Problem:
Define xβy to be β£xβyβ£ for all real numbers x and y. What is the value of
(1β(2β3))β((1β2)β3) ?
Answer Choices:
A. β2
B. β1
C. 0
D. 1
E. 2
Solution:
First observe that 2β3=β£2β3β£=β£β1β£=1 and 1β2=β£1β2β£=β£β1β£=1. Therefore
(1β(2β3))β((1β2)β3)=1β1β1β3=β£1β1β£ββ£1β3β£=β£0β£ββ£β2β£=(A)β2β.
Note: This shows that the β operation is not associative.
The problems on this page are the property of the MAA's American Mathematics Competitions