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On limits

by Organic
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matt grime
#73
Apr7-04, 10:25 AM
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Quote Quote by Organic
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
#74
Apr7-04, 10:30 AM
P: 1,210
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 itís the rigorous agreement.

We need both of them to keep Math alive, no more no less.
Organic
#75
Apr7-04, 10:32 AM
P: 1,210
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
#76
Apr7-04, 10:32 AM
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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
#77
Apr7-04, 10:38 AM
P: 1,210
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
#78
Apr7-04, 10:59 AM
P: 87
Quote Quote by Organic
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
#79
Apr7-04, 11:00 AM
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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
#80
Apr7-04, 11:25 AM
P: 1,210
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
#81
Apr7-04, 11:32 AM
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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
#82
Apr7-04, 11:41 AM
P: 1,210
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
#83
Apr7-04, 11:55 AM
P: 87
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
#84
Apr7-04, 11:58 AM
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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
#85
Apr7-04, 12:00 PM
P: 1,210
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
#86
Apr7-04, 12:04 PM
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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
#87
Apr7-04, 12:11 PM
P: 87
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
#88
Apr7-04, 12:15 PM
P: 1,210
Matt,

What are you talking about?

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

Then you contradict your own statment 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
#89
Apr7-04, 12:20 PM
P: 1,210
|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
#90
Apr7-04, 12:24 PM
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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 dont 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


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