On limits

Organic

Suppose a and b are different real numbers and for any e>0 we know |a-b|<e then a not= b.

proof:

If a is not b then |a-b|>0.

Let d be the difference.

Let e = d/2 then |a-b|=d and |a-b| < d/2 > 0, hence d > 0
and also |a-b|/2 > 0.

therefore non-zero/2 > 0.

matt grime

But your hypothesis is false: if a and b are distinct real numbers then it is not true that for every e>0 |a-b|<e. You do understand what the quantifier for all means?

matt grime

Actually that 'proof' of yours should go down in history: you assume a and b are distinct numbers, make a false claim about them and use that false claim to prove that a and b a different, which is part of the hypothesis... fantastic

matt grime

What are you trying to prove there anyway, now you've edited it? cos looking at it you can't really tell.

Organic

What I say is very simple: Pi is a notation an element, which its exact place in the real line is unknown.

More then thet, any element, that can be represented by infinitely many elements, its exact place in the real line is unknown, for example:

3/9 place is well-known 3/10 place is unknown.

matt grime

Why is 1/3's place known? How do you know where 1 is? Or zero? The real numbers aren't actually physically a line, Organic. You are confusing the representation of something with the something... Oh, no, you're going to talk about x and model(x) again aren't you?

Actually the statement above is trivially true because it is of the form A=>B whre A is false....

Organic

The two different a and b are both < e.

Therefore |a-b| = d < e, but both d and e > 0.

Zurtex

What are you on about?

Do you know what a number line is? It is not something physical...

matt grime

but that isn't deducible from your hypothesis: just because |a-b|<e does not state that a and b are both less than e. (take e=1 a=b=100,000,000). so it's a further pointless assumption.

Try writing out the statement of the lemma again, and its proof making sure all the hypotheses are written correctly and that it is not vacuous (which it was first time)

and seeing as the statement was for all e, then you've just shown a=b=0

Organic

Yes exactly, Math is only a theory therefore x-itself does not exist is its scope, only x-model can be used by Math language.

matt grime

good, then the model of the real numbers that is in mathematics is cauchy sequences, and 0.9999...=1 in that model, and as the discussion started about that model that's the end of the story.

Organic

Thank you for this correction you are right.

When we writing |a-b| < e we mean that d < e.
If a and b are distinct real numbers then for any e > |a-b| = d > 0.

Organic

No, cauchy sequences do not prove that 0.999... = 1 without breaking infinitely many elements to become finitely many elements, by reaching the limit.

matt grime

let e=d>2, then you have

d/2>d and d>0. now think for a second.

d/2>d => d>2d => 0>d ,yet d>0.

Want to rethink that at all.

Organic

This in not the case because e>d always.

matt grime

As real numbers are defined... oh look, circles. Tell you what, why don't you tell us what you think the real numbers are? Since your definition must be equivalent to the one using cauchy sequences where 0.9999 =1 by definition you are in trouble. I think this is because when mathematicians speak of a model, in the sense of something satisfying the axioms, an example, they don't mean what you think the mean. ie a model in the sense of a model of turbulence, or something, which is only an approximation (at the moment). There is no approximation; you are confusing the concrete and the abstract. The Cauchy sequence argument is not some "best approximation" mathematically to the "physical" real numbers, they are the real numbers, in and of themselves, it is the things that you draw on the page using axes that are the approximation, not the other way round.

breaking, infinitely many, finite, that's you wishing something to be true that isn't, you are thinking unmathematically (perhaps intuiitive and physically in your opinion, but that isnt' mathematics).

matt grime

then your initial quantifier, for all e>0, is not correct is it?


Here is your initial post:

Suppose a and b are different real numbers and for any e>0 we know |a-b|<e then a not= b.


No restriction on e>d. at all and it says for any e>0 doesn't it?

Once more you move the goal posts half way through your argument when someone points out where it's gone wrong.

So want to start from the beginning and clearly write out what it is you are trying to prove again?

Because so far you're not doing very well. I mean what was the point of it anyway?