Real numbers are uncountable

We present a variant of Cantor's diagonalization argument to prove the are . This that there exist ^[1].

We use the decimal representation of the real numbers. An overline ( $\overline{9}$ ) 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 , that the real numbers are ; suppose that $f:Z^{+}\twoheadrightarrow R$ is a surjection. Let $r_n$ denote the $n^{\text{th}}$ decimal digit of $r$, so that the fractional part of $r$ is $r_1{r}_2{r}_3{r}_4{r}_5\dots$ Then define a real number $r^{\prime }$ with $0\le r^{\prime }<1$ so that $r_n^{\prime }$ is 5 if $\left(f\right(n){)}_n\ne 5$, and 6 if $\left(f\right(n){)}_n=5$. Then there can be no $n$ such that $r^{\prime }=f\left(n\right)$ since $r_n^{\prime }\ne \left(f\right(n){)}_n$. Thus $f$ is not surjective, contradicting our assumption, and $R$ is uncountable. $\square$

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

1. ^{^︎}

   Since the real numbers are an example of one.