We present a variant of Cantor's diagonalization argument to prove the real numbers are uncountable. This [constructive_proof constructively proves] that there exist [uncountable_set uncountable sets] %%note: Since the real numbers are an example of one.%%.

We use the decimal representation of the real numbers. An overline ( $~$\bar{\phantom{9}}$~$ ) is used to mean that the digit(s) under it are repeated forever. Note that $~$a.bcd\cdots z\overline{9} = a.bcd\cdots (z+1)\overline{0}$~$ (if $~$z < 9$~$; otherwise, we need to continue carrying the one); $~$\sum_{i=k}^\infty 10^{-k} \cdot 9 = 1 \cdot 10^{-k + 1} + \sum_{i=k}^\infty 10^{-k} \cdot 0$~$. Furthermore, these are the only equivalences between decimal representations; there are no other real numbers with multiple representations, and these real numbers have only these two decimal representations.

**Theorem** The real numbers are uncountable.

**Proof** Suppose, for contradiction, that the real numbers are countable; suppose that $~$f: \mathbb Z^+ \twoheadrightarrow \mathbb R$~$ is a surjection. Let $~$r_n$~$ denote the $~$n^\text{th}$~$ decimal digit of $~$r$~$, so that the fractional part of $~$r$~$ is $~$r_1r_2r_3r_4r_5\ldots$~$ Then define a real number $~$r'$~$ with $~$0 \le r' < 1$~$ so that $~$r'_n$~$ is 5 if $~$(f(n))_n \ne 5$~$, and 6 if $~$(f(n))_n = 5$~$. Then there can be no $~$n$~$ such that $~$r' = f(n)$~$ since $~$r'_n \ne (f(n))_n$~$. Thus $~$f$~$ is not surjective, contradicting our assumption, and $~$\mathbb R$~$ is uncountable. $~$\square$~$

Note that choosing 5 and 6 as our allowable digits for $~$r'$~$ side-steps the issue that $~$0.\overline{9} = 1.\overline{0}$~$. %%