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  text: 'We present a variant of Cantor's diagonalization argument to prove the  [4bc real numbers] are [2w0 uncountable].  This [constructive_proof constructively proves] that there exist [uncountable_set uncountable sets] %%note:  Since the real numbers are an example of one.%%.\n\nWe 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.\n\n**Theorem** The real numbers are uncountable.\n\n**Proof** Suppose, for [46z contradiction], that the real numbers are [-6f8 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. $\\square$\n\n\nNote that choosing 5 and 6 as our allowable digits for $r'$ side-steps the issue that $0.\\overline{9} = 1.\\overline{0}$.  %%',
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