conway

Strange considering you can make up what you meant by your made up words, I concede. I would offer however.... if I say to someone......"I am a lover of wisdom." what do they then think? Probably this guys a weirdo, but some will think....this guys a philosopher. while information is not efficiently exchanged, it serves its purposes....weeding out those who have no desire to engage in philosophy. Only a man who says..."this guy must be a philosopher"....will want to engage in philosophy.

John Maybe it is entirely wrong John. I wonder if you would count the number of pure equations you posted in effort to make your argument. While I am sure I have far exceeded you. My point is that semantics is never accurate when talking of mathematics. You only ever suggested semantics. At least I did more than that. You may claim none was needed to make your point. If this was so, then why does the link on #17, through axioms, state with out semantics, why you can't divide by zero. So then if I can add to these axioms with out changing these axioms, then I have clearly solved for multiplication and division by zero. Assuming the axiom holds up. But again John for you to test my axiom you would have to post an equation showing the inconsistency of the axiom. Please do. But truly I tire of the semantic arguments between you and I. All the best to you John. Thanks for your time.

thread Lol, don't get excited guys. Cleary past a certain point "making" up your own words is counter productive. Humpty Dumpty was not being absurd. To him "A nice knock down argument" was glory. There are many ways to actually define it. Further gentlemen and ladies, every word at one time was "made up". No word existed as a "fact" for definition until AFTER it already existed (maybe some exceptions few to be sure). So with in limits, "specifically" exchanging information, then making up words is fine. Sorry to have caused this to go off topic. What I mean my philosophy is "love of wisdom". Regardless of how it is actually defined. Strange I "subconsciously" took your meaning just fine, when you stated "The bananas are explosion to be trumpet". That is the derogatoriness shone through just fine. Therefore information was clearly exchanged. Beside which making me laugh hysterically. +1 sir.

Strange I certainly concede to you, that it may be that I don't really know anything. Clearly I care. I am here posting. But admitting that in the end I do not know is the first rule of philosophy. So clearly I am a philosopher. One who doesn't really know, but cares. What I do "claim" to know is only an educated guess. Is this not also true of scientist. Albeit, their education tends to be much better, and more objective. I like your humpy dumpty story very much. I agree with him as well. I can make words mean what I want, no more, no less. So long as information has been exchanged then I was successful. If however my word meanings do not successfully exchange information, well then...not helpful or useful.

Mmmh, could be I wouldn't really know, or care. Philosophy to me.... does mean "love of wisdom". I find this to be true for most philosophers I have met. While scientist feel the same way, just not towards certain fields. While theologians feel the same way, just not towards certain fields. A philosopher however is attracted to all fields. In any case, the first European scientist were philosophers (by their own accounts), so science is clearly an evolution from this. While it may exist independently now.

It is impossible to ask the questions posed, without first having philosophy. As it means "love of wisdom". Then if there is none of "this", then there is clearly none of "that." No man is a scientist with out first having a love of wisdom. We only nit pick about what is called useful "information".

Strange No.............! lol I understand. Thank you for your time and effort. I tried to explain what I mean. Q and R are unique that Q is always "placed" into an R. There can be varying amounts of Q and R. Sometimes it is that Q is placed subtraction-ally, sometimes it is placed additionally. hopefully this is clearer, if not, again thank you Strange for your time and effort. 1 = (1q,1r) = 1q into 1r = 1 2 = (1q,1q,1r,1r) = 2q into 2r = 2 3 = (1q,1q,1q,1r,1r,1r) = 3q into 3r = 3 so on.... x=2 y=3 X * Y = (Xq,Xr) * (Yq,Yr) = (Xq,Yr) = (Xq into Yr) = (2q into 3r) = (2+2+2) = 6

John Why? Why then is it John that programming a computer this way is the "wrong" way to program it. I have done my best to answer your lawyer question, pie question, and fast car and garage question. It is that you do not like, understand, and or accept these answers. For you and I to continue is not very productive. I hate to say that but....

Strange Awesome were making progress I can concede that your way of representation works perfect. The problem is that each and every result of multiplication leaves us with two sums. And when the variables are not zero, both q and r are the same therefore we can just pick one, and always be correct. This will not be the case with zero. How then do we decided which sum to pick in every case 0 involved or otherwise. I agree we must define q and r. Since clearly they are not numbers but rather compose numbers. Suffice to say that... (q,r) = For any A there is a q divided equally an r number of times then added. 7=(q,r)=(7q,7r)=(1+1+1+1+1+1+1)

Studiot Good stuff....I would have thought they would have had a better way than a program talking about subtraction......! John Here in is a good example......you keep saying just like the computer says "The number is 1, and there is nothing to subtract form" I got to loop out man! Im thinking.......Just reprogram the computer to say ok If I have 1, and nothing to subtract, then I have 1. Strange Ok, I suppose, If this is what it takes to make it work. I am then assuming that (a,a)=a and (0,1)=0 But your suggesting this in not commutative right......let me see..... x * y = (x1,x2)*(y1,y2) = (x1 * y2, x2 * y1) = z y * x = (y1,y2)*(x1,x2) = (y2 * x2, y2 * x1) = z This doesn't work with zero at all.....actually If I make the following statement (x1 * y2 , x2 * y1).....then I have two actually numbers in the parenthesis. This is saying x * y = m,n so that if I multiplied by zero I will always get... (x1 * y2 , x2 * y1).....which is actually 0 and 1. "lol but not undefined" z does not equal (x1*y2,x2*y1) = (qr,rq) z must equal (q,r) or (r,q) addition and subtraction are the only operators used when q and r are put back together to make a number again.

John It would seem I can't help myself.... "How do you divide a pie amongst zero plp. Hint, you can't" I take a pie in my hands. I do nothing. Then I have successfully divided by zero. And when I am done. I have a pie in my hands.

John Lol... That's exactly how I feel about those who claim A/0 = undefined That's not an answer! Strange For every A in S there exist , q and r, such that any A in operation of multiplication or division is only representing q or r in any given equation. Allowing that in division q is always first r is always second. Allowing that.... q for 0 = 0 r for 0 = 1 0 = (0,1) q for A = A r for A = A A = (A,A)

Studiot Yes I was dismissive. Wrongly so. However I still believe that, the "framework" for vectors and tuple's must in the end bend to the framework of "relative mathematics". It is that everything with in it, is in regards to arithmetic. The "base" of mathematics. So that if it is correct it is the limbs that move. So naturally it seems to me the approach is to find flaws in it's logic when applied to vectors and tuples "higher branches of math". Shortest path to this is to ask people that already know, rather than study to the required amount. Requiring much time. I can honestly say as far as arithmetic/algebra is concerned the idea of "relative mathematics" has no contradictions "that I am aware of".

Studiot I must tell you that the link you posted in #17 made this exchange of information between myself and others here possible. I thank you. Also to say that yes, I lacked the proper terminology in my descriptions. It was your link that allowed me to write the axiom, there by putting it in proper terminology....if still not quite pinned down. Also I point out that the axiom I posed was not in my post in "speculations" at all. So truly Studiot, thank you for your time.

John I have come to understand that from your perspective it does not help with dividing by pie, lawyers, and fast cars. But this is not my opinion. As I have stated to you. Any time you write a sentence talking of * and /, the question always is...."how many are in my hand". Such that...... A lawyer has 120k, and no one to give it to, he still has a 120k. A have a pie, I divided it by zero, I do noting, I still have a pie. I have a fast car, I have no garages to divide it into, I still have a fast car.

Strange "For example I thought that 12 = (q,r), where q=12 and r =12 " This is correct. I didn't mean to imply other wise? I gave examples 12 = (12q,12r) 4 = (4q,4r) etc.... I was sloppy when I stated it like this A = (q,r) = (Aq,Ar) I meant A = (Aq,Ar) Also I am prepared to talk of integers and rational and irrationals. Also 0 = (0q,1r) is correct. I will never mention the "nonsense" you spoke of again. Thank you for your time.

Strange Again thank you, I must admit I am also confused on the vectors and tuple examples you give. As I have little experience with these things. Suffice to say that I can see relationships between both ways of representing it. I would have to say that .... N = (q,r)...where q and r are factors of N is a much better representation of this idea than N = (n1,n2) So that at this point I can say the following again. Allowing the first above example to be our continued represented form. 0 = (q,r) = (0q,1r) A = (q,r) = (Aq,Ar) Where (q,r) is defined as a value inside of a space. Where q=value=quantities of values r=space=quantities of dimensions Such that 4=(q,r)=(4q,4r)=(1,1,1,1)=four single values in four single spaces. 1=(q,r)=(1q,1r)=(1)=one single value in one single space. 0=(q,r)=(0q,1r)=(0)=one single undefined value in one single space. so that in multiplication it is that A * X = (Aq,Xr) = Z

Strange The difficulty here is describing what a (5,4) is, or a (4,5). The reality is that x1 for x is value, x2 for x is space. While y1 for y is value, while y2 for y is space. It is that in all cases the value is placed equally into all the spaces then added, if multiplication. It is that in all cases the value is placed equally divisionally into all spaces, than all values subtract but one, if division. A visual example. (5,4)...literally means...... 5+5+5+5...or The value 5 in the space of 4 (4,5)...literally means.... 4+4+4+4+4...or the vlaue4 in the space of 5. It is as you said (4,5) are not numbers, not exactly vectors either. it again is that one is a value, and one is a space. another representation..... x=4 y=5 (x1,x2) * (y1,y2) = (x1,y2) = (4+4+4+4+4) = 20 or commutative (y1,y2) * (x1,x2) = (y1,x2) = (5+5+5+5) = 20

Strange Very well I will use the notions and variables you have suggested. However you still have not quite accurately represented it. So that.... X * Y = (x1,x2) * (y1,y2) = (x1,y2) = Z Y * X = (y1,y2) * (x1,x2) = (x1,y2) = Z X * Y = (x1,x2) * (y1,y2) = (x2,y1) = Z Y * X = (y1,y2) * (x1,x2) = (x2,y1) = Z commutative. X * Y = Z To have a sum other than a variable then yes, I must know if any variables are or are not zero. This is currently the case anyways.

Strange Thank you! I have only a few changes to make to your previous post ((z1,z2)=A), this only exists in operation of multiplication and division. X * Y = (z1,z2) * (z1,z2) = z1(from X) * z2(from Y).....(so then commutative.) = (z1,z2) = Z X / Y = (z1,z2) / (z1,z2) = z1(from X, always first) / z2(from Y, always second )......(not commutative) = (z1,z2) = Z

Ajb, Greg, John I seem to have trouble following you guys here. If... x/0=x x/1=x x/0=x/1 x=x I never arrive at an equation where 0=1. Maybe if I suggest that z2 of zero is =1, and z1 of 1 is = 1, so that in the above equations 0 is only representing z2(of zero), which is = 1. I think on a more important note that if we apply the above equtions and apply current mathematics we do arrive at an inconsistency x/0 = undefined x/1 = x undefined = x This we know is not correct.

Strange x/0 = x x/1 = x I agree that x/0=x/1 1/0=1/1 1=1 Im thinking you made a typo arriving at 0=1, following the current rules of mathematics the equations would have been.... undefined=1.....which we know is not equal. Ajb No equivalency sign was used in the two separate equations given by John "originally". If say... he had of said x/0=x/1......which is true (assuming x does not equal 0). Then we would have to consider the equivalency. But as I pointed out then we must assume that x does not equal zero. So then no contradiction. I believe ajb I read a post somewhere, in reach you replied to a person who suggestd a free energy maching that broke all the laws of physics. And that such a device would have to be very exotic indeed. Any time only variables exist such as x/a, we must at least know if the variables are or are not zero. But isn't this true anyways. That is unless only variables as sums are needed.

John just because in one equation a=0, and in another a=1 does not mean that 1=0. I am still not following your claim to contradiction. "its clear from the first equation a=0" "its clear from the second equation a=1" "thus 1=0" "As I said that's a contradiction." Really? Apologies for the typo

Strange According to the axiom I gave. Any A in operation of multiplication (through extension division) is only representing z1 or z2 in any equation. So that no eqution every posses z1 and z1, or z2 and z2. It is and always must be z1 and z2, one from each A given. John x/0=x x/a=x It is assumed in both equations that x does not equal 0. So then there is no contradiction. Or maybe there is and I am not seeing it. Could you please give me an example via pure mathematics as to the contradiction? To All For every A in S there exist a z1 and a z2, constituting A, such that any A in operation of multiplication, or division is representing only z1 or z2 in any given equation. Allowing that in division z1 is always first z2 is always second. Allowing that... z1 for 0 = 0 z2 for 0 = 1 z1 for A = A z2 for A = A 0=(z1,z2)=(0,1) A=(z1,z2)=(A,A)

MigL Not that it can be proven but.......there is not "nothing" there. Something is. The apples went in. They may no longer be apples, but something is there. Additionally I think a separation should be noted. While it is akin to apples divided by 0. It is also different. That is mass's larger than any apples , and spaces smaller than any 0.