  # conway

Senior Members

261

## Everything posted by conway

1. Oh I take it in strides...ever trying to improve....How do you take the fact your Bigjerk?
2. 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. 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. 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. 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. 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. 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. 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. 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. No number tables...no properties. No axioms change (except) when involving zero. The following projection operators allow for no further axioms...... $0 = \left ( \begin{matrix} 0.z_1 \\ 0.z_2 \end{matrix} \right )$ 0.z1 = 0 0.z2 = 1 $P_1 0 = (1, 0) ~ \left ( \begin{matrix} 0.z_1 \\ 0.z_2 \end{matrix} \right ) = 1 \cdot 0.z_1 + 0 \cdot 0.z_2 = 0.z_1$ $P_2 0 = (0, 1) ~ \left ( \begin{matrix} 0.z_1 \\ 0.z_2 \end{matrix} \right ) = 0 \cdot 0.z_1 + 1 \cdot 0z_2 = 0.z_2$ 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. 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. 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.
15. 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.
16. I have actually. I don't understand you at all on this matter. For example... This is the equation you presented me with... 1*(0 + 0)(as z2) = 1*0(as z2) + 1*0(as z2) This equation does NOT work for the distributive property with ANY number, 0 or otherwise. "change 0 to any other number" "do the math" "equation fails" In NO cases is the equation EVER equivalent (our debate about the adding of the 0's or otherwise). Therefore it is NOT an argument or a representation of the distributive property. This is the OFFICAL equation showing the distributive property NOTE any number OTHER than zero may be chosen and the equation holds true. (with-in this idea)(outside this idea) In the cases involving a,b,c being zero....see op. a ( b + c ) = (a + b) * c Thank you for your consideration. I understand our differences on this matter.
17. Strange, Studiot, Uncool I have tried to take into account the points brought forth by all of you. Perhaps a quick glance....would not be to much to ask? 0 = (0.z1) , (0.z2) (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
19. Please forgive the members of this forum. They can be very rude. Cell phones emit waves not particles. Your thought process was to consider the phones messages like particles (little bullets). But they are waves. While waves can interfere the rules are much different for how waves behave as opposed to particles. Consider if you dropped a stone "your phone message" into a pool. The "wave" or "ripple" would extend in "all" directions. Further here is some info on the behavior of waves v.s. particles. Your post showed a large amount of creative thinking I liked that. +1 http://physics.ucr.edu/~wudka/Physics7/Notes_www/node64.html
20. Studiot It took some time to find this..... 1165 Genius Senior Members 1165 7301 posts Location: Somerset, England Report post Posted May 31, 2015 First the good news to encourage you. This idea is sound, but really in the province of applied maths or physics, rather than pure maths. This is quoting you from two years ago. (x,x) = the first x is value, the second x is space.....you agreed this particular idea was sound. I have merely searched for a better way to communicate it. Perhaps I have failed. Bare with me Studiot....I think in time we can come to some form of understanding and hopefully pin this thing down and make some real progress.
21. Studiot First let me show you that I am fully listening to you. You demand an explanation for HOW (x equates to (x,x)). I have told you as an answer that I do not have to answer. This is an axiom. However I would love to answer......(philosophically) All numbers are composed of a given value inside of a given space. 25 = 25 values each inside of 25 spaces how we arrive at ( x = (x,x)) the value is "placed" into the space observe... value of three = (1,1,1)....NOT the number 3 or the number 1....any symbol here will actually do. space of three= (_,_,_)....Not the number three.....see the "three" empty spaces....this is all abstract of course. each value is "placed" into each "space" so then the number 3 is 3 = (1,1,1) so then the number 1 is 1 = (1) 0 = (0) so then (_) = the space of 1 and the space of 0 if a single value is "placed" into this space... it becomes the number 1. If 0 value is "placed" into this space it becomes the number 0 in binary multiplication "one" symbol represents ONLY value. "one" symbol represents only space. But all this is unnecessary....the table is an AXIOM...I understand you do not agree or like it. This is why I have asked you repeatedly to address the "properties" version. That is why there is two version...to help with communication and axiomatic belief issues. Strange Thank you for going out of your way on this point. I understand you do not agree with me....but I appreciate your help in trying to resolve this particular issue with Studiot...I hope that the two of us might still make progress.
22. Again I did answer..... you just didn't like the answer. x is a number (x,x) are two numbers that arbitrarily compose x This was all in the op. Or (x,x) is a value, and a space.(respectively) Or you can perform this idea without using the table at all....in case you don't like the table....which is your case.....so then how about addressing the properties version of this idea since you don't agree with my answers for x and (x,x)....and the equal sign that is between them. Look....You know very well that I can not define x = (x,x) without using philosophy. Which would get me in greater trouble. Further the table I gave in the op is a axiomatic table. I don't have to define ANYTING inside the axioms. You either take the axiomatic table as self inherently true or not. (I do have to define the math that follows...but NOT the axiom.)(define for my why anything multiplied by 1 is 1, or 0 by 0....they are self inherently true...there is no mathematical answer for these question. Only philosophical answers.) Your post had nothing to do with me and everything to do with the idea thank you. ....+1
23. When the hand that feeds you also slaps you...you may expect a bite. I answered your question...for a fact....you just chose not to accept my answer as valid. Which is fair...but claiming I did NOT answer it is not fair. Return to topic...stay on topic...or cease your reply's...or expect a -1 for a continued tirade against me.
24. If I half listen or not... is not anyone's point. You make this commit to be a troll. I feel I have always listened to others. Just because I don't agree with you on the validity of my table does NOT mean I did not listen to you. In fact it is my judgment that you only half listened to me. While uncool gave it his/her best.....yet had no constructive critique... as far as I was concerned. After all the equation we were discussing was equivalent. Thereby validating the distributive property under my axioms. -1 for making negative passive aggressive needless remarks about me. Further more take the equation uncool gave me... 1*(0 + 0)(as z2) = 1*0(as z2) + 1*0(as z2) Now use any number other than 0 in place of 0 Fact....it fails the distributive property Not sure where uncool even got this equation as an argument for the distributive property...but it clearly fails.
25. If the expression... 1*(0 + 0)(as z2) = 1*0(as z2) + 1*0(as z2) must be equivalent... and if I had it right the "first" time and I must add the two zero's.... thereby making my equation NOT equivalent....and all I have to do is state something to the affect of ... "Let any equality possessing multiple zero's consider the following application of z1 and z2" then it is extremely useful.... I will ponder what you have shared with me....thank you.
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