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- Thread starter LAVRANOS
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dx

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matt grime

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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.

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matt grime

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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.

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matt grime

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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.

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.

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dx

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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 [tex] A \implies A + B [/tex],

[tex] A'A \implies A'(A + B) = A'A + A'B \implies B [/tex].

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.

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matt grime

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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.

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matt grime

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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.

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matt grime

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dx

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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,

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HallsofIvy

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Okay, if you didn't mean "naive", what in the worldI ment above and not naive .

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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 [tex]\forall[/tex]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 [tex]\forall[/tex]x (x=x)

Now using the law of logic called

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.

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

Sorry for the mistake

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dx

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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 [tex]\forall[/tex]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 [tex]\forall[/tex]x (x=x)

Now using the law of logic calleduniversal eliminationwe 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 [tex] \forall{x}, x = x [/tex] 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.

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

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

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Let us start with two simple examples from highschool

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

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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 dont 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

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matt grime

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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".

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