"With" Meaning in Proof, Compared to "Such That"

Question

What does 'with' mean in: For all real numbers x, there is some real number y with y = $x^2$

My thought process so far is that it is similar to: $\forall$ x $\epsilon$ ${R}$, $\exists$ y $\epsilon$ ${R}$ s.t. y = $x^2$

But I thought 'with' and 'such that' would have different meanings because the other questions on my assignment uses the word 'such that' instead of 'with'.

You can say "an $y$ with $x^2=y$, or "an $y$ satisfying $x^2=y$, or "an $y$ such that $x^=y$, or "an $y$ subject to $x^2=y$. — Dietrich Burde, Oct 02 '21 at 16:59

ryang · Accepted Answer · 2023-06-24T07:57:59.420

4

In your given statement, I too would mentally replace the word with with ‘such that’.

There seems to be a tendency in middle- and high-school situations to write with or where in mathematical statements even when doing so introduces ambiguity. This recent example goes

If $p$ is prime, then $\binom p k=np$ where $n\in\mathbb Z$ and $0<k<p$

(typically, with might have been written here in lieu of where). Note that in this case, with/where is not intended to mean ‘such that’.

Apart from the ambiguity of the order of quantifiers, it is not even clear that the two qualifications after the word where are actually intended to refer alternately to existential quantification and universal quantification, respectively! The intended sentence (styled better) is

For each prime number $p$ and each $k\in\{1,2,\ldots,p-1\},$

there exists some $n\in\mathbb Z$ such that $\binom p k=np.$

or

For each $p,k\in\mathbb Z,$

$\quad$ if $p$ is prime and $0<k<p,$ then there exists some $n\in\mathbb Z$ such that $\binom p k=np.$

Related:
What does the word 'with' mean in this theorem?
Is ‘where’ interchangeable with ‘such that’?

edited Jun 24 '23 at 07:57

answered Oct 02 '21 at 17:37

ryang

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(+1) for the ambiguity observations and revisions. I've found it nearly always best to put the variable specifications BEFORE the variables are used, otherwise there is the possibility of having a hanging quantifier. In case you're interested in these sorts of things, see my extensive comments to this question, and this answer, and this answer. – Dave L. Renfro Oct 02 '21 at 18:41
@ryang sorry for commenting an old post, but with "ambiguity of the order of quantifiers" and "it is not clear that the first and second quantifications..." you mean that the example could be interpreted as: "For each prime number $p$ there exists $n\in\mathbb{Z}$ and there exists $0<k<p$ such that $\binom{p}{k}=np$"as well? – ZaWarudo Jun 24 '23 at 05:44
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@ZaWarudo Thanks for pointing out that my reference to the "first and second" quantifications was intended to mean "the sentence's last two quantifications"; I've edited away that mistake. $\quad$ The final 'and' in your suggestion should be replaced with 'such that', in which case yes that's another possible interpretation, and it happens to be true too; notice though that my suggestion (Eddie's intended statement) is a stronger statement. – ryang Jun 24 '23 at 06:20

score 2 · Answer 2 · answered Oct 04 '21 at 17:32

There is nothing wrong with "with" when it is used in place of "such that". See for example this answer. In fact, sometimes it is convenient to use both in sentences like:

For every $k,m∈ℕ^+$ with $\gcd(k,m) = 1$, there are some $a,b∈ℕ$ such that $a·k−b·m = 1$.

"With" Meaning in Proof, Compared to "Such That"

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