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Question: Let \[{R_0}\] denote the set of all non-zero real numbers and let \[A = {R_0} \times {R_0}\]. If ‘*’...

Let R0{R_0} denote the set of all non-zero real numbers and let A=R0×R0A = {R_0} \times {R_0}. If ‘’ is a binary operation on A defined by (a, b)(c, d) = (ac, bd) for all (a, b), (c, d) belongs to A. Find the identity element in A.

Explanation

Solution

Hint: An identity element is an element which when operated with any other element gives the same element. If (e1,e2)({e_1},{e_2}) is the identity element, then it should satisfy the relation (e1,e2)∗(a,b)=(a,b)∗(e1,e2)=(a,b)({e_1},{e_2})*(a,b) = (a,b)*({e_1},{e_2}) = (a,b). Use this to find the identity element.
Complete step by step answer:
A binary operation is a calculation that combines two elements to produce another element. It takes two elements from a set and gives another element which also belongs to the same set, then it is defined as a binary operation on this set.
An identity element is a special type of element of a set with respect to a binary operation on that set, which leaves any element of the set unchanged when it is operated by the identity element.
We can use the above property to find the identity element of * operator.
Let (e1,e2)({e_1},{e_2}) be the identity element of the * operator, then for any element (a,b)(a,b) in the set A, we have as follows:
(e1,e2)∗(a,b)=(a,b)∗(e1,e2)=(a,b)({e_1},{e_2})*(a,b) = (a,b)*({e_1},{e_2}) = (a,b)
Equating the first and the third term, we get:
(e1,e2)∗(a,b)=(a,b)({e_1},{e_2})*(a,b) = (a,b)
Using the definition of * operator, we have:
(e1a,e2b)=(a,b)({e_1}a,{e_2}b) = (a,b)
The corresponding coordinates are equal.
e1a=a;e2b=b{e_1}a = a;{e_2}b = b
Dividing by a and b respectively, we have:
e1=1;e2=1............(1){e_1} = 1;{e_2} = 1............(1)
Equating the second and the third term, we get:
(a,b)∗(e1,e2)=(a,b)(a,b)*({e_1},{e_2}) = (a,b)
Using the definition of * operator, we have:
(ae1,be2)=(a,b)(a{e_1},b{e_2}) = (a,b)
The corresponding coordinates are equal.
ae1=a;be2=ba{e_1} = a;b{e_2} = b
Dividing by a and b respectively, we have:
e1=1;e2=1............(2){e_1} = 1;{e_2} = 1............(2)
Hence, we get the identity element as (1, 1).

Note: The identity element must belong to the set on which the operation is defined. Also, it should satisfy both the conditions (a,b)∗(e1,e2)=(a,b)(a,b)*({e_1},{e_2}) = (a,b) and (e1,e2)∗(a,b)=(a,b)({e_1},{e_2})*(a,b) = (a,b). If you solve only one, then you won’t get full credits.