Please show us how the limit concept is rigorous

[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|.

 

[matt grime]

Do you know what proof by contradiction means? We entertain a silly idea temporarily putting aside our intuition and show that there is a genuine contradiction.

What we do here is to prove A=>B is to show that not(B) => not(A)

we've now omitted the step you find objectionable entirely by proving that it is not possible.

So let us take what I had in my previous post: would have to have. Do you understand what that means?

We are showing that |a-b| not zero implies that (since |a-b| <|a-b| is impossible) that they hypothesis is impossible. That is what we want, because it shows the negation of the conclusion implies the negation of the hypotheses, thus the hypotheses imply the conclusion.

This is very simple logic and every time you post indicating you don't understand the proof you are weakening yet more your stance as being some ultimate arbiter of mathematical correctness.

I wonder, how long do you actually take to try and understand the answers you get?

 

[Organic]

|a-b| <|a-b|

1) a not= b

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

Please show how by (1) and (2) you can prove by contradiction that d = 0

 

[pig]

if d is smaller than any possible number > 0, then d cannot be > 0 because it would have to be smaller than itself. what is so hard for you to understand here?

 

[Organic]

if d is smaller than any number > 0, then d cannot be > 0 because it would have to be smaller than itself

No, abs(a-b)=d < e > 0 is smaller then any number accsept 0, because d approaching but never reaching 0.

I you can't understand that simple thing?

 

[matt grime]

Occam's razor: two seemingly eloquent expositors of mathematics say you are wrong, you insist you are right without being able to explain yourself coherently... hmm, wonder how that's going to work out.

We've omitted any mention of e, and e<d and all the things you found unacceptable so what else are we supposed to do?

We are saying that it is impossible for any real strictly positive number to be less than every other real strictly positive number, that's all.

It's not very hard, and it doesn't even mention e or d.