give a rigorous proof

[Bignose]

The book that Bignose suggested it is difficult to get.

 

The book is available from Amazon, http://www.amazon.com/Elementary-Complex-Analysis-Dover-Mathematics/dp/0486689220/ref=sr_1_1?ie=UTF8&s=books&qid=1246730105&sr=8-1

 

and any good library (university or city) will have an inter-library loan desk where you can request almost anything. My local city asks only a nominal fee to cover the cost of inter-library loan ($1). $1 to get any book available anywhere is truly amazing.

 

----------

 

And, as posted above, the two proofs you have are proofs by contradiction.

 

Since it is already posted on the other forum, there's no point to not copy it here, too.

 

So, assume that there is NOT a unique zero, but they both obey the rules for a zero. Denote them [math]0_1[/math] and [math]0_2[/math].

 

They both obey the rules of a zero so:

 

by the axioms: for any a: [math]a+0_1 = a[/math] and [math]a+0_2 =a [/math]

 

So, start with [math]0_1[/math]

Now [math]0_1 = 0_1 + 0_2[/math] because we can add a zero to it without changing it by the axiom.

 

Now, we can change the order of addition by another of the axioms, so

[math]0_1 + 0_2 = 0_2 + 0_1[/math]

 

Now, we manipulate that RHS again by the axiom involving the zero.

[math]0_2 + 0_1 = 0_2[/math]

 

Putting it all together:

[math]0_1 = 0_1 + 0_2 = 0_2 + 0_1 = 0_2 [/math]

or taking out the middle stuff since they are all equal

[math]0_1 = 0_2 [/math]

 

The statement [math]0_1 = 0_2 [/math] CONTRADICTS the assumption that we made when we started, namely that [math]0_1[/math] and [math]0_2[/math] were not the same, because we derived that they are in fact equal and therefore the same. Therefore, the opposite of the assumption is true; specifically it is false to say that there is not a unique zero; or removing the double negative, that there is a unique zero.

 

Proof by contradiction.

[triclino]

A direct proof for the uniqness of zero consists of two parts:

 

 

.....................................PART A.....................................

 

In part A we prove the existence of zero:

 

From the axiom : for all x ( x + 0 =x) ,we simply say ,there exists a y for all x such that x + y = x .

 

So far we have proved the existence of an identity element

 

 

.......................................PART B...........................................

 

In part B we prove the uniqness of y

 

WE SAY lets bring in ANY other element z and see what happens :

 

For those two elements we have the same axiom holding i.e

 

for all x : x + y = x........................................................1

 

for all x : x + z = x...........................................................2

 

 

Now in (1) we put x=z and in (2) we put x= y and the result is;

 

z + y = z and y +z = y and since z + y = y + z ,we have y=z

 

 

 

THERE IS NOTHING CONTRADICTORY SO FAR.

 

TO SAY THAT:

 

ASSUMING THAT THERE DOES NOT exists one element ,but there exist two is WRONG.

 

HOW do you know ,that if there does not exist one bus in a town there exist two and not a million

 

 

In a contradictory proof you have to negate both parts A and B.

 

SO you can say:

 

O.K , LETS assume that:

 

The statement,that there exists a unique y such that:

 

for all x ( x + y =x ) it is not true what happens ??

 

 

 

 

1) for all y there exists an x such that ( x + y = x).This the negation of PART A

 

 

 

 

......................................OR.............................................

 

 

 

2) there exist a y and there exist a z such that : for all x ( x + y =x) and for all x ( x + z = x) BUT ALSO [math] y\neq z[/math] .This is the negation of PART B.

 

 

And in examining both cases we must end up with a contradiction

[timo]

triclino:

1) Existence is one of your axioms. Axioms are usually (read: always) not proven. You might have meant that with your part A (I see no steps other than noticing it is an axiom) so that's perhaps just semantics.

 

2) To disprove x and y ("and" and "or" being meant as the boolean operations in this point) it is sufficient to disprove x or y ( not( x and y) = (not x) or (not y) ). It is not necessary to disprove x and to disprove y. According to mainstream logic, at least.

 

3) What are you complaining about? You were asking for proofs that you were given quite direct hints how to construct them (see my previous post, for example) and you finally even got the full proof presented. Do you complain about the proof not being a proof or about labeling it as "proof by contradiction"? Note: This is a real question, it is not really clear to me. In the former case: Try being more clear. In the latter case: I see how one could interpret it not being a proof by contradiction (directly showing that all zeros are the same) but I also see how one could be interpreted as such (showing that there cannot be two distinct zeros). I don't think the issue is worth bitching about.

 

4) I am not sure if I understood what you meant with your bus example (I am not even sure if it is a syntactically correct sentence) but perhaps this helps: Assuming we have agreed that if there cannot be a 2nd zero and we ask ourselves if there can be N>2 distinct zeros. You can then just pick two of them and by the same construction show that they are equal which again contradicts the assumption that the N zeros were distinct. Not sure if that's what you meant, though.

 

5) Please do not use write in allcaps and use proper punctation. I do not see a sensible pattern in your capitalization and as a non-native speaker/reader and that really decreases readability.

Edited July 4, 2009 by timo
clarification for "x and y"

[Bignose]

Maybe it is a language issue; whatever it is, I'm done helping. I've done quite, quite enough. I've always had proofs of this kind referred to as "proofs by contradiction". Maybe you are taking that to mean something else, I don't know. I gave you hints, then I gave you a reference to look it up yourself, then I even provided the proof. There is no more help to give.

 

And, maybe it is still a language issue, but almost all your posts directed to me have seemed very rude, so don't expect much more help from me in the future, either.

[triclino]

triclino:

1) Existence is one of your axioms. Axioms are usually (read: always) not proven. You might have meant that with your part A (I see no steps other than noticing it is an axiom) so that's perhaps just semantics.

 

2) To disprove x and y ("and" and "or" being meant as the boolean operations in this point) it is sufficient to disprove x or y ( not( x and y) = (not x) or (not y) ). It is not necessary to disprove x and y. According to mainstream logic, at least.

 

3) What are you complaining about? You were asking for proofs that you were given quite direct hints how to construct them (see my previous post, for example) and you finally even got the full proof presented. Do you complain about the proof not being a proof or about labeling it as "proof by contradiction"? Note: This is a real question, it is not really clear to me. In the former case: Try being more clear. In the latter case: I see how one could interpret it not being a proof by contradiction (directly showing that all zeros are the same) but I also see how one could be interpreted as such (showing that there cannot be two distinct zeros). I don't think the issue is worth bitching about.

 

4) I am not sure if I understood what you meant with your bus example (I am not even sure if it is a syntactically correct sentence) but perhaps this helps: Assuming we have agreed that if there cannot be a 2nd zero and we ask ourselves if there can be N>2 distinct zeros. You can then just pick two of them and by the same construction show that they are equal which again contradicts the assumption that the N zeros were distinct. Not sure if that's what you meant, though.

 

5) Please do not use write in allcaps and use proper punctation. I do not see a sensible pattern in your capitalization and as a non-native speaker/reader and that really decreases readability.

 

me no understand

 

( ANGELO MARGARIS ,first order mathematical logic,page109)

 

By the way,what is the difference between a syntactical and a semantical proof??

 

Why, do you have a secondarystream logic ??

---

Merged post follows:

Consecutive posts merged

Maybe it is a language issue; whatever it is, I'm done helping. I've done quite, quite enough. I've always had proofs of this kind referred to as "proofs by contradiction". Maybe you are taking that to mean something else, I don't know. I gave you hints, then I gave you a reference to look it up yourself, then I even provided the proof. There is no more help to give.

 

Mention one book that labels the uniqness proof as a contradictory one

 

And, maybe it is still a language issue, but almost all your posts directed to me have seemed very rude, so don't expect much more help from me in the future, either.

 

Thats a clever way avoiding the whole issue

[Bignose]

Mention one book that labels the uniqness proof as a contradictory one

 

I gave you the reference (and two different ways to get a hold the book) above. The proof above is good enough for Shilov, one of the most renowned Russian mathematicians (this is a man who has published works with Kolomogorov and Gelfand, also very famous Russian mathematicians). If it is good enough for Shilov, it is good enough for me.

 

Considering that at the top of this thread you didn't even know what a proof by contradiction is, why are you being to belligerent that the proofs you have been given aren't proofs by contradiction? These two are quick and easy examples. The examples can get much more complicated, but what I don't understand is why you keep arguing about what a proof by contradiction is? Why can't you accept that these are indeed proofs by contradiction?!?

Edited July 5, 2009 by Bignose

[the tree]

Since the original problem has been solved in great detail, and with references, can the thread be closed? I rather feel that anything else wont contribute.

[triclino]

When we are in doubt for the nature of a proof ,whether it is contradictory or not ,one must use "the top gun" of logic-mathematics scientific investigation,the formal proof or otherwise called "the stepwise proof"

 

In a formal proof each step of the proof is justified and the laws of logic and axioms,theorem or the definitions involved in the proof ,are explicitly mentioned.

 

And very clearly one ,with no doubts at all , can decide whether the proof is contradictory or not.

 

Hence ,if the forum allows ,I am prepared to produce such a proof

in which the non contradictory nature of the uniqueness proof will triumphantly emerge.

 

In contrast, I am also prepared to produce a complete formal proof using contradiction

[mooeypoo]

Hence ,if the forum allows ,I am prepared to produce such a proof

in which the non contradictory nature of the uniqueness proof will triumphantly emerge.

 

In contrast, I am also prepared to produce a complete formal proof using contradiction

Sure, go ahead, produce it. It will at least give us the opportunity to finally analyze something where we're all on the same page.

[triclino]

Sure, go ahead, produce it. It will at least give us the opportunity to finally analyze something where we're all on the same page.

 

 

O.K ,thanks tomorrow i will produce the 1st of the proofs,because now is very late in the island of Corfu and i must get some sleep

---

Merged post follows:

Consecutive posts merged

 

Here is the direct formal proof for the uniqueness of zero.

 

But before we start the proof let:

 

(A) denote an axiom and (L) a law of logic.

 

 

1) for all ,x : x + 0 = x......................................................................(A)

 

and in quantifier form : [math]\forall x ( x + 0 =x)[/math]

 

 

2)there exists a ,y ,for all , x : x + y = x........................................from (1) and using Existential Introduction (L)

and in quantifier form: [math]\exists y\forall x ( x + y =x)[/math]

 

 

3)for all ,x : x + y = x and for all , x : x + z = x.........................................an assumption to start the conditional proof (L)

and in quantifier form : [math]\forall x ( x + y = x)[/math] and [math]\forall x ( x + z = x)[/math]

 

The above assumption tell us the following:

 

Let us assume that y and z are any two identity elements for which the identity axiom holds

 

 

 

4) for all ,x : x + y =x...................................................................from (3) and using addition elimination (L).Another name for addition elimination is Conjunction elimination.

and in quantifier form : [math]\forall x ( x + y = x)[/math]

 

 

5) z + y = z................................................................................from (4) and using Universal Elimination where we put x=z (L)

 

 

 

6) for all ,x : x + z = x...................................................................from (3) and using again addition elimination (L)

And in quantifier form : [math]\forall x ( x + z = x)[/math]

 

 

7) y + z = y ................................................................................from (6) and using again Universal Elimination where we put x= y (L)

 

 

 

 

8) for all,a for all , b : a + b = b + a....................................................(A)

and in quantifier form : [math]\forall a\forall b ( a + b = b + a )[/math]

 

 

 

 

9) z + y = y + z............................................................................from (8) and using Universal Elimination where we put a = z and b = y (L)

 

 

 

10) y + z = z ...............................................................................by substituting (9) into (5) (L)

 

 

11) y = z ....................................................................................by substituting (10) into (7) (L)

 

 

 

12) [math]\forall x ( x + y = x )[/math] and [math]\forall x ( x + z = x )[/math][math]\Longrightarrow y= z[/math].......................................from (3) to (11) and using the law of conditional proof (L)

 

 

 

13) [math]\forall y\forall z [ (\forall x ( x + y = x)\wedge\forall x ( x + z = x ))\Longrightarrow y = z][/math]..........................................................from (12) and using Universal Introduction for y and z (L)

 

 

This last statement tell us that:

 

for any two identity elements in the set for which the identity axiom applies ,then these two elements are identical

 

SO far we have proved the following:

 

 

[math]\exists y\forall x ( x + y = x )[/math] and [math]\forall y\forall z[(\forall x ( x + y = x )\wedge\forall x ( x + z = x))\Longrightarrow y= z ][/math] ,which is the definition of uniqueness of zero

 

 

As you very well know to have a proof by contradiction ,at least in one of the steps of the formal proof we must have the appearance of a contradictory statement of the form : Q AND NOT Q.

 

Certainly we do not have that in the above proof .

 

 

I may add also that a similar formal proof appears in ANGELO'S MARGARIS book ,named : FIRST ORDER MATHEMATICAL LOGIC ,page 121.

 

In the same book on page 109 ,an excellent definition of the uniqueness appears as well

Edited July 6, 2009 by triclino
Consecutive posts merged.

[Bignose]

As you very well know to have a proof by contradiction ,at least in one of the steps of the formal proof we must have the appearance of a contradictory statement of the form : Q AND NOT Q.

 

Certainly we do not have that in the above proof .

 

You certainly do have that in your proof.

 

You start with the assumption that [math]y \ne z [/math] and yet using the axioms you find that [math]y=z[/math], seems pretty contradictory to me.

 

Your proof is virtually identical to mine. You just used y's and z's instead of [math]0_1[/math] and [math]0_2[/math]. A change in nomenclature is just an aesthetic difference, nothing more nothing less.

 

In my proof, I started with assuming that [math]0_1 \ne 0_2 [/math] and using the same axioms found that [math]0_1 = 0_2 [/math]. Seems pretty contradictory to me.

 

So what again is the issue?

[triclino]

IN LINE (3) of the formal proof, the assumption is not to start a contradiction or otherwise called RAA,But to start a conditional proof.

 

DO you know the difference between ,the law of conditional proof and that of RAA (= Reductio ad Absurdum )

 

A book that explains quite well the difference between the two and gives examples is:

 

THEORY AND PROBLEMS OF LOGIC ,in Schaum's OUTLINE SERIES.

 

I repeat in a formal proof there are not any hidden formulas.

 

everything is explicitly mentioned

 

Your assumption is an imaginary one .

 

IF you think that you can write a formal proof with the assumption [math]y\neq z[/math] then do it .

[iNow]

Since the original problem has been solved in great detail, and with references, can the thread be closed? I rather feel that anything else wont contribute.

 

They should have listened to you sooner. This kid's a real ____ ... erm, piece of work.

[mooeypoo]

People. There is a useful button at the top of every post with a little exclamation mark on it, that is used to report a post or a thread. When you do that, it goes on to the moderator's attention so we can review why you think a certain thread should be closed (or not closed). These type of discussions aren't helping threads continue well.

 

The moderation team isn't everywhere all the time. Reporting a post gets us to notice it, and gives you the stage to explain why you think we should act a certain way, so we can enforce the rules if we must.

 

Seeing as Bignose and the OP are still having a discussion, I'm a bit reluctant to close the thread just yet.

 

triclino, you are urged to read our rules and start reading what others write to you. You seem to be assuming too much in their answers rather than reading about what they're trying to help you with.

 

Please stick to the discussion.

~moo