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

I initially thought that the sentence just means "$\forall x {\in} \mathbb R\, \exists y {\in}\mathbb R$ such that $y = x^2$", however the words 'with' and 'such that' do have different meanings, and the other questions on my assignment have been using the word 'such that' instead of 'with'.

ryang
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czmua
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2 Answers2

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For all real numbers $x,$ there is some real number $y$ with $y = x^2$

In the given sentence above, the word ‘with’ is introducing a condition, so I too would mentally replace it with the phrase ‘such that’. Notice that it is also implicitly the ‘AND’ operator: “....there exists some number $y$ such that $y$ is real AND $y = x^2$.”

On a tangential note, the words ‘with’ and ‘where’ should be avoided when using either of them introduces ambiguity:
What does the word ‘with’ mean in mathematics?
What does the word ‘where’ mean in mathematics?

A recent example of such ambiguity goes

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

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 $\binom p k=np.$

  • For each $p{,}k{\in}\mathbb Z,$

    $\quad$ if $\:p$ is prime and $0<k<p,$

    $\quad$ then  there exists an $n{\in}\mathbb Z$ such that $\binom p k=np.$

  • $\forall p{,}k{\in}\mathbb Z\:\:\exists \color\red{n{\in}\mathbb Z}\:\Big((p$ is prime and $\color\red{0<k<p})\implies\binom p k=np\Big).$

Apart from the ambiguity of the order of quantifiers, observe that that single instance of the word ‘where’ (here, ‘with’ might have been written in lieu) is alternately signalling existential quantification and implication.

Note that here, unlike in the first example, ‘with’/‘where’ is not replaceable with ‘such that’.

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
  • @ZaWarudo That was a typo; I've since rewritten that part. Regarding your suggested interpretation, there are 2 issues: $\quad$ (1) that final instance of 'and' has to be replaced with 'such that', to preserve your intended meaning, $\quad$ (2) your suggestion (containing A(p,k) and B(p,k,n)) is a logically stronger statement than Eddie's intended statement (A(p,k) implies B(p,k,n)), and is in fact untrue. – ryang Feb 09 '25 at 06:41
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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 $a·k−b·m = 1$.

user21820
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