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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.

Complete step-by-step answer:

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 \[A\].

Letâ€™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 .

Note: A binary operation is a calculation that combines two elements of a non empty set to produce another element. A binary operation on a non empty set is an operation whose two domains and co-domains are the same set. Addition, subtraction and multiplication are the common binary operations. As the result of performing an operation on a set lies in the set itself, we say that the given binary operation is closed. Binary operations may or may not be associative and distributive. The inverse of any element of a set under binary operation is the element such that the element operated with its inverse gives the identity element.

Complete step-by-step answer:

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 \[A\].

Letâ€™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 .

Note: A binary operation is a calculation that combines two elements of a non empty set to produce another element. A binary operation on a non empty set is an operation whose two domains and co-domains are the same set. Addition, subtraction and multiplication are the common binary operations. As the result of performing an operation on a set lies in the set itself, we say that the given binary operation is closed. Binary operations may or may not be associative and distributive. The inverse of any element of a set under binary operation is the element such that the element operated with its inverse gives the identity element.

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