Question

Let $A=R\times R$ and $\ast$ be a binary operation on A defined by $(a,b)\ast (c,d)=(a+c,b+d)$. Show that $\ast$ on A. Also, find the inverse of every element $(a,b)\in A$.

Answer 1

Commutativity:
Let $(a,b),(c,d)\in R\times R$
Now, $(a,b)\times (c,d)=(a+c,b+d)=(c+a,d+b)$
[commutativity holds in for $ast$]
$=(c,d)\ast (a,b)$
i.e., $\ast$ is commutative.
Associativity:
Let $(a,b),(c,d),(e,f)\in R\times R$
Now, $\left\{\right(a,b)\ast (c,d\left)\right\}\ast (e,f)=(a+c,b+d)\ast (e,f)$
$=(a+c+e,b+d+f)$....(1)
Also, $(a,b)\ast \left\{\right(c,d)\ast (e,f\left)\right\}=(a,b)\ast (c+e,d+f)=(a+c+e,b+d+f)...\left(2\right)$
From (1) and (2), we have
$(a,b)\ast (c,d))\ast (e,f)=(a,b)\ast ((c,d)\ast (e,f))$
i.e., ${}^{\prime }{\ast }^{\prime }$ is associative.
Existence of identity element:
Let $(e,f)$ be identity element for $\ast$ in $R\times R$
$(a,b)\ast (e,f)=(a,b)=(e,f)\ast (a,b)$
$\Rightarrow (a+e,b+f)=(a,b)=(e+a,f+b)$
$\Rightarrow a+e=a$ and $b+f=b$
$\Rightarrow e=0,f=0$, also $(0,0)\in R\times R$
Hence, $(0,0)$ is the identity element for $\ast$ on A.
(i) for inverse of $(a,b)$
Let $(x,y)$ be inverse of $(a,b)$
$\Rightarrow (a,b)\ast (x,y)=(0,0)$
$\Rightarrow (a+x,b+y)=(0,0)$
On equating, we have
$a+x=0;b+y=0$
$\Rightarrow x=-a;y=-b$
Hence, inverse of $(a,b)$ is $(-a,-b)$