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Let $A=R\times R$ and * be a 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 A. Also find the inverse of every element $(a,b)\in A$ .

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Given $A=R\times R$ and * is a binary operation on A defined by $(a,b)*(c,d)=(a+c,b+d)$

Commutativity :

Let $(a,b),(c,d)\in A$ .

Then $(a,b)*(c,d)=(a+c,b+d)=(c+a,d+b)=(c,d)\ast (a,b)$

Hence, $(a,b)*(c,d)=(c,d)\ast (a,b)$

Therefore, * is commutative.

Associativity :

Let $(a,b),(c,d),(e,f)\in A$ . Then we have $\left [ (a,b)*(c,d) \right ]*(e,f)=(a+c,b+d)*(e,f)$

$=(a+c+e,b+d+f)$

$=((a+c)+e,(b+d)+f)$

$=(a+(c+e),b+(d+f))$

$=(a,b)*(c+e,d+f)$

$=(a,b)*\left [ (c,d)*(e,f) \right ]$

Hence $\left [ (a,b)*(c,d) \right ]*(e,f)=(a,b)*\left [ (c,d)*(e,f) \right ]$

Therefore, A is associative.

Now, let (x,y) be identity element for * on A.

Then $(a,b)*(x,y)=(a,b)\Rightarrow (a+x,b+y)=(a,b)$

$\Rightarrow a+x=a, b+y=b \Rightarrow x=0,y=0$

So, the identity element for the binary operation is (0,0).

Let inverse of element (a,b) be (e,f) so $(a,b)\times (e,f)=(0,0)$

$\Rightarrow (a+e,b+f)=(0,0)$

$\Rightarrow a+e=0,b+f=0\Rightarrow e=-a,f=-b$

$\therefore$ Income od (a,b) is (-a,-b)

Posted by

Ravindra Pindel

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