HW2 Problem 2

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### Problem 2

Consider the binary operation $*$ defined on $S=\{a,b,c,d,e\}$ by the table below.

(1)
\begin{array} {c||c|c|c|c|c} * & a & b & c & d & e \\ \hline a & a & b & c & b & d \\ \hline b & b & c & a & e & c \\ \hline c & c & a & b & b & a \\ \hline d & b & e & b & e & d \\ \hline e & d & b & a & d & c \\ \end{array}

(a) Compute $b*d, c*c,$ and $[(a*c)*e]*a$

(b) Compute $(a*b)*c$ and $a*(b*c)$. Can you say on the basis of these computations whether $*$ is associative?

(c) Compute $(b*d)*c$ and $b*(d*c)$. Can you say on the basis of these computations whether $*$ is associative?

(d) Is $*$ commutative? Why?

Solution

(a)
$b*d = e$
$c*c = b$
$[(a*c)*e]*a = [c*e]*a= a*a =a$

(b)
$(a*b)*c = b*c = a$
$a*(b*c) = a*a = a$
You can't tell if the entire table is associative but $S=\{a,b,c\}$ is.

(c)
$(b*d)*c = e*c = a$
$b*(d*c) = b*b = c$
No, the table is not associative because of the above problems.

(d)
No, the table is not commutative because $b*e \neq e*b$

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