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A real number $n$ is rational iff $\exists p,q \in \mathbb{Z}, n=\frac{p}{q} \:$ and $q \neq 0$. Why do we need that last condition? Sure, I understand it makes it more clear that a fraction is undefined if the denominator is 0, but this is an existential statement, so if a number is rational, I would think this implies that there exists a $q$ that is nonzero. My reasoning is that if a real number is rational, we can find a $p$ and $q$ that satisfies $n=\frac{p}{q}$, and for this to be possible, $q$ must be nonzero. I’ve lost points on a homework for not mentioning that condition in a proof, so I’m curious why it’s so necessary.

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    $\begingroup$ Are you happy for $\frac{1}{0}$ to be a rational number? Are you happy for an indeterminate form to be a number? An indeterminate form is exactly that: "not exactly known, established, or defined." How can we say it's rational or not rational when its indeterminate? $\endgroup$ – Fly by Night Oct 11 '17 at 18:32
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    $\begingroup$ @FlybyNight so if this was restricted to if and not iff, would this still be incorrect? That’s why I say “a real number $n$. $\frac{1}{0}$ isn’t a real number. $\endgroup$ – rb612 Oct 11 '17 at 18:34
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    $\begingroup$ The use of "if and only if" is redundant for a definition, one direction is a tautology... by definition. $\endgroup$ – Fly by Night Oct 11 '17 at 18:38
  • $\begingroup$ @FlybyNight so then if I have this restriction “a real number n” is rational if... then why wouldn’t this exclude all numbers of an indeterminate form? $\endgroup$ – rb612 Oct 11 '17 at 18:41
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    $\begingroup$ "That’s why I say “a real number $n$" Technically, when parsed as (a real number $n$ is rational) IF AND ONLY IF (...), I believe you're correct. As for why this is done --- habit, unintentionally being non-minimalistic, intentionally being non-minimalistic (for pedagogical reasons), etc. $\endgroup$ – Dave L. Renfro Oct 11 '17 at 19:14

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