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Let $A=N \times N$ and $*$ be the binary operation on A defined by $(a, b) *(c, d)=(a+c, b+d)$.
Show that * is commutative and associative. Find the identity element for ${ }^*$ on $\mathbf{A}$, if any.
Show that * is commutative and associative. Find the identity element for ${ }^*$ on $\mathbf{A}$, if any.
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Verified Answer
$\mathrm{A}=\mathrm{N} \times \mathrm{N}$ Binary operation $*$ is defined as $(\mathrm{a}, \mathrm{b}) *(\mathrm{c}, \mathrm{d})$ $=(\mathrm{a}+\mathrm{c}, \mathrm{b}+\mathrm{d})$
(a) $\operatorname{Now}(c, d) *(a, b)=(c+a, d+b)=(a+c, b+d)$
$\Rightarrow(a, b) *(c, d)=(c, d) *(a, b)$
$\therefore$ This operation * is commutative
(b) $\operatorname{Next}(\mathrm{a}, \mathrm{b})^*[(\mathrm{c}, \mathrm{d}) *(e, f)]=(\mathrm{a}, \mathrm{b}) *(\mathrm{c}+\mathrm{e}, \mathrm{d}+\mathrm{f})$
$=((a+c+e),(b+d+f))$
and $[(a, b) *(c, d)]^*(e, f)=(a+c, b+d) *(e, f)$
$=((\mathrm{a}+\mathrm{c}+\mathrm{e}, \mathrm{b}+\mathrm{d}+\mathrm{f}))$
$\Rightarrow(a, b) *[(c, d) *(e, f)]=[(a, b) *(c, d)] *(e, f)$
$\therefore$ The binary operation given is associative (c) Identity element does not exists.
(a) $\operatorname{Now}(c, d) *(a, b)=(c+a, d+b)=(a+c, b+d)$
$\Rightarrow(a, b) *(c, d)=(c, d) *(a, b)$
$\therefore$ This operation * is commutative
(b) $\operatorname{Next}(\mathrm{a}, \mathrm{b})^*[(\mathrm{c}, \mathrm{d}) *(e, f)]=(\mathrm{a}, \mathrm{b}) *(\mathrm{c}+\mathrm{e}, \mathrm{d}+\mathrm{f})$
$=((a+c+e),(b+d+f))$
and $[(a, b) *(c, d)]^*(e, f)=(a+c, b+d) *(e, f)$
$=((\mathrm{a}+\mathrm{c}+\mathrm{e}, \mathrm{b}+\mathrm{d}+\mathrm{f}))$
$\Rightarrow(a, b) *[(c, d) *(e, f)]=[(a, b) *(c, d)] *(e, f)$
$\therefore$ The binary operation given is associative (c) Identity element does not exists.
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