Re: The seven step-Mathematical preliminaries

From: Brian Tenneson <tennesb.domain.name.hidden>
Date: Wed, 03 Jun 2009 09:46:32 -0700

> How do you know that there is no biggest number? Have you examined all
> the natural numbers? How do you prove that there is no biggest number?
>
>
In my opinion those are excellent questions. I will attempt to answer
them. The intended audience of my answer is everyone, so please forgive
me if I say something you already know.

Firstly, no one has or can examine all the natural numbers. By that I
mean no human. Maybe there is an omniscient machine (or a "maximally
knowledgeable" in some paraconsistent way) who can examine all numbers
but that is definitely putting the cart before the horse.

Secondly, the question boils down to a difference in philosophy:
mathematicians would say that it is not necessary to examine all natural
numbers. The mathematician would argue that it suffices to examine all
essential properties of natural numbers, rather than all natural numbers.

There are a variety of equivalent ways to define a natural number but
the essential features of natural numbers are that
(a) there is an ordering on the set of natural numbers, a well
ordering. To say a set is well ordered entails that every =nonempty=
subset of it has a least element.
(b) the set of natural numbers has a least element (note that it is
customary to either say 00 is this least element or say 01 is this least
element--in some sense it does not matter what the starting point is)
(c) every natural number has a natural number successor. By successor
of a natural number, I mean anything for which the well ordering always
places the successor as larger than the predecessor.

Then the set of natural numbers, N, is the set containing the least
element (0 or 1) and every successor of the least element, and only
successors of the least element.

There is nothing wrong with a proof by contradiction but I think a
"forward" proof might just be more convincing.

Consider the following statement:
Whenever S is a subset of N, S has a largest element if, and only if,
the complement of S has a least element.

By complement of S, I mean the set of all elements of N that are not
elements of S.

Before I give a longer argument, would you agree that statement is
true? One can actually be arbitrarily explicit: M is the largest
element of S if, and only if, the successor of M is the least element of
the compliment of S.

If so, then that statement proves that there is no largest element of N:
Letting S be N in particular, note that N is a subset of N (albeit not a
"proper" subset).
Then the statement reads as the following for this particular choice S:
N has a largest element if, and only if, the complement of N has a least
element.

The compliment of N is the empty set. To elaborate: the compliment of N
is the set of all elements of N that are not elements of N. No elements
can both be and not be elements of N, so this set is empty.

The empty set does not have a least element. In fact, it has no
elements at all.

Therefore, N does not have a largest element.

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Received on Wed Jun 03 2009 - 09:46:32 PDT

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