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Topic: Peano's axiom (Read 2222 times) 

Mickey1
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Peano's axiom
« on: Nov 13^{th}, 2010, 3:23pm » 
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I have a problem in relation to Peano’s axioms. I use the first version of the axiom from Wikipedia’s formulation, including having the first number as 0 instead of 1. I am wondering if the definition of “=” allows too much. The additional axiom I want to test is “for all natural numbers n and m different from 0, n=m, and furthermore 0=0” (test axiom to be added to the rest). The first 5 axioms seems to be self evident with my new axiom inserted, since n=m and 0=0. If this is true, can axioms 6 to 9 save the day, falsifying my axiom? Wikipedia mentions the axioms: 1. For every natural number x, x = x. 2. For all natural numbers x and y, if x = y, then y = x. 3. For all natural numbers x, y and z, if x = y and y = z, then x = z. 4. For all a and b, if a is a natural number and a = b, then b is also a natural number. , 5. 0 is a natural number. 6. For every natural number n, S(n) is a natural number 7. For every natural number n, S(n) = 0 is False. 8. For all natural numbers m and n, if S(m) = S(n), then m = n. 9. If K is a set such that: o 0 is in K, and o for every natural number n, if n is in K, then S(n) is in K, o then K contains every natural number. I assume that axioms 16 cannot fulfill this function (saving the day). 6 and 7 together seem powerful in that they rule out that n=m for all numbers, so that S(n) is a natural number (6) but not 0 (7). This rules out a simpler test axiom (n=m for all numbers) but I can’t see how my test axiom is falsified. I can see that “and n=S(n) is also false” being added to (7) might remedy the situation. Other than that, how can we understand the set of axioms? Are the axioms not intended to  uniquely  point to the natural numbers as we intuitively understand them? Or is Wikipedia’s version wrong? Grateful for any help.


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towr
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Re: Peano's axiom
« Reply #1 on: Nov 14^{th}, 2010, 8:15am » 
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Just to be clear, you want to know whether the theorem "for all natural numbers n and m different from 0, n=m, and furthermore 0=0" can be disproved using the Peano axioms? I think axiom 8) poses a big problem, because for any n > m, it brings you back to nm = 0. Which you already had to make one exception for, but you'll have to make an exception for every number.

« Last Edit: Nov 14^{th}, 2010, 8:15am by towr » 
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0.999...
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Re: Peano's axiom
« Reply #2 on: Nov 14^{th}, 2010, 9:25am » 
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Formalizing a specific case of what towr said: The combination of your test statement (call it T), (7) and (8 ) leads to a contradiction: 1. S(0)0 ...... by (7) 2. SS(0)0 ...... by (7) 3. SS(0)S(0) ....... by (T) 4. S(0)0 ..... by (8 )

« Last Edit: Nov 14^{th}, 2010, 9:28am by 0.999... » 
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Mickey1
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Re: Peano's axiom
« Reply #3 on: Nov 17^{th}, 2010, 8:06am » 
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I saw this posibility that towr points out, but I am speculating that my extra axiom might block the steps involving the definition of minus in nm. That is to say: the undecided or  in my suspicion  too open "=" operator allowing my extra axiom (whether I explicitly stated it or not) would perhaps make the minus definition possible, but it would not necessarily be the only consequence of the preceeding axioms regarding the use of a+b=c implying later that a=bc (which is possible but does not necessarily follow from my sets of axiom). So I guess that my question can be rephrased: are the first axioms enough to arrive at a meaningful addiiton or subtraction definition before the introduction of S(n). Or: is + sufficiently limited in its environment of definitions, so that it resembles how we intuitivly understand it? Somewhat unrelated, as I write this I, ask myself if axiom can be orderofappearance sensitive (apart from the trivial case that the use of definitions logically should appear after the definition itself).


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towr
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Re: Peano's axiom
« Reply #4 on: Nov 17^{th}, 2010, 10:06am » 
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on Nov 17^{th}, 2010, 8:06am, Mickey1 wrote:I saw this posibility that towr points out, but I am speculating that my extra axiom might block the steps involving the definition of minus in nm. 
 Axioms only add ways to derive theorems, they can't block derivations. If you can derive something from axiom A to Y, you can derive it from axioms A to Z; at worst adding Z makes the logic inconsistent (as happens in this case). Quote:So I guess that my question can be rephrased: are the first axioms enough to arrive at a meaningful addiiton or subtraction definition before the introduction of S(n). Or: is + sufficiently limited in its environment of definitions, so that it resembles how we intuitivly understand it? 
 I'm not sure what you're asking. n+m is just shorthand for S^{n}(S^{m}(0)) (associativity and commutativity can be proved). And all you need for subtraction is axiom 8. By applying axiom 8 n times, you get S^{n}(S^{x}(0)) = S^{n}(S^{y}(0)) implies S^{x}(0) = S^{y}(0) So then if y = 0, we must have to x = 0 because of axiom 7. And in any case, you only need one specific counterexample, which pex 0.999... has provided.

« Last Edit: Nov 17^{th}, 2010, 1:03pm by towr » 
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0.999...
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Re: Peano's axiom
« Reply #5 on: Nov 17^{th}, 2010, 12:52pm » 
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on Nov 17^{th}, 2010, 10:06am, towr wrote: He figured me out!


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towr
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Re: Peano's axiom
« Reply #6 on: Nov 17^{th}, 2010, 1:04pm » 
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Whoops. Well, they say great minds think alike, so it stands to reason I might confuse them sometimes..

« Last Edit: Nov 17^{th}, 2010, 1:05pm by towr » 
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pex
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Re: Peano's axiom
« Reply #7 on: Nov 17^{th}, 2010, 1:16pm » 
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on Nov 17^{th}, 2010, 1:04pm, towr wrote:Whoops. Well, they say great minds think alike, so it stands to reason I might confuse them sometimes.. 
 Let me thank you from this account


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Mickey1
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Re: Peano's axiom
« Reply #8 on: Nov 17^{th}, 2010, 3:31pm » 
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I still dont see how one arrives at 0.999's last step. S(...n times(0)=S(...times(0) only implies that n=m, as I stated in my extra axiom. How does one arrive at the last step, the forbidden S(0)=0?


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0.999...
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Re: Peano's axiom
« Reply #9 on: Nov 17^{th}, 2010, 3:47pm » 
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I'll do the general case and add some notes (make same assumptions as in my previous postimplicitly adding (5) and (6) ): 1. S^{n+1}(0) 0 and 2. S(0) 0 by (7) The test statement (and n+1 instances of (6)) imply that 3. S^{n+1}(0) S(0) 4. S^{n}(0) 0 This comes from the fact that S^{k} denotes k iterations of the operation S; put in notation, S^{k}(0) denotes S(S^{k1}(0)). So, applying (5) to see that 0 is a natural number, we may use (8 ), substituting m = S^{n}(0) and n = 0 to get our statement 4. Edit: to show that this is allinclusive, use (9).

« Last Edit: Nov 17^{th}, 2010, 3:56pm by 0.999... » 
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Mickey1
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Re: Peano's axiom
« Reply #10 on: Nov 17^{th}, 2010, 4:57pm » 
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Thank you but I am still concerned by your use of minus (or plus) which I don't think is backed by the axioms at this point. Bear with me one more time while I rephrase: Statement 1: The axioms do not contradict that n=m for all numbers different from 0. Statement 2: S(n) can be interpreted as a number different from 0. I guess another way to describe my uneasyness is that you both use a tool to disprove me, consisting of the natural numbers (complete with plus and minus), but these are the very object of our discussion, they cannot be assumed to exist in a certain way at this point other than defined in the axioms. I am not sure you can refer to "k times" or "k1 times". There is no axiom saying that + exists or that x+a=x+b implies that a=b, so I don't think there is a way to go backwards (k1) since k1=k (T) if not 0, or using minus more generally. Another way to rephrase the result: The Peano axioms leads to two elements, one called 0 and the other called "different from 0". An operation can therefore under this interpretation not be done k times, only 0 times or "different from 0" times. I realize that in reality you must start somewhere, perhaps accepting going from an intuitive to a semiintuitive situation and later to a more stringent situation. Perhaps one must accept that.


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towr
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Re: Peano's axiom
« Reply #11 on: Nov 18^{th}, 2010, 12:43am » 
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on Nov 17^{th}, 2010, 4:57pm, Mickey1 wrote:Bear with me one more time while I rephrase: Statement 1: The axioms do not contradict that n=m for all numbers different from 0. 
 Yes, they do. It's contradicted by use of axiom 7 and 8. It's contradicted by the fact that by axiom 8, S(S(0))=S(0) implies S(0)=0, which is contradicted by axiom 7. It's contradicted by the fact that by axiom 8, S(S(S(0)))=S(0) implies S(S(0))=0, which is contradicted by axiom 7. It's contradicted by the fact that by axiom 8, S(S(S(0)))=S(S(0)) implies S(S(0))=S(0), which is contradicted by our derivation two statements up. It's contradicted by the fact that by axiom 8, S(S(S(S(0))))=S(S(0)) implies S(S(S(0)))=S(0), which is contradicted by our derivation two statements up. And so on, and so on. For any specific n and m, if they are not equal, you can construct a proof that leads to a contradiction. And we only need to do it for 1 specific counterexample to disproof the theorem. Quote:Statement 2: S(n) can be interpreted as a number different from 0. 
 Stronger than that, by axiom 7 S(n) can only be interpreted as something different than 0. Quote:I guess another way to describe my uneasyness is that you both use a tool to disprove me, consisting of the natural numbers (complete with plus and minus) 
 No, we aren't. We're only using axioms 7 and 8. There rest is just shorthand, not additional assumptions or tools. If n and m are natural numbers than either n=0 or m=0, or (by 9) n=S(a) and m=S(b), in which case by 8 we have a=b. For the specific case m=1 (i.e. S(0)), this gives an immediate contradiction with axiom 7 if n is different from m. if m > 1, you can just repeat the same step to bring the number down further. Quote:I am not sure you can refer to "k times" or "k1 times". 
 Sure you can, because it's what you do in the proof; it's not something within the logic system, it's a description about the proof. If I apply axiom 8 three times to S(S(S(S(0)))) = S(S(S(0))) that simply means given S(S(S(S(0)))) = S(S(S(0))) by applying axiom 8 we have S(S(S(0))) = S(S(0)) by applying axiom 8 a second time we have S(S(0)) = S(0) by applying axiom 8 a third time we have S(0) = 0 Now by applying axiom 7, we arrive at a contradiction. Therefore S(S(S(S(0)))) = S(S(S(0))) is inconsistent with axioms 19, and therefore the general statement that all natural numbers different from 0 are equal is also inconsistent with the peano axioms. Quote:There is no axiom saying that + exists or that x+a=x+b implies that a=b 
 + is just shorthand. just as 4 is shorthand from S(S(S(S(0)))). Also 0.999...'s counterexample makes no use of it in the first place. Nor does my counterexample in the previous paragraph.

« Last Edit: Nov 18^{th}, 2010, 12:57am by towr » 
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Mickey1
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Re: Peano's axiom
« Reply #12 on: Nov 18^{th}, 2010, 2:23am » 
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Yes, I see that now. Thanks for your comments.


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