triclino Posted April 1, 2010 Posted April 1, 2010 I think every mathematical forum at the very beginning of its existence should produce a definition of the mathematical proof and then according to that definition every mathematical proof should be written down. Also a check process should be mention ,hence to avoid useless duals between the members of the forum whether a proof is correct or not. I think the above should be stipulated as the basic rules of each and every mathematical forum
ajb Posted April 2, 2010 Posted April 2, 2010 A proof is a series of logical arguments or steps from one collection of statements to another. The trouble is that the starting assumptions and what is assumed to be "common knowledge" will in general vary depending on the subject and the people reading the work. Thus, the details, length and emphasis of a proof will vary a lot.
triclino Posted April 2, 2010 Author Posted April 2, 2010 A proof is a series of logical arguments or steps from one collection of statements to another. Can you give an example to justify your definition? Hence to produce a proof consisting of "logical arguments" where those arguments can be isolated and examined for there validity in an indisputable and clear way?? By the way in logic we do not have "logical arguments" ,but valid or not valid arguments
ajb Posted April 2, 2010 Posted April 2, 2010 I mean logical very loosely here, not in relation to mathematical logic really. Your idea of a proof seems correct to me. Going from one starting statements/axioms etc to another statement "the theorem/axiom" etc. in steps that cannot be disputed. A simple example (can be found on Wikipedia in fact.) Theorem The sum of two even integers is itself even. (This is our statement we wish to prove, i.e. end up with this as an infallible truth.) Proof Let [math]x,y[/math] be two even integers. Since they are even, they can be written as [math]x=2a[/math] and [math] y=2b[/math] for integers [math]a[/math] and [math]b[/math]. (First question about rigour and standards of proof. Is the above statement obvious? Can I take it as "given" or should I prove it viz a lemma?) Then the sum [math]x + y = 2a + 2b = 2(a + b)[/math]. (This follows from the rules of real numbers, so I will take this as given.) From this it is clear [math]x+y[/math] has 2 as a factor and therefore is even, so the sum of any two even integers is even. (Given that the sum of integers is an integer.) QED So we have a series of steps from the statement that [math]x,y[/math] are even integers to the statement that the sum of [math]x,y[/math] is also an even integer. The question with proofs is "exactly how much do you need to prove"? This will be a practical stumbling block for any forum rules on what constitutes a proof. In journals etc. exactly what is required will vary from subject to subject.
triclino Posted April 5, 2010 Author Posted April 5, 2010 Why ,when x=2a and y=2b ,then x+y=2a+2b?.What is the axiom ,theorem or definition supporting that statement??. From how many arguments your proof consists of?? Can you isolate each argument??. For example can you say : Argument No 1 : state the argument Argument No 2: state the argument Argument No 3 : state the argument .............e.t.c........e.t.c Merged post follows: Consecutive posts mergedI mean logical very loosely here, not in relation to mathematical logic really. Your idea of a proof seems correct to me. Going from one starting statements/axioms etc to another statement "the theorem/axiom" etc. in steps that cannot be disputed. A simple example (can be found on Wikipedia in fact.) Theorem The sum of two even integers is itself even. (This is our statement we wish to prove, i.e. end up with this as an infallible truth.) Proof Let [math]x,y[/math] be two even integers. Since they are even, they can be written as [math]x=2a[/math] and [math] y=2b[/math] for integers [math]a[/math] and [math]b[/math]. (First question about rigour and standards of proof. Is the above statement obvious? Can I take it as "given" or should I prove it viz a lemma?) Yes this the right question ,why x=2a and y=2b ,what is the axiom ,theorem or definition supporting those statements. The answer here is the following: Definition: for all ,x : x is even => there exists integer ,a such that x=2a Or in quantifier notation : [math]\forall x[/math][ x is even[math]\Longrightarrow\exists a( a\in Z\wedge x=2a)][/math]. The final statement x=2a ,y=2b ,where a,b are integers is concluded in the following way: step 1 : Assume ,x,y to be even step 2 : bring in the above definition of even Nos: [math]\forall x[/math][ x is even[math]\Longrightarrow\exists a( a\in Z\wedge x=2a)][/math] step 3: What do you think step 3 should be???
D H Posted April 5, 2010 Posted April 5, 2010 While you can prove that the sum of two even integers is even by going all the way down to first principles, nobody (well, almost nobody) does proofs that way, triclino. There would be no progress in mathematics if every proof had to be taken down to first principles. The only problem with ajb's proof is this: Then the sum [math]x + y = 2a + 2b = 2(a + b)[/math]. (This follows from the rules of real numbers, so I will take this as given.) We're talking about integers, not reals, so it is best not to invoke rules about reals. This is easily corrected: Then the sum [math]x + y = 2a + 2b = 2(a + b)[/math], the latter step following from the distributivity of multiplication over addition for the integers. There is no need to go deeper than this because that 2a+2b=2(a+b) follows from distributive law is blatantly obvious. Real mathematical proofs skip over the blatantly obvious. What is blatantly obvious depends on the audience, of course. To a novice, very little is blatantly obvious, so beginner's math texts go into a lot of painstaking detail. To a professional mathematician there is no need for that detail. Pick up an advanced (i.e., graduate level) math text or a mathematical journal and look at the proofs. They don't go into the detail you want because that detail would be a hindrance rather than a benefit.
triclino Posted April 5, 2010 Author Posted April 5, 2010 (edited) D.H the word is foundations . Do you know which are the foundations of mathematics ?? What is your definition of mathematical proof?? Do you perhaps know any efficient way of checking why a mathematical proof is correct or not?? Here is a high school mathematical proof (to start with easy one). Prove: |x|<y proof: |x|<y <=> [math]|x|^2<y^2\Longleftrightarrow x^2-y^2 <0[/math] <=> (x+y)(x-y)<0 <=> {[(x-y)<0 and (x+y)>0] or [(x-y)>0 and (x+y)<0]}. IF [(x-y)<0 and (x+y)>0] <=> x<y and x>-y <=> -y<x<y.................1 IF [(x-y)>0 and (x+y)<0] <=> x>y and x<-y...................................2 From (1) and (2) we conclude that : |x|<y <=> -y<x<y Is that proof correct or not??? Edited April 5, 2010 by triclino
Cap'n Refsmmat Posted April 5, 2010 Posted April 5, 2010 http://en.wikipedia.org/wiki/Proof_verification http://us.metamath.org/index.html I think you'd enjoy the latter site.
triclino Posted April 5, 2010 Author Posted April 5, 2010 http://en.wikipedia.org/wiki/Proof_verification http://us.metamath.org/index.html I think you'd enjoy the latter site. I think you would enjoy the high school problem in post #7.
the tree Posted April 6, 2010 Posted April 6, 2010 Do you know which are the foundations of mathematics ?Well, no. There is no universally agreed upon set of postulates. The Peano axioms are fairly standard for that type of stuff but they don't suit all situations and there are other just-as-okay alternatives. Realistically, the deepest you're going to go in matters of arithmetic are the field axioms which are just the definition of a field and no-one feels the need to justify that. Prove: |x|<y => -y<x<y [...] Is that proof correct or not??? Well, no. I mean, obviously there were lots of premises and conditions and whatnot that you left out, but still, ugh. And the line of thought was backwards - your argument started with your conclusion and went on to justify which just, ugh, you know better than that.
triclino Posted April 6, 2010 Author Posted April 6, 2010 . Well, no. I mean, obviously there were lots of premises and conditions and whatnot that you left out, but still, ugh. And the line of thought was backwards - your argument started with your conclusion and went on to justify which just, ugh, you know better than that. A mathematical proof is not an abstract painting where everybody can say whatever he/she likes . A mathematical proof consists of : 1) Axioms,theorems,definitions. 2) The laws of logic . So if you say that the particular proof is wrong ,you have to pinpoint ,the particular axiom or theorem or definition or the application 0f the particular law of logic that is not correct. Otherwise your remarks and comments are of no value at all . But to say for certain whether a proof is correct or not ,one must get into the depths of the proof . This a simple high school problem .One can imagine what actually is happening with other more complicated Analysis proofs e.t.c ......e.t.c
D H Posted April 6, 2010 Posted April 6, 2010 So if you say that the particular proof is wrong ,you have to pinpoint ,the particular axiom or theorem or definition or the application 0f the particular law of logic that is not correct. First and foremost, the hypothesis must be well-formed. This is not a well-formed hypothesis: Prove: |x|<yproof: [proof elided] That statement is obviously not true in general (consider the case y=0). Try again, this time stating your hypothesis correctly.
triclino Posted April 6, 2010 Author Posted April 6, 2010 First and foremost, the hypothesis must be well-formed. This is not a well-formed hypothesis: That statement is obviously not true in general (consider the case y=0). Try again, this time stating your hypothesis correctly. You can easily see that there was o typo in my problem : instead of prove : |x|<y <=> -y<x<y I wrote : prove : |x|<y which has no meaning: But from the proof that followed one could easily concluded that i ment : prove : |x|<y <=> -y<x<y. This a high school theorem and it can be found in any high school books. Since y>|x| and [math]|x|\geq 0[/math] => y>0. Go for the real thing ,is that proof correct or not??
Mr Skeptic Posted April 6, 2010 Posted April 6, 2010 prove : |x|<y <=> -y<x<y. This a high school theorem and it can be found in any high school books. Since y>|x| and [math]|x|\geq 0[/math] => y>0. Go for the real thing ,is that proof correct or not?? Do you consider it incorrect if it is incomplete? You neither proved nor assumed [y>|x| and [math]|x|\geq 0[/math]] => y>0 nor referred to a previous proof for it. It's common sense, yes, but common sense is not formal proof.
triclino Posted April 6, 2010 Author Posted April 6, 2010 Do you consider it incorrect if it is incomplete? You neither proved nor assumed [y>|x| and [math]|x|\geq 0[/math]] => y>0 nor referred to a previous proof for it. It's common sense, yes, but common sense is not formal proof. I simply wanted to show that by writing :|x|<y this automatically implies y>0 Some books even mention that y should be >0. Common sense is common logic .Do you think that a formal proof has not common sense on it?? That would be the end of everything.
Mr Skeptic Posted April 6, 2010 Posted April 6, 2010 Sometimes a formal proof goes against common sense. See for example the Monty Hall problem. Even having a PhD does not mean that your common sense matches the formal proof.
triclino Posted April 7, 2010 Author Posted April 7, 2010 Sometimes a formal proof goes against common sense. Apart from the irrelevant example that you mentioned can you write a small formal proof of any mathematical problem and show us in which way it goes against common sense See for example the Monty Hall problem. Even having a PhD does not mean that your common sense matches the formal proof. To that statement i agree fully. A lot of Ph. Doctors lack even traces of common sense
Mr Skeptic Posted April 7, 2010 Posted April 7, 2010 Right, the problem could just as easily be that common sense is not as common as people like to imagine. The problem I described (which is a mathematical problem) goes against many people's "common sense" -- either their common sense is wrong, or common sense is uncommon.
ajb Posted April 9, 2010 Posted April 9, 2010 We're talking about integers, not reals, so it is best not to invoke rules about reals. For sure, I should have been a little more careful. What is blatantly obvious depends on the audience, of course. This is the real problem and why trying to insist on a tight definition of a proof in the forum rules will not work.
triclino Posted April 9, 2010 Author Posted April 9, 2010 In mathematics sometimes, the blatant ,when we try to prove it becomes blatantly impossibly. For example : The BOLZANO theorem
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