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Equality of two real numbers is undecidable

In the above, it is stated that inequality of two reals is semi-decidable. While this is intuitive (check the digits one by one until one differ), I would like to see a proof of this result.

Specifically, how to deal with the fact that $1.00\dots$ and $0.99\dots$ are not different?

  • If real numbers are different , we can distinguish them by finding accurate enough approximations. But it might be impossible to detect that they are equal because we never can be sure they actually are. – Peter Nov 29 '23 at 13:30
  • @Peter For sure, this is why I said the result was intuitive –  Nov 29 '23 at 15:27

1 Answers1

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We don't use the decimal expansion to define computability of real numbers (because that is a bad representation). A good representation of the real numbers is found in coding a real number $x \in \mathbb{R}$ by sequences of rational numbers $(q_n)$ such that $\forall n \in \mathbb{N} \ |q_n - x| < 2^{-n}$.

Thus, if we are given two real numbers $x, y \in \mathbb{R}$ and are trying to semidecide $x \neq y$, under the hood we have a sequence of rational numbers $(q_n)$ converging quickly to $x$, and a sequence of rational numbers $(r_n)$ converging quickly to $y$. We now search for some $n$ such that $|r_n - q_n| > 2^{-n+1}$, and claim that $x \neq y$ iff we find such an $n$.

If $x = y$ would hold, then for all $n$ it would be the case that $|r_n - q_n| \leq |r_n - x| + |q_n - x| \leq 2^{-n} + 2^{-n} = 2^{-n+1}$, so we never find such an $n$.

If $x \neq y$, then there is some $k$ such that $|x - y| > 2^{-k}$. Since $|x - y| \leq |r_{k+2} - q_{k+2}| + |q_{k+2} - x| + |r_{k+2} - y|$, it follows that $2^{-k} < |r_{k+2} - q_{k+2}| + 2^{-k-2} + 2^{-k-2}$, i.e. that $2^{-k-1} < |r_{k+2} - q_{k+2}|$, which means that $n := k + 2$ meets the condition we are looking for.

Arno
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