Counterexample for ∀x∃y (x=1/y)

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This is not good understanding. The formula says that any $x\in\Bbb Z$ is the inverse of some $y\in\Bbb Z$. This is evidently false. The counterexample is $x=2$. There is no $y\in\Bbb Z$ s.t. $2=\dfrac{1}{y}$ (if should be $y=\dfrac{1}{2}\not\in\Bbb Z$). So, the negation is true: there exists $x\in\Bbb Z$ ($x=2$) s.t. for all $y\in\Bbb Z$ we have $x\ne\dfrac{1}{y}$.

My answer is good, if the domain for $y$ is the set of all integers. Nevertheless, if the domain for $y$ is any subset of raeals, it is enough to take $x=0$ for the counterexample.

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AvenNova
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Updated on August 01, 2022

Comments

  • AvenNova
    AvenNova over 1 year

    So I am working on this math problem which states that for every x (domain: all integers) there is a y (domain: all integers) for which x=1/y is true.

    ∀x∃y (x=1/y)

    If I understand this correctly, it means that there is some y for every x in the set of integers which makes x=1/y always true. This is false, but I don't understand how to show a counterexample for this one. I could try explaining how this is not false but aside from that, I am not sure what to do.

    • Shaun
      Shaun about 6 years
      What is the domain for $y$?
    • Xander Henderson
      Xander Henderson about 6 years
      The statement says that for any integer $x$, there is some $y$ (what kind of $y$? an integer? a rational? and extended real?) such that $x = \frac{1}{y}$. A counterexample would be an integer $x$ that cannot be written as the reciprocal of some $y$ (wherever $y$ is meant to live). $x=0$ might be a good candidate.
    • Shaun
      Shaun about 6 years
    • szw1710
      szw1710 about 6 years
      @XanderHenderson $x=0$ is good for any domain for $y$ :)
    • Xander Henderson
      Xander Henderson about 6 years
      @szw1710 Not necessarily. For example, in the extended reals or the Riemann sphere, $0 = 1/\infty$, right? But in those contexts, there may not be a counterexample. Which is why it is important to know where $y$ lives. :)
    • szw1710
      szw1710 about 6 years
      @XanderHenderson This is the only case. What I wrote in my answer, $x=0$ is good for any subset of reals, when we don't consider extended reals. Thanks for a detail you turned my attention to.
  • AvenNova
    AvenNova about 6 years
    I got it, and I understand that I kind of reversed the reasoning in my initial post which was incorrect. I will keep that in mind next time.
  • szw1710
    szw1710 about 6 years
    @AvenNova It's my pleasure.