Question

Let $A=Q\times Q$ and let $*$ be the binary operation on A defined by $\left( a,b \right)*\left( c,d \right)=\left( ac,b+ad \right)$ for $\left( a,b \right),\left( c,d \right)\in A$. Determine, whether $*$ is commutative and associative. Then, with respect to $*$ on A.
i). Find the identity element in A
ii). Find the invertible element of A.

Answer 1

To solve the above question we will first define the commutative property, associative property and then we will check whether the given relation is commutative and associative or not and then we will find the identity and invertible element of A by using their definition.

Complete step by step answer:
We will first define what is commutative and associative property from group theory.
Commutative property states that we can move variables or numbers in algebra around some defined operation and still we get the same answer.
Let $*$ be the binary relation defined on any set A and let a, b be the two elements of the set, then we can say that $a*b=b*c$ , from commutative property.
Associative property comes from the word ‘associate’ or ‘groups’. It refers to grouping of numbers and variables in algebra. We can re-group the number or variables and we can then also arrive at the same result.
Let $*$ be the binary relation defined on any set A and let a, b, and c be the three elements of the set, then we can say that $a*\left( b*c \right)=\left( a*b \right)*c$, from associative property.
Now, from we know that $A=Q\times Q$, means set A consists of pair of elements and $*$ be the binary relation defined on A as $\left( a,b \right)*\left( c,d \right)=\left( ac,b+ad \right)$, where $\left( a,b \right),\left( c,d \right)$ belongs to A.
Now, we will check whether the binary operation $*$ is commutative or not. For this we have to prove that $\left( a,b \right)*\left( c,d \right)=\left( c,d \right)*\left( a,b \right)$, where $\left( a,b \right),\left( c,d \right)\in A$
Since, we know that $\left( a,b \right)*\left( c,d \right)=\left( ac,b+ad \right)---(1)$, from question.
Then, $\left( c,d \right)*\left( a,b \right)=\left( ca,d+ab \right)---(2)$
Since, $\left( ac,b+ad \right)\not{=}\left( ca,d+ab \right)$
So, we can say from (1) and (2) that $\left( a,b \right)*\left( c,d \right)\not{=}\left( c,d \right)*\left( a,b \right)$, hence, the binary operation $*$ is not commutative.
Now, we will check whether the binary operation $*$ is associative or not.
Let $\left( a,b \right),\left( c,d \right),\left( e,f \right)$ be three elements which belong to A.
Then, to prove associative property for binary operation $*$ , on A we have to prove that:
$\left( a,b \right)*\left( \left( c,d \right)*\left( e,f \right) \right)=\left( \left( a,b \right)*\left( c,d \right) \right)*\left( e,f \right)$
We will at first simplify LHS.
From LHS we have: $\left( a,b \right)*\left( \left( c,d \right)*\left( e,f \right) \right)$
Since, we know that $\left( a,b \right)*\left( c,d \right)=\left( ac,b+ad \right)$
$\Rightarrow \left( a,b \right)*\left( \left( c,d \right)*\left( e,f \right) \right)=\left( a,b \right)*\left( ce,d+cf \right)$
$\begin{aligned} & \Rightarrow \left( ace,b+a\left( d+cf \right) \right) \\\ & \Rightarrow \left( ace,b+ad+acf \right) \\\ & \Rightarrow \left( a,b \right)*\left( \left( c,d \right)*\left( e,f \right) \right)=\left( ace,b+ad+acf \right)---(3) \\\ \end{aligned}$
Now, from RHS of $\left( a,b \right)*\left( \left( c,d \right)*\left( e,f \right) \right)=\left( \left( a,b \right)*\left( c,d \right) \right)*\left( e,f \right)$, we have:
$RHS=\left( \left( a,b \right)*\left( c,d \right) \right)*\left( e,f \right)$
$\begin{aligned} & \Rightarrow \left( \left( a,b \right)*\left( c,d \right) \right)*\left( e,f \right)=\left( ac,b+ad \right)*\left( e,f \right) \\\ & \Rightarrow \left( ace,b+ad+acf \right) \\\ & \Rightarrow \left( \left( a,b \right)*\left( c,d \right) \right)*\left( e,f \right)=\left( ace,b+ad+acf \right) \\\ \end{aligned}$
From (3) we have $\left( a,b \right)*\left( \left( c,d \right)*\left( e,f \right) \right)=\left( ace,b+ad+acf \right)$
$\Rightarrow \left( \left( a,b \right)*\left( c,d \right) \right)*\left( e,f \right)=\left( ace,b+ad+acf \right)=\left( a,b \right)*\left( \left( c,d \right)*\left( e,f \right) \right)$
Hence, $\left( a,b \right)*\left( \left( c,d \right)*\left( e,f \right) \right)=\left( \left( a,b \right)*\left( c,d \right) \right)*\left( e,f \right)$, so, we can say that binary operation $*$ satisfies the associative property.

i). We know that identity element is element which is when combined with any defined operation on any set given the element back as output with which identity element is combined.
Let $*$ be the binary operation defined on any set A and let ‘a’ be an element of the set A and ‘e’ be the identity element for $*$, then $a*e=a$.
Since, we know from question that $A=Q\times Q$, hence set A consists of pair elements and we know that identity element must belong to set A so let the identity element $e=\left( a',b' \right)$ and let $\left( a,b \right)$ be the element of the set A.
So, from definition of identity element we can write:
$\Rightarrow \left( a,b \right)*\left( a',b' \right)=\left( a,b \right)--(4)$
And, we know from question that binary operation $*$ is defined as:
$\left( a,b \right)*\left( c,d \right)=\left( ac,b+ad \right)$ , for $\left( a,b \right),\left( c,d \right)\in A$.
So, we can also write $\left( a,b \right)*\left( a',b' \right)$ as $\left( aa',b+ab' \right)$
$\Rightarrow \left( a,b \right)*\left( a',b' \right)=\left( aa',b+ab' \right)--(5)$
And, from (4) we have $\left( a,b \right)*\left( a',b' \right)=\left( a,b \right)$
Hence, after comparing (4) and (5) we can say that:
$aa'=a$ and $b+ab'=b$
$\Rightarrow a'=1$ and $b'=0$
Hence, (1, 0) is the identity element of A.

ii). We know that an invertible element is defined for every element of a set if the set contains an identity element. So, an invertible element of any element of a set is the element which when combined with the given element with any operation gives the identity element as a result.
Let us assume that $\left( a,b \right)$ be any element which belong to $A=Q\times Q$, and let $\left( c',d' \right)$ be the invertible element which belong to A and we have found previously that (1, 0) is the identity element of set A.
Then, from definition of invertible element we can say that:
$\left( a,b \right)*\left( c',d' \right)=\left( 1,0 \right)---(6)$
We know from question that $\left( a,b \right)*\left( c,d \right)=\left( ac,b+ad \right)$:
$\Rightarrow \left( a,b \right)*\left( c',d' \right)=\left( ac',b+ad' \right)---(7)$
So, after comparing (6) and (7) we can say that:
$ac'=1$ and $b+ad'=0$
$\Rightarrow c'=\dfrac{1}{a}$ and $d'=-\dfrac{b}{a}$
Hence, the invertible element of $\left( a,b \right)$ is $\left( \dfrac{1}{a},-\dfrac{b}{a} \right)$.

Note:
Students are required to memorize the definition of associative property, commutative property, identity element and invertible element otherwise they will not be able to solve the above question.