CTK Exchange Front Page Movie shortcuts Personal info Awards Reciprocal links Terms of use Privacy Policy Cut The Knot! MSET99 Talk Games & Puzzles Arithmetic/Algebra Geometry Probability Eye Opener Analog Gadgets Inventor's Paradox Did you know?... Proofs Math as Language Things Impossible My Logo Math Poll Other Math sit's Guest book News sit's Recommend this site
|Store|

CTK Exchange

 Subject: ""Weighted" Multiplication" Previous Topic | Next Topic
 Conferences The CTK Exchange High school Topic #298 Printer-friendly copy     Email this topic to a friend Reading Topic #298
Ramsey_KJ
Member since Sep-23-04
Sep-24-04, 06:37 AM (EST)

""Weighted" Multiplication"

 Has anyone seen this before? Define the operation A @ B as equal to A*B^2 + (B^2-1)/8 and note that (A @ B) @ C equals A @ (B*C). For instance if A= 1 and B=C = sqrt(3) then we have (1 @ sqrt(3)) @ sqrt(3) = (3 + (3-1)/8) * 3 + (3-1)/8 = 10 = 1*9 + (9-1)/8 = 1 @ 3 Also (N @ B) @ 1/B = N @ 1 = N. This came to me while considering triangular numbers that are squared. For instance 1 = T(1) - 0 and 6^2 = T(8) - 0. 0 @ 3 = 1, thus (1*3)^2 = T(4) - 1 and (6*3)^2 = T(25) - 1 note that 4 = 1*3 + (3-1)/2 and 25 = 8*3 + (3-1)/2. 1 @ 3 = 10, thus (1*9)^2 = T(13)-10 and (6*9)^2 = T(76)-10. note that 13 = 1*9 + (9-1)/2 and 76 = 8*9 + (9-1)/2. 1 @ 5 = 28, thus (1*15)^2= T(22)-28 and (6*15)^2 = T(127)-28 note that 22 = 1*15 +(15-1)/2 and 127 = 8*15 + (15-1)/2 Have a Good DayKJ Ramsey

JJ
guest
Sep-25-04, 00:23 AM (EST)

1. "RE: "Weighted" Multiplication"
In response to message #0

 < note that (A @ B) @ C equals A @ (B*C) >. is FALSE. Example :A=2 ; B=3 ; C=4D = A*B^2 + (B^2-1)/8 = 2*(3^2)+(3^2-1)/8 = 2*9+1 = 19(A @ B) @ C = D @ C = D*C^2 + (C^2-1)/8 = 19(4^2)+(4^2-1)/8 = 305E = B*C = 3*4 = 12A @ (B*C)= A @ E = A*E^2 + (E^2-1)/8 = 2(12^2)+(12^2-1)/8 = 256+143/8 = 273.875Hence (A @ B) @ C NOT equals A @ (B*C)

Ramsey_KJ
guest
Sep-26-04, 11:13 AM (EST)

2. "RE: "Weighted" Multiplication"
In response to message #1