I'm not claiming any original material here, just things that I've gathered over the years that I find pretty interesting.

This is a rather famous proof, so I'm not going to elaborate too much. However, it's a proof by contradiction, so we're going to start by assuming the converse is true and trying to reach a logical contradiction. Let's get started.

Suppose √(2) is rational. Let's say √(2) = *a* /
*b* where *a* and *b* are "natural numbers"– that is,
positive, non-zero integers (1, 2, 3, and so on). Let's also suppose that
*a* and *b* are relatively prime– that is, the fraction is
in its lowest possible terms. *a* shares no divisors with *b*.

With these suppositions made, here's our proof:

- Start with: ∃
*a*,*b*∈**N**where*a*and*b*are coprime and*a*/*b*= √(2). - Get rid of the radical on the right by squaring both sides:
2 = (
*a*/*b*)^{2}= a^{2}/ b^{2} - Multiply both sides by
*b*^{2}: 2*b*^{2}=*a*^{2}. So,*a*^{2}is even. This implies*a*is even. - Since
*a*is even, it can be written as 2*m*, giving us: 2*b*^{2}= ( 2*m*)^{2} - Multiply out the squared term on the right-hand side:
2
*b*^{2}= 4*m*^{2} - Divide both sides by two:
*b*^{2}= 2*m*^{2}. Here's our contradiction:*b*^{2}is even, so*b*is even. - If
*a*and*b*are both even, they could be divided by two, contradicting (1). Q.E.D.

I want to end by saying that this proof originally disappointed me. You could seemingly replace 2 by any other number and have the proof still work. This is kinda touchy, but the simplicity of the odd and even characteristics really make it work well for √(2). After further consideration, it grew on me.

I originally found this next one by searching for further truths about irrational numbers after my √(2) conundrum. I found this proof on Everything2, but the English was pretty bad and it didn't properly explain things.

This proof goes a little beyond my title. It actually proves all numbers
__that are not perfect squares__ are irrational.
This is another proof by contradiction. We start by assuming
*p* is prime and that there exists some *a* and *b* so that
√(*p*) = *a* / *b*. Let's do some math.

- Multiply both sides by
*b*:*b*√(*p*) =*a*. - Let
*c*= ⌊√*p*⌋ ... If the font doesn't display this properly, that's*c*=**floor**[ √(*p*) ], where the floor function rounds down its argument. - Let's set up a new equation and examine
*d*=*b*( √(*p*) -*c*).*d*is an integer since we're multiplying*b*times √(*p*) minus its decimal part*c*. - √(
*p*) -*c*gives you a small integer close to the actual square root.*d*=*b*( √(*p*) -*c*) =*b*√(*p*) -*b**c*(by distribution). - Let's multiply both sides by √(
*p*):*d*√(*p*) =*b**p*- √(*p*)*b**c* - Let's rewrite
*d*once more to factor out the*b*:*d*√(*p*) =*b*(*p*-*c*√(*p*) ) - Since
*c*< 1 and √(*p*) <*p*, then*p*-*c*√(*p*) > 1 and*d*<*b*. - Here's our contradiction:
According to (1),
*b*was supposed to be the smallest number that when multiplied by √(*p*) gave us an integer, but we've proven ∃*d*∈**Z**,*d*<*b*:*d*√(*p*) ∈**Z**... that is to say, there exists an integer*d*that, when multiplied by our square root of a prime number (or non-perfect-square number), yields an integer.

This is a really beautiful proof that makes more sense the more you look
and play with it. It's difficult to describe in plain English and mathematical
notation becomes almost a necessity. (8) is kinda chunky, but it makes lots
of sense. Things like the proof of √(2) might lead you to believe that
these proofs will make things like √(25) drift off into irrationality,
but it doesn't. In fact, if you plug √(25) or any other perfect-square
into the above proof, *c* is equal to √(25) and *d* turns
into zero. Then, of course *d* is less than *b* and *d*
times √(25) is also 0. So, everything falls apart and rationality
is restored.

I found this one in a 1960s textbook on number theory. It's breaking the trend of "prove-this-is-irrational" from the previous two. It's a proof by contradiction that assumes there is a largest prime number.

- Let
*p*be the largest prime number. - Let
*q*= (*p*×*p*- 1 ×*p*- 2 × . . . × 3 × 2 × 1 ) + 1 - Let
*r*= the largest prime number that divides*q*. *r*>*p*. This is in contradiction with (1).

So, *q* - 1 is a hugely composite number. Every number including
*p* divides it. *q*, however, can't be prime since (1) says that
*p* is the largest prime, and every number less than or equal to
*p* divides *q* with a remainder of one. Therefore, there can't
be a largest prime number.

From here, you can either go back to writings, or go back home.