I'm just genuinely curious as to what kind of practical application this subject has.

The first one that comes to mind is the proof that there are more real numbers than natural numbers. The proof goes by showing that there are more real numbers in the interval (0,1) than there are natural numbers (1, 2, 3, ...).

You do this by assuming that there are the same amount, and showing a contradiction ensues. First, note that any number in the interval (0,1) can be written as an infinite binary string of the form "0 . b1 b2 b3 ..." (it can be done as a decimal string as well, but the proof is easier with binary strings). Well, if every such number can be written in this fashion, and there are the same number of such numbers as there are natural numbers, then we can write out every such number with a natural number next to it (we can just label the real numbers "x1, x2, x3, ...") so that no natural number is ever repeated and every natural number is used, like so:

x1 = 0 . b11 b12 b13 ...

x2 = 0 . b21 b22 b23 ...

x3 = 0 . b31 b32 b33 ...

...

It is importat to understand that according to our assumptions,

*every* real number in (0,1) occurs exactly once in this list.

Now I will pick a real number y as follows: y = 0 . y1 y2 y3 ... where yk = 0 if bkk = 1 and 1 otherwise. In other words, you get y by reading off the diagonal digits (starting from the upper-left corner and going down and to the right) of the above list, and then inverting all the digits.

In other words, y = invert_binary_digits(0 . b11 b22 b33 b44 ...).

Clearly 0 < y < 1, so y is in (0,1) by definition. By our assumptions, every real number in (0,1) is written in that list, so y must be written in that list. In other words, y = xk for some natural number k.

So which is it? Is y = x1? No, because the first digit of x1 is different from the first digit of y, by construction. Is y = x2? No, because the second digit of x2 is different fromt he second digit of y, by construction. Similarly, y differs from xk in the kth digit, by construction, so there does not exist any k such that y = xk: a contradiction.

Therefore our original assumption that we could list all the real numbers in (0,1), one after another, is wrong, which means our assumption that there are just as many real numbers in (0,1) as there are natural numbers is wrong. Since I have already produced a list of real numbers, one for every natural number, there cannot be less real numbers in (0,1) than natural numbers, so there must be more, q.e.d.

(this proof is dedicated to BOGWarrior89)