query regarding the uncountability of irrationals

I was discussing infinity with one of my students, and introduced him to the idea of countability as a way of thinking about the size of sets.



We start with set of natural numbers, {1, 2, 3, ...}.

We can then ask the basic question: "Are there more natural numbers than even numbers?" In our normal way of looking at finite sets, it is obvious that there are more natural numbers, since all of the even numbers are contained in the natural numbers.



However, the mathematical notion of countability says that there are actually the SAME number of even numbers as natural numbers, because we can arrange a one-to-one pairing, where 1-->2, 2-->4, 3-->6, etc. For each even number, there is a natural number, and vice versa. Therefore, the size of the sets is the same. (For those of you with problems with this definition, I think it's fair to say that this mathematical notion of "equal sizes" is useful for certain purposes; you may go on thinking that the set of natural numbers is larger.)



We then discussed whether there were more rational numbers (any number that can be expressed as m/n, where m and n are natural integers and n doesn't equal 0) than natural numbers. With a little thought, it can be shown that there is a way of matching them up one-to-one.



I then asked whether irrational numbers, such as pi, e, square roots of various integers, etc. - numbers that when written in decimal form do not terminate or repeat - can also be matched up one-to-one with the natural numbers. We didn't get the chance to go over Cantor's diagonal argument, but I got to thinking about whether it could be done. Now, since it is well accepted in mathematics that it *cannot* be done - and this acceptance has been corroborated in numerous proofs - there are some clear problems with my approach. Anyway, here it is:



To write all the irrational numbers down in some kind of systematic way (thereby leading to a one-to-one correspondence with the natural numbers), do the following:

0.0, 0.1, 0.2... 0.9

1.0... 1.9... 9.9

0.01, 0.02, ... 0.10, 0.11,...0.19,... 0.99, 1.00

1.01, 1.02... 9.99.

10.1, 10.2... 10.9, 10.01, 10.02...10.99

... 99.99



I'm not even sure if I've captured it properly here, but the basic idea is to go up on either side of the decimal point, taking turns and being careful to write each number down. It's clear that in this method, you are going to be able to write down every terminating decimal number.



In writing down my 'proof', it became clear to me that you never really write down any infinite string of decimals, and so you never write down even one irrational number. Yet you are able to write any arbitrarily long rational number (though you never complete writing a repeating decimal expansion), so in some sense it feels to me that all the numbers will be there.



The diagonal argument hinges on having all of the numbers listed, and then changing one of the digits in each of the numbers to create a new irrational number that's not on the list. Yet if it's impossible to write down even one irrational number, to what extent does it make sense to say you can do that? Why doesn't it make just as much sense to say that this process of writing down every number expressible as a decimal will eventually (whatever that means) write down every number, including the repeating decimal expansions of rational numbers like 1/3 and even those with non-repeating non-terminating decimal expansions that are infinitely long?