conway

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conway last won the day on August 11 2015

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About conway

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  1. Is this a new number ?

    Oh I take it in strides...ever trying to improve....How do you take the fact your Bigjerk?
  2. Is this a new number ?

    https://www.math.hmc.edu/funfacts/ffiles/10005.3-5.shtml https://brilliant.org/wiki/what-is-00/ The point being that "whatever" it is.... All representations of (-0) as exponents and logarithms work like 0 and without change from 0 thank you for your time and reply
  3. Is this a new number ?

    In every R there exists an integer zero element ( -0 ) ( -0 ) =/= 0 |0| = |-0| ( -0 ) : possesses the additive identity property ( -0 ) : does not possess the multiplication property of 0 ( -0 ) : possesses the multiplicative identity property of 1 The zero elements ( 0 ) and ( -0 ) in an expression of division can only exist as: (0)/( -0 ) 0 + ( -0 ) = 0 = ( -0 ) + 0 ( -0 ) + ( -0 ) = 0 1 + ( -0 ) = 1 = ( -0 ) + 1 0 * ( -0 ) = 0 = ( -0 ) * 0 1 * ( -0 ) = 1 = ( -0 ) * 1 n * ( -0 ) = n = ( -0 ) * n Therefore, the zero element ( -0 ) is by definition also the multiplicative inverse of 1 . And as division by the zero elements requires ( - 0 ) as the divisor ( x / ( -0 )) is defined as the quotient ( x ) . 0 / n = 0 0 / ( -0 ) = 0 n / ( -0 ) = n 0 / 1 = 0 1 / ( -0 ) = 1 1 / 1 = 1 ( 1/( -0 ) = 1 ) The reciprocal of ( -0 ) is defined as 1/( -0 ) 1/(-0) * ( -0 ) = 1 (-0)^(-1) = ( 1/( -0 ) = 1 (-0)(-0)^(-1) = 1 = ( -0 )^(-1) Any element raised to ( -1 ) equals that elements inverse. 0^0 = undefined 0^(-0) = undefined 1^0 = 1 1^(-0) = 1 Therefore, all expressions of ( -0 ) or ( 0 ) as exponents or as logarithms are required to exist without change. Therefore, division by zero is defined. Therefore, the product of multiplication by zero is relative to which integer zero is used in the binary expression of multiplication.
  4. A Brand New Approach

    Well nice...but maybe you should have read all the pertinent information before assaulting me... 0.z1 * 1 =/= 0.z1 0.z2 * 1 =/= 0.z2 yes you may divide through by 1 but... 0.z1 * 1 = 0 0.z2 * 1 = 1 therefore after your cancelation of 1 you will have 0 = 0 = 0 1 = 1 = 1 It was unfair of you to have read so little before making such a negative reply. Pzkpfw... please note I made a mistake here I meant to say 0.z1 * 1 =/= 0.z2 * 1 0.z1 * 1 = 0.z2 * 0 Apologies...
  5. A Brand New Approach

    No one is debating that the old approach works John. Just because something works doesn't make it right John. Just because something is different doesn't make it wrong John. You have a history with me of passive aggressiveness and thread de-railing. I have NO desire to communicate with you any further. Further replies from you on this thread will not be responded to by me... and will receive a -1
  6. A Brand New Approach

    0 * 1 = ( 0.z1 ) * 1 = 0 0 * 1 = ( 0.z2 ) * 1 = 1 0.z1 = 0 0.z2 = 1 No inconsistences....however the binary expression ( A * 0 ) is RELATIVE to what projection operator for zero was used. If I use the projection operator 0.z1 in a binary operation for zero the product is 0 If I use the projection operator 0.z2 in a binary operation for zero the product is X You MUST CONVERT 0 to a projection operator BEFORE you can solve for the binary equations involving zero. If you do not understand this...or agree...or I am a lousy at explaining myself.... that is fine. I thank you for your time. Perhaps in a year or so I will find a better way to communicate with you. Thank you.
  7. A Brand New Approach

    NO.... I never stated 0 = 1.....why don't you post a quote of this I said 0.z2 =1..... clearly your mistake was to replace my "wiggildybop" (0.z2) with 0...therefore 0 = 1 you MUST replace my "wigglildybop" 0.z1 "with" 0 you MUST replace my "wiggligdybop" 0.z2 "with" 1 If you replace it in the way in which I suggest what you end up with is 0 = 0 * 1 = 01 = 1 * 1 = 1 AS I STATED 0.z1 = 0 0.z2 = 1 therefore when you replace them correctly you have 0 = ((0z1)/1) * (1/(0z2)) = 0 * 1 = (0z1) * 1 = 01 = ((0z1)/1) * (1/(0z2)) = 0 * 1 = (0z2) * 1 = 1 0 = 0 * 1 = 01 = 1 * 1 = 1 notice 0 * 1 = (0.z1) * 1 notice 0 * 1 = (0.z2) * 1 but... 0.z1 * 1 =/= 0.z2 * 0 again please note the projection operators from the original post.... 0.z1 is a projection operator FOR 0 but = 0 0.z2 is a projection operator FOR 0 but = 1
  8. A Brand New Approach

    Clearly you did NOT read my op in this thread did you Studiot. This is more of the same typical behavior from you and frankly I would rather not hear from you again. FACT 1. NO new terms where introduced (or used unorthodoxly) in the newest version of this. (only new notations referring to CURRENT real numbers) while I had a 0.z1 and a 0.z2 it was only a facilitator for 0 and for 1.....the thing is studiot I went OUT of my way here to do EXACTLY as you asked. THERE IS NO NEW terminology here. please reread the op in this thread. copy/quote and past ANYTING that is NEW to mathematics. Even the projection operators were typical. Until you can ACTUALLY post a quote from THIS thread showing ANYTHING NEW .....-1....should you do so....I will withdraw this. pzkpfw I seem to think so. Most do not. 0 and 1 are not equal. No where is that equivalency in my equations. It does however imply that there is a similarity between 0 and 1 yes. But that would be philosophy and I would rather stick to the mathematics. thank you Strange ...I have done so already.
  9. A Brand New Approach

    mathematic 1. Relative binary multiplication by zero. 2. Defined division by zero. 3. Create varying amounts of zero. 4. Unify semantics, and physics with theoretical mathematics. 5. Offer a new approach on the continuum theory. 6. Suggest solutions for the physics regarding the unification of quantum and classical mathematics. Also this might help you. This is from a previous closed thread. As I was asked by a moderator to come up with something new here...in order to continue.(As I have done) *NOTE this is edited from the previous post.. because I have learned a great deal form the members of this community...and would like to reflect an ability to learn and grow. (0.z1) = in a binary expression of multiplication yields the product 0 : in a binary expression of division is the numerator and yields the quotient 0 : if both numbers are 0 in an expression of binary multiplication the binary product is 0 (0.z2) = in a binary expression of multiplication yields the product x : in a binary expression of division is the denominator and yields the quotient x : if both numbers are 0 in an expression of binary division the binary quotient is 0 0 = ((0z1)/1) * (1/(0z2)) = 0 * 1 = (0z1) * 1 = 0 1 = ((0z1)/1) * (1/(0z2)) = 0 * 1 = (0z2) * 1 = 1 x = x/0 = x/(-1 + 1) = ( x/-1 + x/1 ) + x = (x/0) * (1/0) = 1 * x = x 0 = x * ( 0 + 0 ) = x * (0z1) = (0z1) * x = ((0z1)/1) * (1/(0z2)) = (0z1) * x = 0 x = x * ( 0 + 0 ) = x * (0z2) = (0z2) * x = ((0z1)/1) * (1/(0z2)) = (0z2) * x = x thank you for your time.
  10. A Brand New Approach

    No number tables...no properties. No axioms change (except) when involving zero. The following projection operators allow for no further axioms...... [math]0 = \left ( \begin{matrix} 0.z_1 \\ 0.z_2 \end{matrix} \right ) [/math] 0.z1 = 0 0.z2 = 1 [math] P_1 0 = (1, 0) ~ \left ( \begin{matrix} 0.z_1 \\ 0.z_2 \end{matrix} \right ) = 1 \cdot 0.z_1 + 0 \cdot 0.z_2 = 0.z_1[/math] [math] P_2 0 = (0, 1) ~ \left ( \begin{matrix} 0.z_1 \\ 0.z_2 \end{matrix} \right ) = 0 \cdot 0.z_1 + 1 \cdot 0z_2 = 0.z_2[/math] The distributive property (all combinations of a, b, and c as zero) a * (b + c) = a * b + a * c a = 1, b = 0 , c = 0 1 * ( 0 + 0 ) = 1 * 0 + 1 * 0 1 * (0 + 0) = 1 * (0.z1) = 1 * (0.z1) + 1 * (0.z2) a = 1, b = 1 , c = 0 1 * (1 + 0 ) = 1 * 1 + 1 * (0.z1) a = 0, b = 0 , c = 0 0 * (0 + 0) = 0 * 0 + 0 * 0 a = 1, b = 0 , c = 1 1 * (0 + 1) = 1 * (0.z1) + 1 * 1 1 = 0, b = 1, c = 0 (0.z1) * (1 + 0 ) = (0.z1) * 1 + 0 * 0 (0.z2) * (1 + 0 ) = (0.z2) * 1 + 0 * 0
  11. A Request for Peer Review

    pzkpfw Well said...+1...I did finally after much exhaustion see why uncool chose this equation. I chose not to reply on the matter as uncool has said some very rude things. I also stated in my reply with uncool that ALL this shows is that a simple axiom of the nature that the distributive property remains without change...except zero....and then provided equations and expressions showing what I mean by this. You put much time into this reply....put a little more into it. Use the original post....apply any number other than zero to the equation brought forth by uncool...and the equations remains equivalent. As I was already changing other axioms regarding the nature of 0....and as I was frustrated I rushed my last replies to him and poorly chose my expressions and equations. please note this has only happened with him. please note I gave him a +1 for pointing this out please note he then attempted a system of circular logic regarding the "order of operations" in an attempt to "trip" me up. The math on the matter... a * (b + c) = a * b + a * c 1 * (2 + 3) = 1 * 2 + 1 * 3 uncool gave... 1 ( 0 + 0 ) = 1 * 0 + 1 * 0 uncool made a = 1, b = 0 , c = 0 then he said some mean things...successfully frustrating me...and tripping me up... so here you are then....with HIS equalities for a, b and c 1 * ( 0 + 0 ) = 1 * 0 + 1 * 0 1 * (0 + 0) = 1 * (0.z1) = 1 * (0.z1) + 1 * (0.z2) I hope you can understand the nature of my mistakes with him. I even told him this was all that was necessary. He would have seen it if he hadn't been so focused on being a troll. Which is why no one else brought it up. Or supported him. again thank you...and a well earned +1
  12. A Request for Peer Review

    So then....this "idea" is sound....in whatever application you wish So then...I did for a FACT do my best to answer your questions.....you denied this Uncool gave an equation that is NOT a valid representation of the distributive property (this can be proven).(I thanked him many times, verbally and with +1) I believe the "idea" to be AS basic as the underlying math....I understand that you do not agree with this. I thank you for your time and peer review You did put and extraordinary amount of effort into it....towards the end....I mean you finally looked up my quote of you....to bad you didn't read it back when we where both happy and getting along. But in all seriousness thank you.
  13. A Request for Peer Review

    It certainly appears to me that you are stamping your feet and showing your temper. I don't think it will get you anything. I have given a link with a quote of YOU....agree with me... on the explanations of the definitions your continuously asking for. I am waiting for you to click on it and read for yourself....yourself...answering these questions. Please leave this alone. This is dangerously close to getting closed. If your done just leave me be.....please. I apologize for BOTH our tempers flaring.....but please just leave me be. +1 to get us back on the peace trail
  14. A Request for Peer Review

    Studiot Do you pick and chose the post you want to read? Why do you blatantly make stuff up? challenge.... Post a quote of me saying that you where right about my number table and I was wrong. I NEVER said this. I did say that I could not list all of them on a table. That is obvious and has nothing to do with the table it's self. case in point: YOUR PEER STRANGE does NOT agree with me on this idea. But he does agree your point here is "unfair".....exact words. Facts are...I did answer your questions...you didn't like the answers I DEMAND you address my link and quote of you from two years ago.....AGREEING with me about space and value as definitions for z1 and z2...the "question" you insist I didn't answer. Why did you not address this? Mad that I caught you in a PROVEABLE lie? If you did not like my latest rendition, why did you bother to reply? Your negative posts are what make this thread "unproductive".....-1 for replying without merit. Admit you where wrong 2 years ago...or admit your wrong now....otherwise I will not continue to address your replies as they are (as you point out) unproductive Uncool Excluding zero is not a problem if you provide solutions. 1*(2 + 3) = 1*4 + 1*7 mmh that's funny I just picked three arbitrary numbers and the equation failed to be equivalent.....lol give this one up buddy it's embarrassing. you are human that makes you my peer....even if you don't like it. yes it is clear you are done' as you have nothing further to add....please note...not one other person complained about the distributive property. I have posted links to other websites....many different one's in fact NO ONE HAS ever complained that it fails the distributive property....so yes Im done with your peer review...thank you. -1 for not claiming another human being is your peer. ALL HUMAN BEINGS are PEERS....ALL HUMAN BEINGS are equal...including me and you. Yes yes I still had it wrong...I was in a hurry....but clearly there was NO NEED for precision as I provided links for you......
  15. A Request for Peer Review

    LOL....LOL...no really....LOL sure I'll give you that if you work hard enough you can find the right combination of numbers to "force" the equation to be equal But then it would be NO good as an example of the distributive property..... Yeah I messed up here... a ( b + c ) = (a + b) * c yeah I mixed up the associate and the distributive....I meant a( b + c) = a*b + b*c sorry bout that.... As I said chose any number other than zero.....apply to equation...distributive property holds Any number other than zero see op. Here you go... http://mathworld.wolfram.com/FieldAxioms.html Let there be NO FURTHER argument about the distributive property. The given link gives our definition of the distributive property. The op follows the linked definition excluding zero. The op gives rules for 0.