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### Problem 2

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

(1)(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 $b*e \neq e*b$