Prove that p\sqrt p is irrational for all prime pp

Woah! What the hell happened to me, I’m all serif and academic ! … huh.

Well anyway, today I thought it’d be fun to share this little proof, plus try to pull off the TeX look just for the hell of it! :p

I’m going to slow this off using proof by contradiction, by the way.


Assume that p\sqrt p is rational - that is p=ab\sqrt p = \frac a b, where aa and bb are coprime integers.

Author’s note: coprime means that they do not share any common factors.

It is a sensible part of the definition of rational, as it enforces that rational numbers should be given in their simplest form, which should be simple-ish to do for all true rational numbers.

No rational number must necessarily have coprime numerator and denominator.

p=a2b2p=\frac{a^2}{b^2}pb2=a2p b^2=a^2

If aa divides by pp:

aa is of the form kp,kZkp, k\in\mathbb Z

a2=(kp)2=k2p2a^2=(kp)^2=k^2 p^2pb2=k2p2pb^2=k^2 p^2b2=k2pb^2=k^2 p

For this to be true, b2b^2 must divide by pp, b\therefore b must divide by pp.

This would mean that aa, bb share factor pp, so are not coprime! ↯

If aa does not divide by pp:

aa is of the form kp+n,kZ,nkp+n, k\in\mathbb Z, n does not divide by pp.

pb2=(kp+n)2=k2p2+2nkp+n2pb^2 = (kp+n)^2 = k^2 p^2 + 2nkp + n^2b2=k2p+2nk+n2pb^2 = k^2 p + 2nk + \frac{n^2} p

nn does not divide by pn2p \therefore n^2 does not divide by pp.

n2p\therefore \frac{n^2} p is not an integer, b2\therefore b^2 is not an integer, b\therefore b is not an integer! ↯


aa cannot divide by pp, but also cannot not divide by pp.

This is impossible, so p\sqrt p cannot be rational!

p\therefore \sqrt p is irrational for all prime pp \square

So, uh, back to normal!

That was a proof that I managed to find earlier after a lot of playing, and I was very pleasantly surprised with how the second half pretty much just fell into place once I had the first half sorted.

This was very fun to think about and devise, hopefully its not too dense / different to what people were maybe expecting me to post about!

Seems I’m drifting from tech and stuff towards math…

Anywho, hope to cya back here soon!
— Yellowsink