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Logical Systems+new Principles+attempt To Solve Collatz Conjecture

  1. Feb 11, 2008 #1
    LOGICAL SYSTEMS+NEW PRINCIPLES+ATTEMPT TO SOLVE COLLATZ CONJECTURE.

    First i would like to say that am honoured to share my thought with great people in here who always provide help.I will not say iam right or wrong.I hope this post will be aspark to good mindes.

    I have viewed the laws of logic and I think that there is asort of aconfusion about how the logical system works? and how the logical process conducts the conclusion?there is also aconfusion about the limits of the logical laws, specially the mixing between the law of contradiction and the law of noncontradiction.
    In this post I will be trying to show that all laws of logic are serving two simple prima principles and we will call them "principle of interior contradiction”and”principle of exterior contradiction”.
    Now let, A, be acase we want to prove,the logical process that followed by any logical or mathematical system to prove A,is
    1=(A,exists,end of proof),
    2=(A,doesnot exist),
    we know that the pair(1,2) is called"contradiction law".we have to notice that the whole idea to build aproof start with that pair.
    now,the pair(1,2)leads to alogical processes depend on aseries of steps of "the noncontradiction law "until we end to astate we cannot step over with anew contradiction or astep of afinal contradiction which is the conclusion,(A,exists,end of proof),the problem is ,whenever the noncontradiction law stops working during the process then we have to create or find anew contradiction other wise the statmment,A,stills open or unsolvable or we get aparadox.
    We will call the relation between the law of contradiction=pair(1,2)and the law of noncontradiction that works in the logical process as"EXTERIOR CONTRADICTION PRINCIPLE "or the ECP.
    Now the essential question,(I) is the contradiction law exists?(II)or it doesnot exists?if it exists then we will never be able to proof or know any judgement that the logical statement,A,exists or true or false .if it doesnot exists then why we need the proof in the first place?because we will know for sure any judgement about the statement,A.now we will call the the pair(I,II)”THE INTERIOR CONTRADICTION PRINCIPLE”or the ICP.


    The ICP. and the ECP.are the two principles. that all the logical and mathematical systems and processes depend on.
    In asipmle word,(ICP) →(ECP) →(conclusion).or in another word,(contradiction law) →(prosseses of steps of noncontradiction law) →(final contradiction,or what we call it the conclusion).

    Law of identity.
    P ≡ P
    The proof,
    P,exists

    1- P ≡ P,end of proof
    2- P no ≡ P
    (1,2) → if (P no ≡ P)→ (P ≡anything ,anything includes, P)→( P ≡ P),end of proof. (1,2) → if (P no ≡ P) →( P ≡nothing) → (P,doesnot exist) →(final contradiction=conclusion).as amatter of facts all logical laws are eventually serving the LIC.

    law of the excluded middle.
    Eventhough , some systems of logic have different but analogous laws, while others reject the law of excluded middle entirely. We will trying here to prove that this law is serving the LIC.
    The law states,
    "P ∨¬P"
    1-p,exists.
    2-p,doesnot exist.
    (the pair (1,2)=contradiction law).it sems that law of the excluded middle is identical to law of contradiction or another forming to the law of contradiction.


    THE LOGICAL SPACE.
    The logical spaces is asystem of operations ,relations,groups,variables,constants and operators where the logical statement exists.logical space is not aring or field.it is agroup of tools that we can use to create either ATRUE logical statement or FALSE logical statement.ofcourse we can shrink or expand the logical space depend on what we have or create of tools.

    For example,the statement ,
    1/x ‹x ,x≠0
    .this statement exists in the logical space (1,=,/,x,‹,G)where ,G,is the real numbers› 1.

    assume that ,A, is alogical statement exists in the logical space Q, and K,is adifferent logical space
    1-A,is true in,Q
    2-A,false in,K
    The pair(1,2) is not acontradiction.
    This means that we will not be able to prove the existance of,A,is true. in the example above. we either testing all the numbers of,G,which is impossible,to prove the statement, 1/x ‹x, x≠0, or we start with the contradiction
    1-A,is true in,Q
    2-A,false in,Q
    (1,2) →assume 1/x ‹x ,x≠0,is true in another space,such as, (1,=,/,›,x,G) →(1/x›x)→(1›x)→(1›1)
    → contradiction of identity law=final contradiction=conclusion.

    We also can find anothers spaces to serve the case.
    One could say ,1/x ‹x ,x≠0 , in (‹,G)is so obvious and doesnnot need any proof,even if it does,we can prove it by ashorter way like,assume, (1/x ›x)→(1/2›2)→(contradiction).or (1/x =x)→(1/2=2)→contradiction of law identity.
    The answer is yes,it is the same above but all the mathematical process in the argument is going intuitively while we want it to goes more axiomatically by defining what logical space the statement exists?and if it is atrue? or false?.when we are unable to know the limits where contradiction and noncontradiction lwas work,when we are unable to define the logical statement and in which logical space it exists,then we will get an open problems or paradox.

    The important thing here is to notice that the time is not actually amthematical operator we depend on.instead of time we use more well defined operators.for example
    1 ≠0 , in the space(G,›),where ,G,is any group.not now not any time,but it is simple to notice that ,(1,could=0),in any space hase the operator,d/dx,i.e,d/dx(1)=0.

    THE ONE DIRECTION INFINITY PRINCIPLE,THE STEADY STATEMENT
    Assume he statement ,(a1,exists)→(a2,exists)→ …….an,exists).the qustion is there astatement,a∞ exists?
    Obviousely the possiblity of existance of any of the statements,an, is,1,which leads us to say the possibility of existance , a∞,is also 1.we will call, a∞,asteady statement.


    THE TWO DIRECTIONS INFINITY.

    Assume he statement ,(a1,exists)→(either,a2,exists,or,a3,exists). (a2,exists)→(either,a4,exists,or,a5,exists). (a3,exists)→(either,a6,exists,or a7,exists)……etc.
    Obviousely the possiblity of existance , a∞,is,1\2.

    THE MULTI DIRECTIONS INFINITY.

    Assume he statement ,a1,exists→(either,a1,or,a2,or,………. an,exists)
    Obviousely the possiblity of existance , a∞,is,0.

    COLLATZ CONJECTURE AND STEADY STATEMENT.
    Post:the staedy statement exists in collatz formula and it equals,1.
    collatz formula,
    A-F(n)=n/2……if,n≡0(mod2).
    B-F(n)=3n+1…if,n≡1 (mod2).

    Proof.

    (some,A→ A),(some,A→ B),BUT(all,B→A,in the space(1,=,2k,/,N),Where,k,apositive integer and,N,is the natural numbers set.by using the one direction infinity principle,there exists asteady statements,k∞,that makes 2k=1,is atrue statement in the space above ,namely,2k.
    end of proof.
     
  2. jcsd
  3. Feb 15, 2011 #2
    Iff n ≡0(MOD2) then n will be divided by 2 any number of times until an odd number, m[/1SUB], is obtained. Iff n ≡1(MOD2) let n=m[/1SUB].

    Iff m[/1SUB]≡1(MOD4) then 3m[/1SUB]+1≡0(MOD4). Let 3m[/1SUB]+1=a, then either a/2 ≡2(MOD4) or a/2 ≡0(MOD4). In the latter case, either a/2 ≡2(MOD4) or a/2 ≡0(MOD4) and successive divisions by 2 will yield a number ≡2(MOD4). Another halving will give an answer m[/2SUB]≡1(MOD4), and the process is repeated. Since 3m[/nSUB]+1 is always a multiple of 4, m[/n+1SUB]=(3m[/nSUB]+1)/(4*2r), where r>/=0. Therefore, if [/nSUB]>1 then m[n+1/SUB]<m[/nSUB]. Then each term is less than the previous, and because no term can be <1, the result must converge to 1.

    Iff m[/1SUB]≡3(MOD4) then 3m[/1SUB]+1≡2(MOD4) and a/2 ≡1(MOD4), and from there the argument continues as in the case for[/1SUB] m≡1(MOD4).
     
  4. Feb 18, 2011 #3
    Ahhh, very insightful. This is such agood approach to aproblem that is having much interesting. I think asort of alogical conclusion would hat you are acrank. Of course, I couldbecompletlywrong. but i wold have no. idea given that your post is impossible to aread. If you are going to create anew system to unconfuse aconfusion, then why not use it to solve say, Goldbach's.
     
  5. Feb 20, 2011 #4
    husseinshimal,

    If English is not your first language, I suggest that you let members of this forum know now. Otherwise, I would tend to concur with Robert's "first thought, best thought." That you are a thinker is clear. That you think in rational manner, less so.

    I am also an unorthodox thinker with unorthodox approaches. But I have also, in tandem with numerical observations and mathematical constructions, posited specific mathematical statements upon this forum; statements that, at least at times, are eminently verifiable or falsifiable and that arose from those observations and constructions.
     
    Last edited: Feb 20, 2011
  6. Feb 20, 2011 #5

    micromass

    User Avatar
    Staff Emeritus
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    Hmm, I fear that this is original research, and I'm quite sure that this is quite forbidden in thsi section of the forum...

    Why not post it in the section "Independent research", which is a subforum of General Physics? That's the right place to post such things.

    As for your post itself: I'm sorry, but it seems that I don't follow your reasoning at all. Can you phrase it in such a way that dumb people like me can understand??
     
  7. Feb 20, 2011 #6
    Forgive me, Micromass, but, in fairness, you are not offering a realistic alternative here to the potential OP. The "Independent Research" Section of this site requires moderator approval.

    - RF
     
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