>> Genuine question then, why isn't the set of real numbers countable,
>> given that you could represent each one with two integers (one for the
>> digits before the decimal point, one for the digits after the decimal
>> point)?
>
> The set of all strings that contain all possible combinations of two
> or more digits (in other words, the set of all possible decimal
> representations) is uncountable. In other words, it's not possible to
> enumerate them all. This might be a bit surprising (it was to me...)

This is surprising to me too, I would have thought that because every 
string of digits (without a decimal place) represents an integer then it 
would be countable.

> Let's assume that it is possible to enumerate every single possible infinite
> string of digits. Thus we have a(n infinite) list of numbered strings (its
> order doesn't really matter, as long as it contains *all* of them.) For
> example like:
>
> 1: 1001101110...
> 2: 0010110111...
> 3: 1100011110...
> 4: 0111001101...
> ...
>
> If the assumption is true, that means that this list is complete, ie. it
> contains every single possible string using those two digits. Which would
> mean that it's impossible to construct a string that does not appear in
> that list.
>
> However, we can construct such a string: Take the first digit of the first
> string and invert it, the second digit of the second string and invert it,
> and so on. In other words, with the example above the string would be
> 0110... etc.

I understand that, but don't understand why that proves it is 
uncountable. Because you can do a similar thing with a list of all 
natural numbers, ie calculate the sum and then you get a new natural 
number that wasn;t in the list to start with.

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