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For any integers $m$ and $n$, if $7m+5n=147$, then $m$ is odd or $n$ is odd.

$$Q(m,n) \equiv 7m+5n=147$$

$$∀m∀n: Q(m,n) → \bigl(m \not\equiv 0 \!\!\pmod 2 \lor n \not\equiv 0 \!\! \pmod 2\bigr)$$

Am I right in assuming $\forall$ means "any" in this case? It doesn't seem to make sense to me ($\exists$ to me means "at least one, many, one, all but one, etc; anything less than all but more than none"), but Wikipedia states that $\forall$ can also mean "for any".

Is this correct?

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    $\begingroup$ Yes, you are correct. Normally mod is done $m\not\equiv 0\mod 2$. $\endgroup$
    – TheNumber23
    Oct 9, 2013 at 6:20
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    $\begingroup$ $\forall$ means "for all" or "every". $\exists$ means "exists" and could include any possibility except "does not exist". In particular, $\exists$ includes "all". $\endgroup$
    – abiessu
    Oct 9, 2013 at 6:22
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    $\begingroup$ "Any" can mean "for all" or "there exists" depending on the sentence. It takes some experience to tell, in a particular sentence, which meaning is intended. See math.stackexchange.com/questions/509520/… $\endgroup$
    – Carl Mummert
    Oct 9, 2013 at 10:50
  • $\begingroup$ Also: math.stackexchange.com/questions/402020/… $\endgroup$
    – Carl Mummert
    Oct 9, 2013 at 11:08

3 Answers 3

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Yes, "for any" means "for all" means $\forall$. "Any" implies you pick an arbitrary integer, so it must be true for all of them.

As for mods: usually, it's not expressed as an operator, but instead as a kind of equivalence relation: $a \equiv b \pmod{n}$ means that $n$ divides $a - b$. So you would write "m is odd" as $m \equiv 1 \pmod 2$.

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  • $\begingroup$ In fact there is a binary mod operation: $x\bmod y$ is defined to be $x-y\lfloor x/y\rfloor$ for $y\ne 0$. The $\LaTeX$ code for it is \bmod. It is perfectly correct to write $m\bmod 2=1$ for ‘$m$ is odd’. $\endgroup$
    – Brian M. Scott
    Oct 9, 2013 at 6:26
  • $\begingroup$ For the sake of simplicity, do mathematics persons generally recognize % as the modulo operator? It doesn't take much to change my m%2=1 to m=1 (mod 2), but I'd rather remain consistent. $\endgroup$
    – gator
    Oct 9, 2013 at 6:33
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    $\begingroup$ @BrianM.Scott That is true, but I think it's correct to say the usual expression is a relation? And @ riista, no, I'd bet most would know what you're talking about, but it's not considered correct, as far as I know. $\endgroup$
    – Henry Swanson
    Oct 9, 2013 at 6:36
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    $\begingroup$ @riista: In my view the only thing that makes it look odd to a mathematician is the use of % instead of $\bmod$. $\endgroup$
    – Brian M. Scott
    Oct 9, 2013 at 6:41
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    $\begingroup$ "any" can also mean "there exists"; it depends on the particular sentence. $\endgroup$
    – Carl Mummert
    Oct 9, 2013 at 10:49
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$\forall m$ is usually read for all m or for each m; for any m is a possible paraphrase in some contexts. $\forall x\big(\varphi(x)\big)$ means that for all possible values of $x$ in the domain of discourse, the statement $\varphi(x)$ about $x$ is true. If you think of $\varphi(x)$ as saying that $x$ has some particular property, then $\forall x\big(\varphi(x)\big)$ says that every element in the domain of discourse has that property.

Your understanding of $\exists$ isn’t quite correct, however; $\exists m\big(\varphi(m)\big)$ means precisely that there is at least one thing in the domain of discourse that has the property in question. It does not exclude the possibility that every element of the domain of discourse has it. For example, if we’re talking about integers, $$\exists n(n\text{ is even }\lor n\text{ is odd}\}$$ is a true statement, even though $$\forall n(n\text{ is even }\lor n\text{ is odd}\}$$ is also true.

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If a statement is true "for all", it implies that it is true for a subset or in this case for any one element.

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