lever numbers tommy1729 Ultimate Fellow Posts: 1,924 Threads: 415 Joined: Feb 2009 03/20/2023, 12:16 AM This idea might have occured here before and it also relates to many known ideas. In particular : mediant and weighted mediant. center of mass. mass point geometry. " extended reals " and "hyperoperator numbers " and such terms , which are not universally agreed upon hence the " ". I might have mentioned it before. It also related to numerical methods and averaging. and some number theory. So forgive me if I am repeating myself or others. Often we only see its sum defined but it can be extended. And it has nice properties ! commutative , distributive , associative , solvable in a sense etc although it may not agree with the generalized binomium theorem. ( it works for the binomium case of squares for sure see later ) I call them lever numbers because they compute the balancing point of levers for weights and distances. That is imo the best definition. With special thanks to Archimedes. ( I also considered them as norms for other numbers but lets ignore that complication for now ) ( do not confuse with my spiral numbers or hypercomplex ideas ) Ok time to define and rationalize : For cases where we do not devide by zero : (a,p) + (b,q) = ( a+b , (ap + bq)/(a+b) ) a,b are called weights and p and q positions. examples : (1,0) + (2,1) = (3, (1*0 + 2*1)/3 ) = (3,2/3) (1,1) + (2,2) = (3, (1*1 + 2*2)/3 ) = (3,5/3) = (3, 1 + 2/3) (notice the distance shift, this is why I picked this example. Basically the lever and weights are all placed a unit further ) conjecture |p,r|/|q,r| = b/a [p - (ap + bq)/(a+b)]/[q - (ap + bq)/(a+b)] = [pa + pb - ap - bq]/[qa + qb - ap - bq] = [pb - bq]/[-pa + aq] = b(p-q)/[a(q-p)] = b/(-a) so |p,r|/|r,q| = b/a Qed ( this shows the mediant and mass point geometry interpretations agree ) *** (a,x) * (b,q) = (ab,q) (a,x) = a *** c [(a,p) + (b,q)] = ( c(a+b) , (ap + bq)/(a+b) ) c [(a,p) + (b,q)] = ( c(a+b) , (c/c)(ap + bq)/(a+b) ) *** (c,z) * [(a,p) + (b,q)] = ( c(a+b) , z<(ap + bq)/(a+b)> ) z<(ap + bq)/(a+b)> = (a z

+ b z)/(a+b) *** (c,z)*(a,p) = (ca,zp) (1,1)*(a,p) = (a,p) (c,1)*(a,p) = (ca,p) (c,p) + (a,p) = (a+c,p) *** This is how the pairs are defined. you can take the variables real or complex or ... or increase the variables or the dimensions etc. But I wanted to start simple with a couple of 2 variables. Notice the product is designed to be distributive over addition. And also it follows (a,p)^0 = (1,1) what is the multiplicative unit. ( the idea of adding mod to the party occured ) ok now how about squares and the (a+b)^2 = a^2 + 2ab + b^2  formula ? ( binomium square case ) binomium : [(a,p) + (b,q)]^2 = ( (a+b)^2 , (ap + bq)^2/(a+b)^2 ) = (a^2,p^2) + (2ab,pq) + (b^2,q^2) = (a^2 + 2ab, (a^2 p^2 + 2abpq)/(a^2 + 2ab) ) + (b^2,q^2) = ( (a+b)^2 , [(a^2 + 2ab)*(a^2 p^2 + 2abpq)/(a^2 + 2ab) + b^2 q^2]/(a+b)^2 ) so (a^2 p^2 + 2abpq) + b^2 q^2  must = (ap + bq)^2 = a^2 p^2 + 2 apbq + b^2 q^2. and it does ! And probably this ( binomium ) works for higher powers too ( by induction ). How about taylor or fourier ? analytic analogues ? Iterations ? hyperoperators ? Defining exp , log , sine etc ? What do you think ? Regards  tommy1729 Tom Marcel Raes « Next Oldest | Next Newest »

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