Please show us how the limit concept is rigorous

[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.

 

[pig]

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?

wait a minute. you say it is smaller than any number except 0, but is "never reaching 0". what is it then? santa claus? i am wasting my time here.

 

[matt grime]

but d isn't approaching anything because d is fixed, organic.

 

[Organic]

Matt,

You are looking on your impossible d from the wrong side.

Look at d from 0 side and see how d always > 0, because there are infinitely many of them between any given d and 0.

Shortly speaking, if d=0 then we have the smallest d, and as we know, this collection of infenitely many elements > 0 has no smallest element.

 

[pig]

please demonstrate this on the example of a=2, b=2. what is there between d and 0?

 

[matt grime]

that could be a goody...

d is |a-b| and a and b are fixed not mention the fact thta card((0,d)) isn't important in the slightest, really (not that we expect you to realize that and not that you even know what card means either).

 

[Organic]

please demonstrate this on the example of a=2, b=2

a not= b, this is the first rule.

 

[pig]

a not= b, this is the first rule.

sorry, i didn't realize we were still stuck there, remind me what exactly are you trying to prove? just please be clear and exact.

i mean, i now have a feeling you're trying to prove that if d>0, then d cannot be 0, which is self-evident.

 

[Hurkyl]

Since we've left the domain of mathematics, we're also leaving the domain of the mathematics forum.

 

[matt grime]

were we (well organic) ever in the realm of mathematics?

 

[Organic]

Matt,

not that we expect you to realize that and not that you even know what card means either

This is the whole idea.

a not= b --> 1=|{d}|

a = b --> 0=|{}|

So as you see to reach 0, d has a phase transition form 1(exists) to 0(does not exist).

By this phase transition we are breaking the rules by eliminating d existence.

And why we are breaking the rules? because our case is the Math universe where a not= b.

When a=b we made a phase transition to another Math universe where d does not exist, therefore cannot be used in our proof as the impossible d<d.

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?

1) a not= b

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

d is always smaller than e where e>0, and we never comparing d to itself but only to e.

Therefore we can never conclude that d<d.

 

[matt grime]

crank the crackpot index by a couple more points.

 

[pig]

Therefore we can never conclude that d<d.

we conclude that a=b, because otherwise d would be < d which is impossible. what do you find wrong with my statement that you quoted? i am trying to avoid mentioning "e" because you clearly do not understand what it represents from the very beginning.

 

[Organic]

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?

1) a not= b

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

d is always smaller than e where e>0, and we never comparing d to itself but only to e.

Therefore we can never conclude that the impossibility of d<d --> a=b.

 

[matt grime]

But that isn't what we conlcude at all

 

[Organic]

Therefore we can never conclude that the impossibility of d<d --> a=b.