# Why does Cantor's Proof (that R is uncountable) fail for Q?

Why doesn't the "diagonalization argument" used by Cantor to show that the reals in the intervals [0,1] are uncountable, also work to show that the rationals in [0,1] are uncountable?

To avoid confusion, here is the specific argument.

Cantor considers the reals in the interval [0,1] and using proof by contradiction, supposes they are countable. Since this set is infinite, there must be a one to one correspondence with the naturals, which implies the reals in [0,1] admit of an enumeration which we can write in the form x$_j$ = 0.a$_{j1}$ a$_{j2}$ a$_{j3}$... (for j $\in$ $\mathbb{N}$).
Now Cantor constructs a number x* where the jth digit of x* is (a$_{jj}$+2) mod 10 (I know there are other schemes; this is the one my professor used). The observation that x* $\neq$ x$_j$ $\forall$ j $\in$ $\mathbb{N}$, leads us to the conclusion that x* is not in this list, and hence the reals in [0,1] cannot be enumerated, and so [0,1] is not countable (which implies that the real numbers are not countable).
I asked my professor and she was unable to tell me why this same argument couldn't be used to prove that the rationals in [0,1] are also uncountable. It seems the argument would have to somehow show that the number you constructed using Cantor's method must be either a terminatingor repeating decimal, but I can't see how to prove this

Matt

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$x^*$ might not be rational. –  littleO Sep 20 '13 at 13:44
The elements in the list are rationals now. –  Matt Brenneman Sep 20 '13 at 13:47
Yes, the elements in the list are rationals, but $x^*$ is not. –  littleO Sep 20 '13 at 13:50

Why do you think that number must be in the list? You can list the rationals, you can construct $x^*$, and $x^*$ isn't in the list. There's no contradiction. –  littleO Sep 20 '13 at 13:47