How is the "diagonal proof" valid? Years ago I was taught a proof that there are more irrational numbers than rational numbers, i.e. that it is a bigger infinity, that went like this:
Take the set of all rational numbers. Now construct a new number as follows: Take the first rational number, and choose a digit for the first digit of our constructed number that is different from the first digit of this number. Then make the second digit different from the second digit of the second number. Make the third digit different from the third digit of the third number. Etc. If you did this for an infinite number of digits, you would then have a number that is different from all the numbers in the original set. Therefore, there are numbers that are not in the set of rational numbers. Therefore, there are more irrational numbers than there are rational numbers.
I see the proof up to the last sentence. How does this prove that there are more irrational numbers? Just because I can find a number in set B that is not in set A doesn't prove that set B is bigger than set A.
I mean ... suppose I take the set of all even integers. There are an infinite number of such numbers. (I assume you agree with that and I don't need to go into a proof.) Now consider the set of all integers, odd and even. There are members of the set of all integers that are not in the set of integers. But there are not more integers than there are even integers. It is easily provable that there are the same number: I can establish a one-to-one correspondance between the two sets.
So how does the diagonal proof prove that there are bigger infinities? I presume I'm missing something. What?
 A: Short version: that's not actually the diagonal proof.

That argument is indeed incorrect, for precisely the reason you say - as written, all it shows is that there is an irrational number, and so $\mathbb{R}\supsetneq\mathbb{Q}$.
However, it does contain the germ of the right idea. Note that in the first part we didn't actually use anything about rationals: what we really showed was the following.

Suppose $(a_i)_{i\in\mathbb{N}}$ is any sequence of real numbers. Then there is a real number $b$ such that $b\not=a_i$ for any $i$.

This does ultimately imply that $\mathbb{R}$ cannot be put in bijection with $\mathbb{Q}$. We already know that there is a sequence $(q_i)_{i\in\mathbb{N}}$ such that every rational appears in the sequence. Now suppose $f:\mathbb{Q}\rightarrow\mathbb{R}$ were a bijection. Then letting $a_i=f(q_i)$ we would get a counterexample to the result above.
So basically the argument in the OP has the right idea, but doesn't implement it correctly.
A: It sounds like the version of the argument you gave here is slightly different from the standard version. The standard version says: suppose that we had an enumeration $a_1, a_2, a_3, \ldots$ not just of the rationals, but of every real number. Then the diagonal argument constructs a new real number that couldn't have appeared in your enumerated list, yielding a contradiction.
Note that this argument, as written, doesn't directly talk about the rational numbers at all -- the contrast between irrational numbers and rational numbers comes from observing that we can enumerate the rational numbers in this manner, so since we can't enumerate $\mathbb{R}$, the irrationals must be making up the gap.
