Re: Cantor's circular "proof" that evens = integers
- From: "Jesse F. Hughes" <jesse@xxxxxxxxxxxxx>
- Date: Sat, 05 May 2007 09:00:26 -0400
Phil <toob-headman@xxxxxxxxxxxxx> writes:
Okay, first I will hopefully correctly state Jesse's proof, and then
show why it is actually just a circular argument.
Preliminary definitions:
Set N is an infinite set containing all the natural numbers (for these
proofs, I still prefer the "old" naturals, which begin with 1).
Set E is an infinite set containing all the even natural numbers,
meaning every element e in E = 2n for some n in N.
And that 2n is in E for each n in N.
Set N x E is the Cartesian product of N and E, the set of all the
ordered pairs that can be formed from sets N and E, where the first
element in each pair is from N.
Set f is a subset of N x E containing every ordered pair from N x E
with the form <x,y> where y = 2x.
And *only* containing such ordered pairs.
Observations and deductions:
For any n in N, there exists an element 2n in N. Therefore, set N x E
contains an ordered pair of the form <n,2n> for EVERY n in N.
Set f is:
1. A function, since for any two pairs <x,y> and <x,y'>, y = y'. In
other words, for any x in f, there is one y.
2. One-to-one, since it is also true that for any two pairs <x,y> and
<x',y>, x = x'.
3. Onto, since for any y, there exists one x, and (by 1. above) for
any x, there exists one y.
Therefore, there is a bijection between the infinite sets N and E,
making sets N and E equinumerous.
Quite right, and you're done now.
[...]
Now, if anyone can PROVE that whenever x is in N, then 2x is in N,
then you have a true PROOF that N and E are equinumerous. However,
simply claiming that "this is obvious" is not adequate, because we are
working with inherently incomprehensible objects, namely INFINITE
sets.
You're right that we skipped this step. Before you saw the proof that
|N| = |E|, you did not doubt that if n is in N, then so is 2n.
Indeed, you doubt it now only because you are clinging to your
original claim that this theorem is circular.
Nonetheless, if we merely claimed without proof that this fact holds,
then it would not be a proof. But there is a proof that each element
of N can be doubled. It is a bit technical and to be honest, I don't
want to waste time giving it. It is not all that illuminating.
Anyway, someone with more patience will surely offer it.
Instead, let's talk about some simpler questions.
(1) Do you agree that whenever n is in N, then so is n+1?
(2) And hence, whenever n and m are in N, then so is n + m?
(3) And do you agree that 2n = n + n?
Which of these are not obvious?
If you agree to (1), (2) and (3), we can see that 2n is in N whenever
n is in N. This is not the usual proof, since the usual proof
requires a digression into recursion. But perhaps this simpler
argument will convince.
--
Jesse F. Hughes
"LOL. How arrogant you are. Now when you realize that I DID prove
Goldbach's conjecture and that I proved Fermat's Last Theorem as well,
how are you going to feel then?" -- James Harris
.
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