Why aren't rationals suspect to Cantor's diagonal proof? That is, can anyone give a proof that they aren't suspect? And can you try to prove this without taking the method of proving that they are countable first?

Let's just take rationals between 0 and 1 in binary to simplify things this way.

 

 

Cantor's diagonal proof for the real numbers goes as follows:

Proven: The real numbers between 0 and 1 are uncountable (they have no one-to-one correspondence with the integers).

Proof:

Let us assume the real numbers between 0 and 1 are countable. Then write them down in any order in a table. Put them in binary for simplicity.

Then, take the 1st number after the decimal place in the first number, the second digit after the decimal place in the second, etc. and create a number such that the digits are switched from the number created by the earlier method. This number must be different from all numbers in the table because for the nth number, the nth digit is different. Even if you add this number, there will be another number for the new table which will be different than every element.

For example:

Table:

.0

Element not in the table: .1

 

Table:

.00

.10

Element not in table: .11

.000

.100

.110

Element not in table: 0.111

etc.

-Uncool-

because the resulting expansion after applying the diagonal argument is not necessarily a rational number. generally it is better to write Cantor's proof as a constructivits proof rather than a contradictory one - there is no need to claim that the list in the 'table' is complete; just assume it is any injection from N to R, or [0,1], or some subset of [0,1] and it follows it is never surjective. also, you should be careful here - it is better to assume these strings of 0s and 1s are a subset of the base 3 expansions. this negates any issues with numbers that possess two different binary expansions.