Please show us how the limit concept is rigorous

[matt grime]

Here's a little thing you need that you don't seem to know, Organic.

Suppose a and b are real numbers and for any e>0 we know |a-b|<e then a=b.

proof: if a is not b then a=b is nonzero. let d be the difference let e = d/2 then |a-b|=d and |a-b| <d/2, contradiction, hence d is zero.

 

[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]

How do you mean it is unknown?

I'm fairly sure it is at ... If you let be your base unit then it is really easy to mark it on.

Or do you just mean there is no given ratio between 1 and in terms of decimals?

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]

if a and b are distinct real numbers then it is not true that for every e>0 |a-b|<e.

The two different a and b are both < e.

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

 

[Zurtex]

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.

What are you on about?

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

 

[matt grime]

The two different a and b are both < e.

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

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]

Oh, no, you're going to talk about x and model(x) again aren't you?

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]

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.

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 it is not true that for every e>0 |a-b|<e

If a and b are distinct real numbers then for any e > |a-b| = d > 0.

 

[Organic]

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.

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]

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.

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]

let e=d>2,

This in not the case because e>d always.

 

[matt grime]

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

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]

This in not the case because e>d always.

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?

 

[Organic]

you are thinking unmathematically

There is no objective thing like mathematics that we can compare our way of thoughts to it.

Math language is only a rigorous agreement between people, no more no less.

I have found that the current agreement includes lot of weak point in it, where one of them is the infinity concept.

 

[Organic]

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

If a and b are distinct real numbers then for any e > |a-b| = d > 0.

proof:

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

Let d be the difference.

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

therefore non-zero/2 > 0.

 

[matt grime]

No, the idea of infinity is well understood, but apparently not by you. Neither, it seems, is the idea of axioms and definition.

All of the 'problems' you've come across have been because of your own refusal to accept the definitions that are there. (Cantor, Natural numbers, axiom of infinity, convergence, real numbers).


There are some deep and troubling issues in mathematics that we don't understand and have to live with. They cause no practical problems. Your findings aren't these, though. If you want to say things like 'there is no objective thing like maths' at least take the time to learn some of it, you might, well, learn something.

 

[Zurtex]

There is no objective thing like mathematics that we can compare our way of thoughts to it.

Math language is only a rigorous agreement between people, no more no less.

What you think of maths is not relevant to how maths works. Maybe you should go and argue about the philosophy of maths rather than fail to attempt to argue in maths...

 

[matt grime]

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

so you are saying that for any real positive number |a-b|<e Got it? That's what that quantifier for all means. Notice that you desired conclusion is part of the hypothesis. Note also that by the proof immediately preceding the first appearance of this claim, there are no pairs of real numbers satisftying both hypotheses, and thus the statement is vacuous ((AandB)=>A is true tautologically as well)

If a and b are distinct real numbers then for any e > |a-b| = d > 0.

What's th point of this line above? there is an 'if' , but no 'then'

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.

but that doesn't show anything other than your errors. In fact you said that e = d/2 wasn't allowed a couple of posts ago. And now it is? You've just shown d<d/2 ie d is negative. And you've proved d>0 allegedly after assuming d =|a-b| which was already greater than zero by assumption. So the best you#ve done is prove that if you assume X you can deduce X from that assumption. Howeever the logic contains so many errors that even that is in doubt. (X=>X is tautologically true again)

This makes no sense, Organic.

 

[Organic]

Behind any rigorous agreement there is a meaning.

During the time, people forgetting the meaning and using only the technical tools of the agreement.

When this is happens, it means that we are dealing with a dieing system.

The soul of Math is based on philosophy, is body is it’s the rigorous agreement.

We need both of them to keep Math alive, no more no less.

 

[Organic]

I used your original proof, also a corrected it:

If a and b are distinct real numbers then for any e > |a-b| = d > 0.

proof:

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

Let d be the difference.

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

therefore non-zero/2 > 0.

 

[matt grime]

As you evidently don't know what or where the body is (see the above expanded repudiation of your 'proof') how do you even know there is a soul?

 

[Organic]

I used your original proof, also I corrected it:

If a and b are distinct real numbers then for any e > |a-b| = d > 0.

proof:

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

Let d be the difference.

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

therefore non-zero/2 > 0.

 

[pig]

If a and b are distinct real numbers then for any e > |a-b| = d > 0.

i am having trouble understanding this, could you please rephrase it in a clearer way?

 

[matt grime]

But my proof that you thoughtfully corrected (ha!) wasn't of the statement you made. I showed that if, for all e>0 |a-b|<e then a=b.

The statement you've made above:

If a and b are distinct numbers then for all e>|a-b|=d>0

are you attempting to say that if e>|a-b| and a and b are distinct that e is greater than zero? But that is trivially true and doesn't require a proof, and doesn't even require that a and b are distinct.

As it stands your initial statement doesn't even have an obvious conclusion, it appears there is nothing to prove. It doesn't make sense.

It looks vaguely mathematical but there's nothing in what you've just written.

You cannot let e=d/2 since e is strictly greater than d, you even said I wasn't allowed to let e=d/2 in an earlier post.

 

[Organic]

Matt,

First let us write your rigorous proof:

Suppose a and b are real numbers and for any e>0 we know |a-b|<e then a=b.

proof: if a is not b then a=b is nonzero. let d be the difference let e = d/2 then |a-b|=d and |a-b| <d/2, contradiction, hence d is zero.

You clime that if for any e>0 we cen find |a-b|=d<e then a=b.

Your proof is:

1) a is not b

2) |a-b|=d>0

3) e=d/2, but since we know that |a-b|=d<e, then e=d/2 breaking our on rules, because |a-b|=d<e does not exist anymore.

Instead we have to use now |a-b|=d/2<e.

Shortly seaking, this part:

let e = d/2 then |a-b|=d and |a-b| <d/2, contradiction, hence d is zero.

breaking the rules, therefore this proof doesn't hold because e=<e.

 

[matt grime]

I will remove the 'then' that Organic objects to so that instead of being a statement in inverted commas (ie incorrect) it is a quote. It doesn't alter the meaning of the sentence though


"You clime that if for any e>0 we cen find |a-b|=d<e then a=b."

That is not what I claim.

a and b are given to you at the start. You do not "find" an a and b with |a-b|=d

Now d is a non-zero positive number,and thus so is d/2, so given the hypothesis that |a-b| <e for any e>0 it must be that it is true if I let e = d/2
thus d<d/2 which is impossible so my assumption that a is not equal to b is incorrect, hence a=b.

0.99999... and 1 satisfy the hypotheses of the lemma, hence they are equal.

Note it should be a-b is non zero not a=b is nonzero in the statement of the lemma

 

[Organic]

Matt don't add words that I did not use (for example: "then")!

Here it is again, and now give your answer step by step, according to what I write:

First let us write your rigorous proof:

Suppose a and b are real numbers and for any e>0 we know |a-b|<e then a=b.

proof: if a is not b then a=b is nonzero. let d be the difference let e = d/2 then |a-b|=d and |a-b| <d/2, contradiction, hence d is zero.

You clime that if for any e>0 we have |a-b|=d<e then a=b.

Your proof is:

1) a is not b

2) |a-b|=d>0

3) e=d/2, but since we know that |a-b|=d<e, then e=d/2 breaking our on rules, because |a-b|=d<e does not exist anymore.

Instead we have to use now |a-b|=d/2<e.

Shortly seaking, this part:

let e = d/2 then |a-b|=d and |a-b| <d/2, contradiction, hence d is zero.

breaking the rules, therefore this proof doesn't hold because e=<e.

The problem is in e=<e, therefore your proof does't hold water.

 

[pig]

if |a-b|=d>0, d/2 is still > 0 so e=d/2 doesn't break the rule that e>0.

the contradiction which appears is not the result of the fallacy of the proof, it is the whole point of the proof.

 

[matt grime]

Do you understand what proof be contradiction means? By the very fact that we are have the non-sensical assertion that d<d/2 (which is "allowed" by hypothesis on a and b) we have shown that the assumption that a does not equal b is incorrect, and thus a=b.

As I now see Pig has eloquently stated too.

 

[Organic]

But it breacks the rule that |a-b|=d<e, which means that e always MUST be greater than d.

When you write e=d/2, you break the rules.

 

[matt grime]

And this is the contradiction that shows the assumption that a=/=b is false, hence a=b as we were require to prove. We have two "rules" that by assumption must both be satisfied, yet this is nonsensical so it must be that our assumption is incorrect. That is what proof by contradiction does.

 

[pig]

i will try to restate this in a way you will understand it.

if |a-b| is smaller than ANY number larger than 0, then |a-b| cannot be larger than 0, or it would by definition "smaller than any number > 0" have to be smaller than itself.

if |a-b| < any number larger than 0, then

a!=b -> |a-b|>0 -> |a-b|<|a-b|

 

[Organic]

Matt,

What are you talking about?

You start with this statement |a-b|=d<e

Then you contradict your own statement by writing that e=d/2.

The result is e=<e, which is a contradiction.

Therefore e is logicaly meaningless and you can't use it to prove anything about |a-b|.

Edit: The result is e not= e, which is a contradiction.

 

[Organic]

|a-b| is smaller than ANY number larger than 0

|a-b| is a general notation to say that the gap between d and 0 is always d and not 0.

 

[matt grime]

I think the easiest way to point out that you haven't got a clue is to state

e<=e is not a contradiction.

any real number is less than or equal to itself.

I have a property that a and b satisfy, if a and b are distinct then I can show there is a strictly positive number d that satisfies d<d/2 (in fact d<d)

If you don't like me setting e=d/2 then how about let e be any non-zero positive number less than d, then by hypothesis |a-b|<e and simultaneously e<|a-b| contradiction, hence a=b.

You do see the word contradiction don't you? The only rule the e must satisfy is that it is greater than zero. It is not that e satisfies the rule that it must be greater then |a-b| but that |a-b| must be less than e, that is subtly different, this is not a result about e, which is allowed to be any positive real number, but about a and b.

Look at where the quantifiers come in the construction of the proposition

 

[Organic]

let e be any non-zero positive number less than d

It does not change anything because, when you say |a-b|=d < e then e is alway greater than d.

Threfore:

"let e be any non-zero positive number less than d" --> e < d

"|a-b|=d < e" AND "e < d" --> false, and cannot be used in your proof.

I made a miskate in my pervious post, because by your proof we get:

e not= e which is a contradiction, therefore e is meaningless.

 

[matt grime]

No, no, no, no. Look at which points the quantifiers and the their referents occur.

let a and b be two real numbers. Suppose for all e>o that |a-b|<e, then this implies a=b.

IF |a-b| is not zero then we have a contradiction, because |a-b|= d is a real number >0, thus a and be cannot satisfy the hypothesis since no real number satisfies d<d. Is that better for you? I've not said e=d/2 or anything else that can't be true. Thus we can safely say the proposition is true.

 

[pig]

this is my last attempt. i think you are confused with what the number e represents. let's try this way:

1. let a and b be real numbers.
2. let S be a set of all real numbers > 0.
3. let |a-b| < all members of that set.

if a!=b, then |a-b|>0 therefore |a-b| is a member of S* and therefore |a-b|<|a-b|*. this is impossible.

if a=b, |a-b|=0 therefore |a-b| is not a member of S*.

notice that the contradiction doesn't result from wrong reasoning, but from the impossibility of a!=b if the first 3 statements are true.

 

[Organic]

let a and b be two real numbers. Suppose for all e>o that |a-b|<e, then this implies a=b.

Yes,yes,yes,yes.

I am talking about abs(a-b)=d therefore d is always a positive value greater than 0.

When you say: abs(a-b)=d < e, it means that no matter how d is small, e is always bigger than d.

Threfore:

"let e be any non-zero positive number less than d" --> e < d

"|a-b|=d < e" AND "e < d" --> false, and cannot be used in your proof.

e not= e which is a contradiction, therefore e is meaningless.

 

[matt grime]

I'm afraid saying things like 'all' when the set is infinite is only likely to provoke Organic to even greater heights of crankiness.

 

[Organic]

let |a-b| < all members of that set

It means that S is not complete, no more no less.

Shortly speaking, statement 2 is meaninless.

 

[matt grime]

So you agree the proposition is true, but that the proof is not correct? Well what's your proof then? (I can think of at least two more) But the problem isn't the result it's the idea of proof by contradiction, isn't it?

I know that the "and" is false ok, so on of the 'inputs' is false in some way agreed the only way for that to happen is if d=0 which leads us to the conclusion that a=b.


Suppose I prove sqrt(2) is irrational, by saying suppose it's p/q p and q both not even... and then get a contradiction thus sqrt(2) is not rational. But the first step was to assume that it's rational, therefore as it's irrational my proof can't be valid because I've got two mutually exclusive things happening!

This is how contradiction works. I am not saying e must be greater than d, but that if the hypotheses are true that we get a contradiction. Anyway, I've rewritten the proof to omit entirely the mention of e a couple of posts back does that ease your worried mind?

 

[Organic]

if the hypotheses are true that we get a contradiction

Why?

1) a not= b

2) abs(a-b)=d < e > 0

So, where is the proof by contradiction?

 

[matt grime]

If a and b satisfied |a-b|<e for all e>0 then it impossible for a not be equal to b since if a were not eqaul to b then |a-b| is some positive number and we would have to have BY HYPOTHESIS that |a-b|<|a-b| which is impossible hence if the hypotheses are true it implies that a=b (as we've seen that the assumption otherwise invalidates the hypothesis ***A CONTRADICTION***).

 

[Organic]

and we would have to have BY HYPOTHESIS that |a-b|<|a-b|

You never get to this HYPOTHESIS because:

1) a not= b

2) abs(a-b)=d < e > 0

e cannot be both litte than and greater than d, and you have no hypothesis that shows the contradiction that we get from |a-b|<|a-b|.