Define * on Z by a * b = a – b + ab


Define $*$ on $Z$ by $a * b=a-b+a b$. Show that $*$ is a binary operation on $Z$ which is neither commutative nor associative.


* is an operation as $a * b=a-b+a b$ where $a, b \in Z$. Let $a=\frac{1}{2}$ and $b=2$ two integers.

$\mathrm{a} * \mathrm{~b}=\frac{1}{2} * 2=\frac{1}{2}-2+\frac{1}{2} \cdot 2 \Rightarrow \frac{1-4}{2}+1=\frac{-3+2}{2} \Rightarrow \frac{-1}{2} \in \mathrm{Z}$

So, $*$ is a binary operation from $Z \times Z \rightarrow Z$.

For commutative,

$\mathrm{b}^{*} \mathrm{a}=2-\frac{1}{2}+2 \cdot \frac{1}{2}=\frac{4-1}{2}+1 \Rightarrow \frac{3+2}{2}=\frac{5}{2} \in \mathrm{Z}$

Since a*b ≠ b*a, hence * is not commutative operation.

Again for associative,

$a^{*}\left(b^{*} c\right)=a^{*}(b-c+b c)$

$=a-(b-c+b c)+a(b-c+b c)$

$=a-b+c-b c+a b-a c+a b c$

$(a * b) * c=(a-b+a b) * c$

$=a-b+a b-c+(a-b+a b) c$

$=a-b-c+a b+a c-b c+a b c$

As $a *(b * c) \neq(a * b) * c$, hence $*$ not an associative operation.


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