Quick question about Cantor's diagonal argument Regarding the number we create from the diagonal of the hypothesized listing of all real numbers, could we just add 1 to each diagonal digit to create it, instead of the usual more complicated method? 
What I mean is, suppose the diagonal of the list forms the number 
a.bcdefg.... , where a, b, c... are digits. 
Could we form the new number not on the hypothesized listing of all the reals by adding 1 to each of the digits a, b, c, d, e, ...?
I can't see any reason why this wouldn't work, which leaves me wondering why don't we use such a simpler method? 
 A: You have to deal with the fact that the decimal representation is not unique: $0.123499999\ldots$ and $0.12350000\ldots$ are the same number. So you have to mess up more with the digits, for instance by using the permutation $(0,5)(1,6)(2,7)(3,8)(4,9)$ - this is safe since no digit is mapped into an adjacent digit.
In the hypothetical listing every real number is represented only once, but maybe by perturbing the diagonal in a soft way you may get an equivalent representation for a number already listed, that is the issue. Let us consider the hypothetical listing
$$ \begin{array}{rcl}1&:& 0.\color{red}{1}06752\ldots \\ 2&:& 0.1\color{red}{9}9999\ldots \\ 3&:& 0.20\color{red}{9}652\ldots \\ 4&:& 0.346\color{red}{9}33\ldots\end{array} $$
By taking the diagonal and adding $1$ to each digit you get $0.2000\ldots$, which is already in the list, just in a equivalent form.
A: Adding one would be fine,  i believe.  but I don't really believe this is any simpler:  it's merely a particular way of changing the diagonal entry.  Any way will do:  you can change the diagonal entry to any different digit;  and you will produce a number not in the list...
