6

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?

gator
  • 1,835
  • 2
  • 31
  • 48
  • 1
    Yes, you are correct. Normally mod is done $m\not\equiv 0\mod 2$. – TheNumber23 Oct 09 '13 at 06:20
  • 1
    $\forall$ means "for all" or "every". $\exists$ means "exists" and could include any possibility except "does not exist". In particular, $\exists$ includes "all". – abiessu Oct 09 '13 at 06:22
  • 2
    "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 http://math.stackexchange.com/questions/509520/the-use-of-any-as-opposed-to-every/509828#509828 – Carl Mummert Oct 09 '13 at 10:50
  • Also: http://math.stackexchange.com/questions/402020/whats-the-difference-between-any-all-and-some/402073#402073 – Carl Mummert Oct 09 '13 at 11:08

3 Answers3

10

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 $m \equiv 1 \pmod 2$.

Henry Swanson
  • 12,972
  • 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’. – Brian M. Scott Oct 09 '13 at 06:26
  • 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. – gator Oct 09 '13 at 06:33
  • 1
    @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. – Henry Swanson Oct 09 '13 at 06:36
  • 1
    @riista: In my view the only thing that makes it look odd to a mathematician is the use of % instead of $\bmod$. – Brian M. Scott Oct 09 '13 at 06:41
  • @Henry: Usual seems to me a bit too strong; the binary operation is increasingly common, and in this context it’s more convenient. – Brian M. Scott Oct 09 '13 at 06:42
  • Cheers, you've both been immensely helpful. I would have thought there'd be some consistency between mathematics and compsci, so I was confused by % is not the universal operator. However, upon further recollection, a lot of languages also use mod in place of a % so it makes sense. Thanks again. – gator Oct 09 '13 at 06:43
  • 2
    "any" can also mean "there exists"; it depends on the particular sentence. – Carl Mummert Oct 09 '13 at 10:49
  • But "for any" is unambiguously "for all". – Henry Swanson Oct 09 '13 at 17:05
5

$\forall m$ 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.

Brian M. Scott
  • 616,228
2

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

sraok
  • 121