On limits

Organic

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

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

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)

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


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

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 does'nt 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:
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:
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:
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:
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.