Define * on Z by a * b = a + b - ab. Show that


Define $*$ on $Z$ by $a * b=a+b-a b$. Show that $*$ is a binary operation on $Z$ which is commutative as well as 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.

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

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

For commutative,

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

Since $a * b=b * a$, hence $*$ is a commutative binary 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 $a *(b * c)=(a * b) * c$, hence $*$ an associative binary operation.


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