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
Best Answer
Why should the decimal expansion constructed from the diagonal of the list yield a rational number? (I.e., why would that decimal expansion be repeating?)