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Can it constructively be shown that for all $x,y \in \mathbb{R}$, $xy = 0 \rightarrow x = 0 \vee y = 0$?

e: The reason that I am interested in this is that I recall hearing something about how the roots of polynomials in a constructive setting aren't generally as nice as you would expect. For instance, for $\alpha \geq 0$ we can show that the polynomial $x^2 - \alpha$ does indeed factor as $(x-\sqrt\alpha)(x + \sqrt\alpha)$ but we cannot conclude that an arbitrary root $r$ is either $\sqrt\alpha$ or $-\sqrt\alpha$ without appealing to the integral domain property, which seems suspect for reals.

user181407
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    That depends entirely on your definition of $\Bbb R$. If you have defined it as a (the) Dedekind-complete, totally ordered field, then this is true by definition of field. If you have defined it as the set of Dedekind cuts (or equivalence classes of Cauchy sequences) on the rationals, then it's not that simple. I would think the latter falls more in line with constructive mathematics, though. – Arthur Nov 13 '17 at 11:22
  • @Arthur It doesn't seem obvious to me that "every field is an integral domain" is valid constructively. The standard proof requires that you can decide whether a given element is equal to $0$. – user181407 Nov 13 '17 at 11:35
  • @user181407: could you please edit the question to include more context and motivation, such as the remarks you made about integral domains in the previous comment? I see there is a vote to close, which is understandable for a post which only states a problem without any discussion. – Carl Mummert Nov 13 '17 at 11:45
  • @user181407 you're right. It depends on how you define "field". In my book, it's an integral domain where each non-zero element has a multiplicative inverse. But I am aware that there are others. – Arthur Nov 13 '17 at 11:46
  • @user181407: have you looked at Bishop's book to see how he treats the field of real numbers? I don't have it at hand here, but I suspect he must make some comments about the algebraic structure of the reals. – Carl Mummert Nov 13 '17 at 11:49
  • @CarlMummert I've added a bit on why the question interests me. I have not looked into what Bishop has to say on this but I'll look into it, thanks. – user181407 Nov 13 '17 at 12:27
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    Thanks. In the particular case of $x^2 - a$, where $a > 0$, because we know $\sqrt{a}$ and $-\sqrt{a}$ are separated, if we know $c$ is a root of the polynomial then we can determine effectively whether $c = \sqrt{a}$ or $c = -\sqrt{a}$. – Carl Mummert Nov 13 '17 at 12:51

2 Answers2

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This is equivalent to the nonconstructive axiom, the lesser limited principle of omniscience (LLPO) (as long as we assume countable choice).

To show that $\mathbb{R}$ being an integral domain implies LLPO, let $\alpha \colon \mathbb{N} \rightarrow 2$ be an infinite binary sequence such that $\alpha(n) = 1$ at most once. To prove LLPO is to show that either $\alpha(2n) = 0$ for all $n$ or $\alpha(2n + 1) = 0$ for all $n$. Define $x := \sum_{n = 0}^\infty 2^{-\alpha(2n)}$ and $y := \sum_{n = 0}^\infty 2^{-\alpha(2n + 1)}$. Then $xy = 0$ (because for all $\epsilon > 0$ either $x < \epsilon$ or $y < \epsilon$ and certainly $x, y \leq 1$, so $xy < \epsilon$), but $x = 0$ implies $\alpha(2n) = 0$ for all $n$, and $y = 0$ implies $\alpha(2n + 1) = 0$ for all $n$.

To show that LLPO implies $\mathbb{R}$ is an integral domain, let $x, y$ be such that $xy = 0$. We construct a sequence $\alpha \colon \mathbb{N} \rightarrow 2$ as follows. For any $n$, if $\alpha(m) = 1$ for some $m < n$ take $\alpha(n)$ to be $0$. Otherwise split into cases depending on whether $n$ is even or odd. If $n$ is even, then either $x < 2^{-n}$ or $x > 2^{-n -1}$. Take $\alpha(n)$ to be $0$ if $x < 2^{-n}$ and $1$ if $x > 2^{-n -1}$. If both hold, $\alpha(n)$ can be either $0$ or $1$, so we need to use countable choice to pick a value. When $n$ is odd, we do the same but for $y$ instead of $x$. Then if $x = 0$ then $\alpha(2n) = 0$ for all $n$, and if $y = 0$ then $\alpha(2n + 1) = 0$ for all $n$.

aws
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I don't immediately see a good way to show how nonconstructive it is, but the principle $$ (\forall x,y \in \mathbb{R})[xy = 0 \to (x = 0 \lor y = 0)] $$ has no computable witness for the disjunction, and so it will not be constructively provable. Here I am taking $\mathbb{R}$ to be the set of quickly-converging Cauchy sequences of rationals, in the style of Bishop's Constructive Analysis.

In this situation, a computable witness for the disjunction is a (partial) function $F(x,y)$ so that if $x y= 0$ then $F(x,y) \in \{0,1\}$ and $F(x,y) = 0 \to x= 0$ and $F(x,y) = 1 \to y= 0$. In other words $F$ identifies one of two disjuncts which is true, in the case $xy = 0$. A general principle of constructive provability in this setting is that we would expect a computable witness to exist in this case if the formula were to be constructively provable. This is because a constructive proof of the overall principle would give us the algorithm needed to compute $F$.

The proof that there is no computable witness for the disjunction is by contradiction (we are using a classical proof to show that the original formula is not constructively provable.) Start by building both $x$ and $y$ as the sequence $1/2^n$. Because $F$ is computable, at some point it must return $0$ or $1$ just looking at long enough initial segments of $x$ and $y$. As soon as it does, we change the construction of whichever sequence $F$ selects ($x$ or $y$) so that this sequence becomes constant and nonzero. (If $F$ never returns $0$ or $1$ on the unmodified sequences $1/2^n$, we have a contradiction already.)

This process gives two computable sequences $x$ and $y$ so that $xy = 0$. However, $F$ does not select from $x$ and $y$ the sequence that converges to zero, and so we see there is no computable witness for the disjunction.

A "better" proof would show how nonconstructive the original principle is, rather than just showing that the principle is not constructively provable.

Carl Mummert
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