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# Let $A={{R}_{0}}\times R$ where ${{R}_{0}}$ denotes the set of all non-zero numbers. A binary operation $'O'$ is defined on $A$ as follows: $\left( a,b \right)O\left( c,d \right)=\left( ac,bc+d \right)$ for all $\left( a,b \right),\left( c,d \right)\in {{R}_{0}}\times R$. Find the identity element in $A$.  Verified
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Hint: Take any general element of ${{R}_{0}}$ and find the identity element of $R$ by forming the equations using properties of the identity element and then solve them. Similarly take any general element of $R$ and find the identity of ${{R}_{0}}$ by forming equations using properties of the identity element.

We have the set $A={{R}_{0}}\times R$ where ${{R}_{0}}$ denotes the set of all non-zero numbers. We have a binary relation on $A$ defined as $\left( a,b \right)O\left( c,d \right)=\left( ac,bc+d \right)$ for all $\left( a,b \right),\left( c,d \right)\in {{R}_{0}}\times R$. We have to find the identity element in $A$.
Let’s assume that the identity element of $A$ is of the form $\left( x,y \right)$.
We know that any identity element has the property that any element operated with identity element returns the element itself, i.e., for any $\left( a,b \right)\in A$, we have $\left( a,b \right)O\left( x,y \right)=\left( x,y \right)O\left( a,b \right)=\left( a,b \right)$.
We know that for all $\left( a,b \right),\left( c,d \right)\in {{R}_{0}}\times R$, we have $\left( a,b \right)O\left( c,d \right)=\left( ac,bc+d \right)$.
Thus, we have $\left( a,b \right)O\left( x,y \right)=\left( ax,bx+y \right)=\left( a,b \right)$.
Comparing the terms on both sides of the equation, we have $ax=a,bx+y=b$.
We can clearly see that the solution of the above equations is $x=1,y=0$.
Hence, we have $\left( x,y \right)=\left( 1,0 \right)$ as the identity of the given set $A={{R}_{0}}\times R$ where ${{R}_{0}}$ denotes the set of all non-zero numbers .