Discussion of 'Valid method of proof'.

[LAVRANOS]

definitely you can assume the premis and arrive at a truth and then consider your theorem truth only if the way from the premises to the truth result is connected by double implications.
see 2 examples in the attached file

 

[dx]

In fact, you can prove anything by assuming a false statement. This is also the reason behind Godel's theorem which says that there is no theorem deducible from a set of axioms that can prove the consistency of the axioms. The fact that a particular theorem T can be deducible from the axioms cannot tell us anything about the consistency of the axioms because if the axioms were inconsistent then T could certainly be deduced from them.

 

[LAVRANOS]

< originally posted by dx In fact,you can prove anything by assuming a false statement> Can you be so kind as to justify your nabeve doctrine by a couple of examples or example i.e assume a false statement and then prove a theorem.

 

[matt grime]

But that is very easy - I don't know what you think is 'naive' (assuming nabeve was supposed to mean naive) about it..

Suppose that 1=-1, then (1)^2=(-1)^2, or 1=1. So from a false presumption I have deduced a true statement. Although I don't really see the point of teaching logic to beginning mathematicians, it is one of the first things that they teach in a course on propositional logic: false implies true is true.

 

[LAVRANOS]

I ment above and not naive .Is 1=1 atheorem? I asked you from a false statement to prove atheorem because you said and I quote ((In fact,you can prove anything by assuming afalse statement)) As to how much I know logic and particularly how logic is involved in aproof I CHALENGE you here to produce a proof of any theorem and then explisitly mention the laws of logic involved apart from anything else Any way if 1=-1 then 1=1 is a theorem congratulations you have invented new maqthematics .For your information what you said is used in proving a theorem by contradiction.

 

[LAVRANOS]

dx and mattgrime I am waiting for your answer the chalenge is still there I WILL BE GLAD to compet with you in anything and everything involving mathematical proof and proof in general

 

[matt grime]

Define theorem, please. That you dismiss 1=1 as not a theorem is just one if taste. OK, it's not deserving of the title, really, but that doesn't make it less of an example.

 

[matt grime]

For what it's worth, several erroneous proofs of FLT were given on the assumption that all rings of integers in number fields are UFDs.

NB - the proofs may have had other errors too, but I'm sure that one solid proof was given (modulo the false premise), even if only for some n.

 

[dx]

Is 1=1 atheorem?

Yes, it is a theorem.

I asked you from a false statement to prove atheorem because you said and I quote ((In fact,you can prove anything by assuming afalse statement))

I don't think you understand what a theorem is. A theorem is any statement that can be logically deduced from a set of axioms.
Lets say A is a theorem. Now if you assume that its negation A' is true, you can prove any statement. Since $$A \implies A + B$$,

$$A'A \implies A'(A + B) = A'A + A'B \implies B$$.

So by assuming a contradiction we have proved an arbitrary proposition B.

For your information what you said is used in proving a theorem by contradiction.

No, its not. In proof by contradiction, an assumption is shown to be false by deducing a proposition that you know to be false from it. In matt grime's example, a true statement was derived from a false statement.

 

[matt grime]

Oh, and what's all this nonsense about a competition? There's no competition at all, nor challenge. I'm sure you're very good at maths, but there's no need to flex those muscles and to posture about it. It is simply the case that (F=>T) is T, i.e. false implies true is true.

 

[LAVRANOS]

To extent the catastrophic implications of the said doctrine.( you can prove anything by assuming a false st/ment) follow the next reasoning: 1=-1 then 1=-1or(ANY THEOREM) this we can do by using a law which is called addition introduction the above is logicaly equivalent to if 1=/-1 then ANY THEOREM BUT 1=/-1(1is not equal to -1) hense by using M. ponens we get ANY THEOREM Therefor using the above reasonig we can prove any theorem at least in the field of real Nos.Now coming to your answer dx: Ist of all if 1=1 is a theorem what is( for all x,x=x)?? in which if you substitute for x=1 you get 1=1.For information again this is an axiom in equality and hense not provable and hense not a THEOREEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEM I will congratulate you should you by using ANY SET OF AXIOMS WHATSOEVER PROVE 1=1 since you said and I quote :A theorem is any statement that can be logicaly deduced from a set of axioms: A S for the rest of your writings i will let you think them over again and i will not ask you what a logical deduction is etc etc BUT remember the chalenge is still there particularly for mattgrim.YOU must also learn ,and this is for mattgrim ,not to call your opponent names like (begining mathematician) you must always criticize of what he knows and not who he is. And to make the chalenge better let us first define what is proof a thing that should had been done a long time ago in that forum to clear a lot of mistakes occurring in some proofs and get away with a lot of stupid arguments.

 

[LAVRANOS]

Well did you not chalenge me by calling me (beginning mathematician) The chalenge is still there unless you apology .Who said i am agood mathematician,but i happen to know afew things about logic that i do not throw arround to cause impressions to other people that are ignorant of Also you must realize that titles do not imply straight thinking.In my life i have come across professors and doctors so stupid that made me ready to vomet many times .In ancient Athens every body used to go to a place called the MARKED and there in the open discuss everythimg known in the planet then.That is where civilazation was born.For your information the GREAT Aristoteles never earned any titles for hemself but he was known for what he wrote and said the following is the heart of the two valued logic that we use in maths and our ordinary life ((ει γαρ αληθες ειπειν οτι λευκον η οτι ου λευκον εστιν αναγκη ειναι λευκον η ου λευκον,και ει εστιν λευκον η ου λευκον,αληθες ην φαναι η αποφαναι.Και ει μη υπαρχει,ψευδεται,και η ψευδεται,ουχ υπαρχει.ωστε αναγκη η την καταφασιν η την αποφασιν αληθη ειναι η ψευδη)) Αριστοτελης Περι Επμηνειας,ΙΧ.18β,1-6 Υes F impies T IS TRUE Tommorow we will talk about contradiction.

 

[matt grime]

Well did you not chalenge me by calling me (beginning mathematician)

No, I didn't. I said I don't particularly like the teaching of logic to beginning mathematicians; I was thinking of a particular course I have dealt with for first year undergraduates where precisely this idea, (F=>T is T) causes some head scratching. I'm sorry if you took that aside as a personal slight.

I don't see your point: you admit (F=>T) is T, which is all that either of the two people you seem to wish to "challenge" have said. That you don't like calling the statement 1=1 a theorem is just one of taste. It was not chosen because it was all one can prove, but for the fact it was undeniably true. Do you want something more complicated? How about I prove the theorem that all finite groups have a 1-dimensional representation?

Assume all groups are abelian.

Since the sum of the squares of the simple chars of |G| equals |G|, and there are |G| conjugacy classes of an abelian group, then all reps of G are 1-dimensional, and hence at least one is 1-dimensional.

Now, why is that catastrophic? It is silly, but since the premise was false it is not important.

 

[matt grime]

Oh, and Lame's proof of Fermat's Last theorem still remains the example I'd have in mind of this kind of thing. He assumed that rings of integers were UFDs and proved FLT. They aren't UFDs, but FLT is still a theorem.

 

[dx]

Now coming to your answer dx: Ist of all if 1=1 is a theorem what is( for all x,x=x)?? in which if you substitute for x=1 you get 1=1.For information again this is an axiom in equality and hense not provable and hense not a THEOREEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEM I will congratulate you should you by using ANY SET OF AXIOMS WHATSOEVER PROVE 1=1 since you said and I quote :A theorem is any statement that can be logicaly deduced from a set of axioms:

Ok, I will prove that 1 = 1 from the Peano axioms for the natural numbers. The relevant axioms are Peano's first, fifth and sixth axioms.

1. For every natural number x, x = x. That is, equality is reflexive.
5. 0 is a natural number.
6. For every natural number n, S(n) is a natural number. (S(n) is the successor of n)

It follows from (5) and (6) that 1 = S(0) is a natural number. Since 1 is a natural number, it follows from (1) that 1 = 1. Therefore, 1 = 1 is a theorem of the Peano system.

 

[HallsofIvy]

< originally posted by dx In fact,you can prove anything by assuming a false statement> Can you be so kind as to justify your nabeve doctrine by a couple of examples or example i.e assume a false statement and then prove a theorem.

I ment above and not naive .

Okay, if you didn't mean "naive", what in the world is a "nabeve doctrine"?

 

[LAVRANOS]

First of all we must decide "what is proof" and give a working definition about it. Because a proof that is right for you it might be wrong for me and the opposite.
Coming now to your "proof" that 1=1 from the Peano's axioms, to me this is not a proof.
Or at least is a wrong proof.
Because when you put 1=S(0) that is mere definition of 1 and nothing else. Again $$\forall$$x (x=x) it is an equality axiom that you will find nearly in every mathematical system and not only in the Peano axioms.
Also the symmetric and transitive properties are axioms concerning equality. And again you will find in any mathematical system that uses the equality predicate.
But to help you i will prove for you that 1=1 is a theorem.
We have $$\forall$$x (x=x)
Now using the law of logic called universal elimination we can say for x=1 then 1=1.
Since we use a general axiom and a law of logic you might say that 1=1 is a theorem.
That law of identity emanates from the natural world surrounding us because the mountain will be itself for eternity.

 

[LAVRANOS]

Okay, if you didn't mean "naive", what in the world is a "nabeve doctrine"?

above doctrine
Sorry for the mistake

 

[dx]

First of all we must decide "what is proof" and give a working definition about it. Because a proof that is right for you it might be wrong for me and the opposite.

No, a proof is a proof. It can't be right for me and wrong for you (unless you disagree with standard logic). If A is a set of axioms, and if you can deduce a proposition T from A by the rules of logic, then T is a theorem of A.

Because when you put 1=S(0) that is mere definition of 1 and nothing else.

Of course it is. If you don't have a definition of what 1 is how the hell can you prove a theorem about it?

Again $$\forall$$x (x=x) it is an equality axiom that you will find nearly in every mathematical system and not only in the Peano axioms.
Also the symmetric and transitive properties are axioms concerning equality. And again you will find in any mathematical system that uses the equality predicate.

Which has nothing whatsoever to do with my proof of 1 = 1. I'm not concerned with any other mathematical system other than the natural numbers. All I did was define the natural numbers by the peano postulates, and then prove that in that system 1 = 1.

But to help you i will prove for you that 1=1 is a theorem.
We have $$\forall$$x (x=x)
Now using the law of logic called universal elimination we can say for x=1 then 1=1.
Since we use a general axiom and a law of logic you might say that 1=1 is a theorem.

Now you claim that 1 = 1 is a theorem? Didn't you say repeatedly that "1=1 is not a THEOREEEEEEEEEEEEEEEEEEEEEEEEEEEEEEM!". Also your proof is wrong. When you say $$\forall{x}, x = x$$ you must say what x is. The correct axiom is "for all natural numbers x, x = x". Then you must show that 1 is a natural number, or take it as an axiom.

That law of identity emanates from the natural world surrounding us because the mountain will be itself for eternity.

Not necessarily. Mountains can be destroyed. In the distant future, the Earth will be swallowed by the sun, and the mountain will no longer be a mountain.

 

[LAVRANOS]

Yes but even destroyed will always be itself...It will be a destroyed mountain

 

[uman]

That law of identity emanates from the natural world surrounding us because the mountain will be itself for eternity.

Classic.

 

[LAVRANOS]

dx said and i quote:A proof is a proof: No. A proof is something . In a proof we have the involvment of two things ,knowledge and the laws of logic.Knowledge is the axioms the theorems the definitions. WE ALL SAY THAT WE ALL KNOW THAT.But CAN ANYONE in the forum write down a simple proof and then point out the laws of logic involved,the theorems ,the axioms or the definitions.

Let us start with two simple examples from high school
1) IXI<Y <====> -Y<X<Y where x,y belong to the real Nos
2) (-1).x = -x xεR By the way this theorem might be the answer to the thread or section ((how can you multiply two negative Nos))

Remember the proof must be in steps every step must be justify
Let us now take two examples from soft core analysis
3) Prove that the emty set Φ is both open and closed
4)The f:(0,1]---->R where f(x)=1/x is not uniformly continuous over (0,1]
And an example from hard core analysis
Let (S,Ds) (T,Dt) be two metric spaces where Ds,Dt denote the respective metrics
Let A be a subset of s
Let f:A---->T be afunction from A to T
Let p be an accumulation point of A
Let bεT
Then limf(x)=b as x----> to p if and only if limf(Xn)=b as n---> infinity,for every sequence {Xn} of points in A-{p} which converges to p

 

[LAVRANOS]

Now let us see how dx's doctrine(( In fact,you can prove anything by assuming a false statement)) can be used in contradiction By the way dx can you find me a book or a page in the INTERNET where 1=1 is proved as theorem?
Now we must be very careful to distinguish between an implication and a logical implication. We say P implies Q noted as P---->Q and this can be true or false depending on the values of p and q. And if p is false and q true then P--->Q is true A thing that mattgrime so much insisted on
And we say that P logicaly implies Q denoted by P===>Q(DOUBLE ARROW) ,If the implication P---->Q IS ALWAYS TRUE no matter what are the values of p and q
Now let us see how contradiction works
Suppose we want to prove P===>Q BY the power invested on us by the rule in logic called conditional we can assume P and if prove Q we can say we have proved P===>Q
So let P

NOW it hapens some time that we don't know how to arrive at Q
hence we reason by contradiction,hence we assume
notQ
THEN along down the steps of proof we come across two statement which are contradictory i.e R and notR A statement which is always false
And here now the doctrine (false LOGICALY IMPLIES everything) can be used so we can get Q But why? AND HOW ?
Let us see why:from R AND notR we can get notR (The law is called additio elimination)
Now from notR we can get (notR or Q) using the law addition introduction.
But (notR or Q) is logicaly equivalent to ( R--->Q) The law is called material implication
But from( R and notR) we can get R.
Hence now from (R--->Q)and R by using the law called M.Ponens we get Q

HENCE P===>Q

Note T is logicaly equivalent to S Iff they logicaly imply each other

 

[Vid]

I hope no one is going to suffer through that incoherent wall to bait the troll with a meaningful reply.

 

[matt grime]

I am going to post one thing to him/her.

Lavranos it is considered *by definition* that the axioms of an axiomatic system are theorems of that system. Sorry you don't agree with the convention - that is not our problem, though. And 1=1 I say for, what the fourth time, was chosen for simplicity. You have ignored this point, and the other points made about what constitutes a theorem repeatedly. You asked for a real "theorem" that was proven from a false premise - surely FLT is a very good example, and a point that you repeatedly ignore.

The convention, by the way, is quite justified. For example in ZFC would you say that Zorn's Lemma as a theorem (or lemma)? It is equivalent to one of the axioms...

Finally, I've think I've guessed where the confusion lies. You believe that when we say (X=>Y) is true if X is false, that this means we have a proof of Y. This is not what we're saying. We're saying that we have a proof that X=>Y: a series of correct logically justified steps. The steps themselves, if I'm allowed one pathetic fallacy, do not 'care' if X is true or false.

I can only presume that is the problem, since you readily seem to accept that "F=>T is T".

 

[Diffy]

dx said and i quote:A proof is a proof: No.

But you said $$\forall$$ x (x=x) So therefore, a proof is a proof! :-)

 

[LAVRANOS]

Do you mean that from now on the axiom in real Nos (For all x and y x+y=y+x) we should call it a THEOREM?
when you have let's say two sta/nts P and Q which are logicaly equivalent (P===>Q),if you start from P to prove Q P is called axiom and Q theorem,and since are equivalent when we start from Q to prove P Q is called axiom and P THEOREM
Now the axiom of choice,the Well ordering theorem,Zorn's Lemma are all equivalent and depending where you start from you call it an axiom and the rest thoerms

 

[matt grime]

Yes, for the umpteenth time it is conventionally taken that an axiom is *trivially* a theorem - it can be deduced from itself since (X=>X) is always true. If you don't like that then you're just arguing against the convention that the mathematical community has adopted, which is going to be a fruitless endeavour.

And the point is irrelevant anyway since I've given you plenty of other examples of showing how to deduce a (true) non-vacuous statement from a false one. That I've been able to do so doesn't prove that this non-vacuous statement is true, obviously, which is what you still appear to think we're telling you.

 

[LAVRANOS]

A THING that implies itself ((X==>X)<===>(X OR notX)) IT does not necesseraly mean that it looses its name
Now if the mathematical community decides to call the axioms theorems ,and since X IMPLIES X and not Y the thoerems axioms i will be the last person on Earth to say otherwise
But inform the mathematical community to change the books too
At this point please let me take arest because i have given the Forum couple of problems and since nobody has responded yet i must at least give a sample solution

 

[LAVRANOS]

Let me now prove IxI<y<===> --y<x<y the way i said
NOTE (L) will mean law of logic ,(T) A theorem,(D) a definition,(A) an axiom (x>=y
) x> or =toy
So
1) IxI<y an assumption
2) for all x x>=0 or x< 0 (A)
3) x>=0 or x<0 from line2 and using Universal elimination (L)
4) x>=0 A hypothesis
5) for all x x>=0 ------> IxI=x (D) in apsolute values
6) x>=0-------> IxI=x from 5 and Univ. elimin. (L)
7) IxI=x from 4 and 6 and using M.Ponens (L)
8) x<y By substituting 7 into 1 (L)
9) for all a,b,c a<=b and b<c -----> a<c (T) or a result coming out from iniqalities
10) 0<=x and x<y-----> 0<y from 9 and Univ.Elim.where we put a=0,b=x,c=y
11) 0<=x and x<y from 4 and 8 and using Conjuction Introduction (L)
12) 0<y from 10 and 11 and using M.Ponens (L)
13) for all a<b<----->-a>-b (T)
14) 0<y<-----> 0>-y from 13 and Univ. Elim. where a=0,b=y
15) -y<0 from 12 and 14 and using M.Ponens
16) for all a<b and b<=c------> a<c (T)
17) -y<0 and 0<=x------> -y<x from 16 and Univ. Elim.where we put a=-y,b=0,c=x
18) -y<0 and 0<=x from15 and 4 and using conjuction introduction (L)
19 -y<x from 17 and 18 and using M.Ponens
20) -y<x and x<y ( -y<x<y) from 8 and 19 and conjuction introduction (L)
21) x>=0-----> -y<x<y from steps 4 to 21 and using the rule of conditional proof (L)
Now in a similar way and using the definition of apsolute values x<0----> IxI=-x we will come to the result x<0-----> -y<x<y ( 1a)
22) -y<x<y from 2, (1a),21 and Disjuction Elimination (L)
23) IXI<y-----> -y<x<y from steps 1 to 23 and using the rule of conditinal proof
the converse i,e -y<x<y----> IxI<y can be done in a similar way

 

[LAVRANOS]

The laws of logic used in this proof are:
1) M.Ponens ( P--->Q and P)====>Q (Where a single arrow means an implication and the double one a LOGICAL IMPLICATION)
2) Conjuction introduction P,Q =====> P and Q
3)Disjuction Elimination ( P---->S and Q---->S and P or Q)=====>S
4)The law
of Universal Elimination where if aproperty holds for a set it will hold for one of its members
5)The law of substitution
6)The law of conditional proof in some books is called the deduction theorem an it is considered some times as ametatheorem
The theorems the definition and the Trichotomy Law are well known to every one they are high school stuff

 

[LAVRANOS]

This type of proof leaves no room for imagining things in mathematics or arguing for ever and ever or a teacher in every level to pass wrong proofs to the students because he likes to.For people involved in hard core analysis where a lot of quantification is involved that machinery can solve most of your problems. Of course it takes a little pactise but it is worth while

 

[Crosson]

It is simply the case that (F=>T) is T, i.e. false implies true is true.

Hi Matt, I have a subtle disagreement on using the explanation (F=>T) is T involving the material conditional =>. The statement that theorem A can be deduced from premises S is written as $S\vdash A$, or in some other way than a material conditional. In other words, saying that A is a consequence of B is not the same as saying A implies B.

Here is the proof that with one false statement as premise one can derive any conclusion.

Mc = I'm posting from the moon. (This is false)
(x) ~Mx = No one can post from the moon. (A true fact)
B = anything you want!

1p. Mc
2p. (x) ~Mx
---------
2. Mc v B 1, (valid logical addition, if Mc is true then Mc v B is true)
3. ~Mc 2, (universal instantiation of a specific fact)
4. B 2,3 Disjunctive syllogism

 

[LAVRANOS]

And now i will quote couple of proofs from different books and if they are right or wrong how can we establish that
So
proof No 1
Let a' mean a to the square
IxI<y<===> IxI'<y' <===> x'-y'<0 <===> (x-y).(x+y)<0 <===> { x-y<0}
{x+y>0} or
{x-y>0}
{x+y<o}
If {x-y<0} {x<y}
{x+y>0} <===> {x>-y} <====> -y<x<y
If {x-y>0} {x>y}
{x+y<0} <====> {x<-y} not true



{x+y<0} <====>

 

[LAVRANOS]

I am sorry that did not come out all right let me try again