Need example from number theory

[Mario Carneiro]

Whether it is obvious that this use is unnecessary probably depends on your
disposition / background, but one example that comes to mind is the proof
that Q is countable because it surjects from Z x N. This is in some sense
the "obvious" proof, but if you use the lemma that surjections decrease the
cardinality then you are invoking choice. In this case, you can get around
it by noting that there is a way to make canonical choices, namely taking
only coprime numerator/denominator instead of sticking with the equivalence
class formulation.

Mario

On Sat, Aug 28, 2021 at 6:48 PM JOSEPH SHIPMAN <joeshipman at aol.com> wrote:

> I don’t want a necessary use of AC. I want an unnecessary use, that isn’t
> obviously unnecessary!
>
> — JS
>
> Sent from my iPhone
>
> > On Aug 28, 2021, at 4:40 PM, Gyorgy Sereny <sereny at math.bme.hu> wrote:
> >
> > 
> > The most elementary example of the necessary use of AC I know
> > (of course, it is not number theory):
> >
> > (5.3) The union of a countable family of countable sets is countable.
> > (By the way, this often used fact cannot be proved in ZF alone.)
> >
> > To prove (5.3) let A_n be a countable set for each n \in N.
> > For each n, let us _choose_ an enumeration (a_nk: k \in N) of A_n. [...]
> >
> > (Thomas Jech: Set Theory
> > Second Corrected Edition, p.39)
> >
> >
> > Gyorgy Sereny
> >
> >
> >> On Sat, 28 Aug 2021, JOSEPH SHIPMAN wrote:
> >>
> >> What is an example of a theorem of number theory which has a
> >> well-known proof *suitable for undergraduates*,
> >> that uses the Axiom of Choice in a way that is not obviously
> unnecessary?
> >>
> >> We know that AC can be eliminated from the proof of any arithmetical
> >> statement, but I’d like an example I can easily explain.
> >>
> >> — JS
> >>
>
>
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