One proof of the number's irrationality:
Assume that √2 is a rational number, meaning that there exists an integer a and an integer b in general such that a / b = √2.
Then √2 can be written as an irreducible fraction a / b such that a and b are coprime integers and (a / b)^2 = 2.
It follows that a^2 / b^2 = 2 and a^2 = 2b^2.
Therefore a^2 is even because it is equal to 2b^2. (2b^2 is necessarily even because it is 2 times another whole number and even numbers are multiples of 2.)
It follows that a must be even (as squares of odd integers are never even).
Because a is even, there exists an integer k that fulfills: a = 2k.
Substituting 2k from step 6 for a in the second equation of step 3: 2b^2 = (2k)^2 is equivalent to 2b^2 = 4k^2, which is equivalent to b^2 = 2k^2.
Because 2k^2 is divisible by two and therefore even, and because 2k^2 = b^2, it follows that b^2 is also even which means that b is even.
By steps 5 and 8 a and b are both even, which contradicts that a / b is irreducible as stated in step 2.
released January 1, 2013
Thank you to pat, who recorded several of these. I have no idea how to get in touch with you, nor do I even remember your last name. Also, thanks to Indiana at Secret Pennies Records for releasing the cassettes!