# [FOM] Mathematical vs Logical Finitism

Dean Buckner Dean.Buckner at btopenworld.com
Sat May 3 09:40:21 EDT 2003

```>With 4, I was deliberately trying to bring in the plural
>quantification "there are some integers > 1 such that ..." where this
>allows or forces one into infinite many.

Apologies for the mistake.  For natural numbers, we deal with being > than n
differently than being < than n.  Broadly, for > n we must use predicates,
and say, for EVERY m such that ....  In the other case, we can validly speak
of (the set of) ALL numbers < n.  This means rewriting where necessary.  My
textbook "Introduction to Abstract Algebra" has a proof that begins:

Consider the set S of numbers expressible as sa + tb
Let d be the lowest member of S

But (in this case at least) we can re-write this as

Consider F() such that  F(x) iff x = sa + tb

Instead of an "infinite set" we have a predicate F() that may be true of
infinitely many objects.  Now we want to consider "the lowest number that is
F".  Let m be any number such that F(m).  Then a plural quantification
approach allows us to speak of "the set S of all numbers" less than or equal
to m.  Then, given other assumptions which I'll talk about below, we can
infer the existence of some number d which is the lowest number x in S such
that F(x).  We allow sets down, so to speak, but no sets "up".

Skimming through the "very elementary" bits of my textbook, it seemed to me
that all the proofs were of a form that could be either left as they were,
or rewritten so as to remove all "sets of numbers such that ." where what
follows "such that" uses predication rather than reference to an "infinite
set".

>I thought your idea might be that this sort of plural quantification
>is OK, but "there is a set of integers > 1 such that ..." is not.

That is correct.  But we can still talk about "being greater than 1".

>I should point out that there cannot be a finite set of integers > 1
>such that every integer > 1 is divisible by at least one of them,
>simply because there are infinitely many primes.

But we can still talk about "being a prime number".  (Being an x such that
any y that divides x is such that y=1 or y=x).

>Any very elementary book on number theory has a treatment of the
>fundamental theorem of arithmetic, not paying any attention to plural
>quantification. Is your idea that you want finite sets of integers,
>in a treatment of the fundamental theorem of arithmetic, replaced by
>plural quantification over the integers?  Is your position that you
>accept only what might be called "finite plural quantification"?

Yes.  Or rather that plural quantification, correctly understood, is
inherently finite. It depends on "George and Ringo were musicians" being
just shorthand for "George was a musician and Ringo was a musician".
Suppose we are given

p | (a & b) implies p | a  v  p | b

which we can read as we like, for example as "p divides the product of a &
b", and where "b" may be a plural term.  We want to prove

p | (a1 & a2 ... & an) implies p | x, where x is oneof (a1 & a2 ... &
an)

On the plural quantification approach, we can infer

p|a1 v p|a2 ... v p|an

since operations on "compound" proper names are understood to be shorthand
for a series of propositions involving simple proper names.  From the long
disjunction above we can now infer

p|x where x=a1 v x=a2 ... v x=an

Finally we define "x is oneof (a1 & a2 ... & an)" as "x=a1 v x=a2 ... v
x=an", giving

p|x where x is oneof (a1 & a2 ... & an)

So plural expressions have a rule of induction built into them.  Hence they
are inherently finite.  Or rather, we could define "finite" in terms of
these inductive expressions.

Harvey:
>It does appear that you are accepting quantification over natural
>numbers ... without
accept "for all natural numbers x,...", and "there exists natural
numbers x,...",

Correct, with the reservation that we read "ALL" as ANY or EVERY.

>but not "for all sets of natural numbers S,...", and
>not "there exist sets of natural numbers S,..."

Incorrect.  We can write "for every set of natural numbers S,...", with the
qualification that S is a finite set (or rather, that "S" is a plurally
referring term with the inductive properties I talked about above.  We also
allow that there are infinitely many such sets.  Ie for any set of natural
numbers we "choose", there will always be other natural numbers left over.
(Another way of saying there are infinitely many of "them").

Dean Buckner
London
ENGLAND

Work 020 7676 1750
Home 020 8788 4273

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