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A natural number $n$ is even if it has the form $n = 2k$ for some $k$ $\in$ $\mathbb N.$

I'm confused by the usage of 'some' in this definition. Isn't "exactly one" more suitable for this definition? I mean, when you choose one value of $n,$ just one value of $k$ matches that.

How do we know whether 'some' means at least one or just one? Depending on the context?

Anyway, why do we say "some", rather than "all" or "any", to mean at least one? I mean, it does sound like they all suggest “at least one”.

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    $\begingroup$ This is not a question of mathematics but rather English. $\endgroup$
    – Zhen Lin
    Jul 2 at 2:20
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    $\begingroup$ Well, we definitely couldn't use all. $\endgroup$
    – J.G.
    Jul 2 at 6:35
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    $\begingroup$ @ZhenLin: this question is not about mathematics per se I would say that it is a mathematical question: OP is trying to understand a mathematical statement where the meaning of the natural language used in the statement is unclear to OP. $\endgroup$ Jul 20 at 12:45
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    $\begingroup$ Here is my perspective on why the OP's question is a natural (thus common) one for someone unschooled in Mathematical English. $\endgroup$
    – ryang
    Jul 20 at 12:54
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    $\begingroup$ This question is very clearly not a duplicate of the suggested question since neither that question nor any of its answers say anything at all about the meaning of "some". I would not be surprised if this is a duplicate of some other question but voters should pay attention to what questions actually say and not just assume two questions that are kind of similar are the same. $\endgroup$ Jul 21 at 1:32

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How do we know whether 'some' means at least one or just one?

For some $k,\, P(k)$ is true”, “there exists at least one $k$ such that $P(k)$ is true” and “there exists one $k$ such that $P(k)$ is true” all have the same meaning.

This remains the case even when we separately deduce that only/exactly/just one value of $k$ actually satisfies $P(k),$ as in this example:

A natural number $n$ is even if it has the form $n = 2k$ for some $k$ $\in$ $\mathbb N$

when you choose one value of $n,$ just one value of $k$ matches that.

yes.

isn't "exactly one" more suitable for this definition?

In formulating the above definition, “for exactly one $k$” and the less specific “for at least one $k$” are equally logical and correct. But there is a subtle but important difference (EDIT: thanks, Surb, for contributing the following point):

  1. if the definition of $n$ being even is the existence of exactly one $k$ such that $n=2k+1,$ then, to verify that $17$ is even, we must find $k=8$ AND show that there is no other $k$ satisfying the property;
  2. whereas if the definition of $n$ being even is the existence of (at least one) $k$ such that $n=2k+1,$ then, to verify that $17$ is even, we JUST need to find $k=8.$

However, when using a theorem, the former formulation is stronger and more convenient: if $n$ is even, then we know that there exists exactly (rather than merely at least) one $k$ such that $n=2k+1.$

Depending on the context?

We often default to “at least one” because it is a direct translation of the commonly-used quantifier $\text‘\exists\text’.$ If more precision is necessary (and justifiable),we can just say, for example, “exactly one” or “at least two” or “three”, instead of “some”.

Why do we say "some", rather than "all" or "any", to mean at least one? I mean, it does sound like they all suggest “at least one”.

All three indeed imply “at least one”—but only “some” is equivalent to “at least one”.

After all, why should some (i.e., at least one) value of $k$ satisfying $P(k)$ imply that all values of $k$ satisfy $P(k)\,??$

And, letting $k\in\mathbb R,$ why should $(k-3)(k-7)=0$ and $14=2k$ being true for $k=7$ (some value of $k$) imply that they are true for every and any value of $k,$ including $13\,?$

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