Fornumber r, n exists st n > r.any

Let r be any member of the set of all integers.

Then there is an integer greater than the set of all integers.

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- January 3rd 2013, 06:34 AMHartlwArchimedes Paradox
For

number r, n exists st n > r.__any__

Let r be any member of the set of all integers.

Then there is an integer greater than the set of all integers. - January 3rd 2013, 06:47 AMPlatoRe: Archimedes Paradox
- January 3rd 2013, 07:29 AMDevenoRe: Archimedes Paradox
4 > 3, but this by no means implies that 4 is greater than all integers.

specifically, for all integers r, we have r+1 > r. however, r+1 depends on r. there is nothing that says that we must pick "the same n" for every r.

in fact, there IS no such integer n, for if there were, then:

n+1 ≤ n

1 = (n+1) - n ≤ n - n = 0.

there is a subtle difference between the statements:

A) there exists an n such that for all r in N, n > r <---one n for EVERY r (one n in all)

-and-

B) for all r in N, there exists an n such that n > r. <---one n for EACH r (infinite n's)

one cannot exchange the order of: "for all" and "there exists" willy-nilly, no matter how similar the resulting english sentences may sound. - January 3rd 2013, 08:02 AMHallsofIvyRe: Archimedes Paradox
And please don't accuse

**Archimedes**of saying such a thing! - January 4th 2013, 06:52 AMtopsquarkRe: Archimedes Paradox
- January 4th 2013, 10:20 AMHartlwRe: Archimedes Paradox Math
To begin with, please note that my version of Archimede’s Postulate is the same as that in Birkhoff and McLean.

I wish to challenge the postulate by giving Deveno a number for which the postulate is not true. I give hin n, he gives me n+1. Am I allowed to respond? According to the postulate I can, because I can give any r. So I give him n+2 and he gives me n+3. I give him n+4 …………

By the principle of mathematical induction we both win. Because for every n I can give n+1 and so can Deveno. This is truly a paradox.

COMMENT: I have a serious disadvantage here because others can say what they want while I have been given an infraction because of this post and threatened with possible permanent banning because of this post, and other unfair infractions. Is a serious challenge to a 2300 yr old fundamental law of mathematics not appropriate for this peer review? - January 4th 2013, 10:40 AMHartlwRe: Archimedes Paradox
- January 4th 2013, 11:39 AMtopsquarkRe: Archimedes Paradox Math
Just to clarify I am asking you to read what the others are saying and respond to them as if, I don't know, they were giving you constructive criticism. I have yet to see you do that. And if I'm not mistaken most or all of the threads you've created have that flaw.

As far as this being in the wrong forum, what journal are you trying to publish this in?

-Dan - January 4th 2013, 12:23 PMHallsofIvyRe: Archimedes Paradox
And several people have pointed out that you are interpreting Archimedes incorrectly. His principle does NOT say what you seem to think it does. Saying that, given any specific integer n, there exist larger integers (for one thing, n+ 1> n for all n) it does NOT follow that "there exist an integer larger than all integers".

- January 4th 2013, 02:08 PMHartlwRe: Archimedes Paradox Math
I have read all posts and, in my opinion, responded to them. If others think not, so be it.

Perhaps this clarifies my postion:

I agree:

There is an integer greater than pi.

There is an integer greater than 37.5

There is an integer greater than the number of atoms in the universe.

I do not agree:

There is an integer greater than any number.

Your comments and letter of infraction make it perfectly clear you do not consider my post suitable for this forum. - January 4th 2013, 04:36 PMtopsquarkRe: Archimedes Paradox Math
Yes.

And no.

I have no problem with your posts, only that you start from the assumption that you are correct, make an open challenge for someone to disagree, then end up not answering the objection except to say "I'm right and you're wrong."

Devano's comment is quite well made. Work with his comment for several minutes to make sure you understand it. If you don't understand it or think that it does not fully address the problem, then make a new post addressing the issue. Work on it together. This is how people learn.

-Dan - January 4th 2013, 04:50 PMHartlwRe: Archimedes Paradox Math
Any number r can be finite or infinite. If r is qualified as finite (it isn’t), Archimedes Postulate is correct, otherwise it is incorrect.

The trouble with Deveno's argument is that he is, in effect, assuming r is finite.

I didn't simply say "Archimedes postulate is incorrect" (a paradox) and throw it out to see what happens. I repeatedly prove it and no one proves me wrong.. - January 4th 2013, 05:17 PMDevenoRe: Archimedes Paradox
Archimedes' postulate is (was?):

given a magnitude A, and another magnitude B, there is some integer n such that nA > B (Euclid actually phrases this in terms of ratios, but this is close enough). magnitude means "positive amount" in this context.

if an ordered field (F,<) possess this property for any x,y in F, the field is said to be archimedean. for example, the rational numbers Q possess this property:

given x = a/b, and y = c/d (we may assume without loss of generality that both a/b,c/d > 0...why?)

then we can pick n = bc, if one of a,d > 1:

nx = bc(a/b) = ac = (acd)/d = (c/d)(ad) > c/d.

if a = d = 1, then we have x = 1/b, y = c, and we can use n = b(c+1), so that:

nx = b(c+1)/b = c+1 > c.

*********

the whole idea is that (in Q, and other archimedean ordered fields) "integers grow unboundedly". the real numbers inherit this property from the rationals, because the order on the rationals induces a (similar) order on the reals. one consequence of this is:

the rationals are dense in the reals (there's always a rational number "nearby", no matter how small a distance away you mean by "nearby").

the archimedean property of the reals is often used to prove "R doesn't have any infinitesimals". for suppose A is the set of all positive infinitesimal elements of R. by "infinitesimal", we mean: "infinitesimal with respect to 1", that is, for any x in A, there is no integer n with:

nx > 1.

if A is non-empty, then sup(A) = s exists, since A is bounded above by 1. since s > 0, we have:

s/2 < s < 2s.

since 2s > s, 2s cannot be an infinitesimal. since 0 < s/2 < s, s/2 must be an infinitesimal (for if not, s is not the least upper bound of A).

but 4(s/2) = 2s is not an infinitesimal, so there is some integer n with n(2s) > 1, whence (4n)(s/2) = n(2s) > 1, a contradiction.

so A must be the empty set. a similar analysis shows R cannot have any negative infinitesimals (compared to -1). - January 4th 2013, 07:27 PMzzephodRe: Archimedes Paradox
- January 5th 2013, 07:08 AMHartlwRe: Archimedes Paradox
Thanks for addressing my original post. That's what I said, there is no integer greater than the set of all integers. If there were, it would go into the set of all integers, the paradox.

But another question arises here:

A disproves a theorem. B proves it without finding a flaw in A's proof. Is the theorem right or wrong?

I note it is often the case that in response to A's short transparent proof, B's is long, abstract, tortuous, and often irrelevant. An example of this might be paraphrasing a few pages from an advanced analysis text as a counter claim and then challenging the original propounder to find fault with it. I addressed this once in a post called "the irrelevant premise proof," you can google it.

I also experienced this in a thread where I questioned a post and was answered with a page of abstract analysis, replete with all the symbology, and challenged to find a flaw in it. Eventually I found that buried in all the abstraction was an assumption of what was to be proved. But by then the thread was closed so my time and effort was wasted, and in the process I got some letters of infraction. Perhaps the intent of a long paraphrase from an analysis text is to somehow become associated with the original proof.

Thanks again for your clear, transparent post.