Therefore if x2 is divisible by a prime p
x is divisible by p (1)

Then the solution is just a generalisation of the proof of irrationality of sqrt(2).

a/b = sqrt(p) where p is prime, a & b have no common factor
=> a2/b2 = p
=> a2 = p.b2
from (1) a = pc where c is some constant integer
=> p2.c2 = p.b2
=> p.c2 = b2
from (1) b = pd where d is some constant integer
a/b = pc/pd => a & b have a common factor

reductio ad absurdium

The square root of any prime cannot be rational.
The square root of any prime is irrational.

Does anyone know who did this (or alternative proof) first?

By Dan Goodman (Dfmg2) on Wednesday, July 12, 2000 - 03:04 am:

Yup, that looks right to me. See if you can generalise this to get the proof that the square root of any integer which isn't a perfect square is irrational, you've already got the basic method of proof. Then, to really test yourself, see if you can prove the fundamental theorem of arithmetic (it's not too difficult, but certainly a challenge). The most important step in proving this theorem (if you want to have a go) is proving that p|ab implies p|a or p|b (p|a means p divides a).

By Dan Goodman (Dfmg2) on Thursday, July 13, 2000 - 11:27 pm:

You might also want to take a look at the message I posted for a similar discussion, called Proving the nth root of X is irrational; it has a much more general result.