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”.
“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:
yes.
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):
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 $n=2k+1.$
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”.
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 $13\,?$