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Sep 16, 2021 at 22:04 vote accept ToucanIan
Sep 16, 2021 at 16:59 answer added Arno timeline score: 3
Sep 16, 2021 at 16:55 comment added Noah Schweber Now asked at MSE.
Sep 16, 2021 at 16:52 comment added ToucanIan @MattF. Thanks for your feedback! I will post the question else where.
Sep 16, 2021 at 16:09 comment added user44143 @Toucanlan, since you have only provided one example, and resolving it is not close to research-level, I have voted to close this question as not representing research-level mathematics.
Sep 16, 2021 at 15:55 comment added ToucanIan @NoamD.Elkies thanks for the clarification!
Sep 16, 2021 at 15:52 history edited ToucanIan CC BY-SA 4.0
added 153 characters in body
Sep 16, 2021 at 15:46 comment added ToucanIan @MattF. The question that is posed stands with or without an example. The example may be removed if you insist that it is not neccesary, but the substance of the question is not dependent on any one example.
Sep 15, 2021 at 20:55 comment added Noam D. Elkies @ToucanIan ouch, typos: $x<1$ if $x' \leq 0$ (not $\epsilon \leq 0$), and $x > -1$ if $x' \geq 0$ (not $\epsilon \geq 0$) -- sorry! (and too late to edit the originally comment)
Sep 15, 2021 at 20:33 comment added user44143 @Toucanlan, what is the point of more general techniques when you have only provided one example and it is resolved well already?
Sep 15, 2021 at 18:59 history edited ToucanIan CC BY-SA 4.0
added 303 characters in body
Sep 15, 2021 at 13:36 comment added ToucanIan @NoamD.Elkies If you let $\epsilon = 1/2$ how are there cases where $\epsilon \leq 0$?
Sep 15, 2021 at 13:30 comment added ToucanIan @LSpice I would disagree but I guess it will be up to the moderators.
Sep 15, 2021 at 13:27 comment added ToucanIan @AndreasBlass great point. Adding the assumption of apartness seems reasonable for many situations.
Sep 15, 2021 at 3:23 comment added Noam D. Elkies The example of $x^2 - 4 = 0$ does not really need the "zero product property": even for a constructivist, it is true that every real $x$ satisfies $x < 1$ or $x > -1$, and then $x = -2$ or $x = 2$ respectively. (A constructive real number $x$ is basically a black box that inputs rational $\epsilon > 0$ and outputs rational $x'$ such that $|x-x'| < \epsilon$, with the outputs for different $\epsilon$ consistent with the triangle inequality. So, let $\epsilon = 1/2$, and then $x<1$ if $\epsilon \leq 0$ while $x>-1$ if $\epsilon \geq 0$.)
Sep 15, 2021 at 2:14 comment added user44143 The zero-product property fails for continuous functions too: $fg=0$ does not imply $f=0$ or $g=0$. But one version of the @AndreasBlass comment is that if continuous functions $f,g$ satisfy $fg=0$ and $g-f>1/n$ then $f=0$ or $g=0$.
Sep 15, 2021 at 1:54 comment added Andreas Blass Unless I'm overlooking something, it seems constructively valid that, if $xy=0$ and $x$ is apart from $y$ (meaning there's a positive rational number smaller than $|x-y|$) then $x=0$ or $y=0$. That's enough to handle your example, but I'm not sure whether there are tougher examples or methods to handle them.
Sep 15, 2021 at 0:25 review Close votes
Sep 29, 2021 at 11:09
Sep 15, 2021 at 0:07 comment added LSpice This seems like a good question for our sister site MSE; but, since it does not concern research-level mathematics, it will probably be closed here.
Sep 15, 2021 at 0:07 history edited LSpice CC BY-SA 4.0
Name of article; proofreading
Sep 14, 2021 at 23:58 history asked ToucanIan CC BY-SA 4.0