CTK Exchange
Front Page
Movie shortcuts
Personal info
Awards
Reciprocal links
Terms of use
Privacy Policy

Interactive Activities

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

Manifesto: what CTK is about |Store| Search CTK Buying a book is a commitment to learning Table of content Things you can find on CTK Chronology of updates Email to Cut The Knot Recommend this page

CTK Exchange

Subject: ""Weighted" Multiplication"     Previous Topic | Next Topic
Printer-friendly copy     Email this topic to a friend    
Conferences The CTK Exchange High school Topic #298
Reading Topic #298
Ramsey_KJ
Member since Sep-23-04
Sep-24-04, 06:37 AM (EST)
Click to EMail Ramsey_KJ Click to view user profileClick to add this user to your buddy list  
""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 Day
KJ Ramsey


  Alert | IP Printer-friendly page | Reply | Reply With Quote | Top
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=4
D = 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 = 305
E = B*C = 3*4 = 12
A @ (B*C)= A @ E = A*E^2 + (E^2-1)/8 = 2(12^2)+(12^2-1)/8 = 256+143/8 = 273.875
Hence (A @ B) @ C NOT equals A @ (B*C)


  Alert | IP Printer-friendly page | Reply | Reply With Quote | Top
Ramsey_KJ
guest
Sep-26-04, 11:13 AM (EST)
 
2. "RE: "Weighted" Multiplication"
In response to message #1
 
   Check your math. In each case the answer should be 305.875


  Alert | IP Printer-friendly page | Reply | Reply With Quote | Top

Conferences | Forums | Topics | Previous Topic | Next Topic

You may be curious to have a look at the old CTK Exchange archive.
Please do not post there.

|Front page| |Contents|

Copyright © 1996-2018 Alexander Bogomolny

71493653


Google
Web CTK