Asked by Sinachi
A binary operation * is defined on the set of rational numbers by m*n=m^2-n^2/2mn, m, n is not equal to 0. Find -3*2. Is * associative?
Answers
Answered by
oobleck
m*n = m^2 - n^2/(2mn)
so
-3*2 = (-3)^2 - 2^2/(2(-3)2) = 9 - 4/-12 = 9 1/3
a*(b*c) = a^2 - (b*c)^2/(2a(b*c))
= a^2 - (b^2 - c^2/(2bc))/(2a(b^2 - c^2/(2bc)))
= a^2 - 1/(2a)
(a*b)*c = (a*b)^2 - c^2/(2(a*b)*c)
= (a^2 - b^2/(2ab))^2 - c^2/(2(a^2 - b^2/(2ab))*c)
which is certainly not the same as
(a*b)^2 - 1/(2(a*b)) = (a^2 - b^2/(2ab))^2 - 1/2(a^2 - b^2/(2ab))
so, * is not associative
so
-3*2 = (-3)^2 - 2^2/(2(-3)2) = 9 - 4/-12 = 9 1/3
a*(b*c) = a^2 - (b*c)^2/(2a(b*c))
= a^2 - (b^2 - c^2/(2bc))/(2a(b^2 - c^2/(2bc)))
= a^2 - 1/(2a)
(a*b)*c = (a*b)^2 - c^2/(2(a*b)*c)
= (a^2 - b^2/(2ab))^2 - c^2/(2(a^2 - b^2/(2ab))*c)
which is certainly not the same as
(a*b)^2 - 1/(2(a*b)) = (a^2 - b^2/(2ab))^2 - 1/2(a^2 - b^2/(2ab))
so, * is not associative
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