Let o be a binary operation on the set $Q_{0}$ of all non-zero rational numbers defined by $a o b = \frac{a b}{2}$ , for all $a, b \in Q_{0}$ . Show that operation $^{'} o^{'}$ is commutative.

Question

Let o be a binary operation on the set ${{Q}_{0}}$ of all non-zero rational numbers defined by $aob=\dfrac{ab}{2}$, for all $a,b\in {{Q}_{0}}$. Show that operation $'o'$ is commutative.

Vedantu Content Team · Accepted Answer

Hint: In the question, we want to show that

^{'} o^{'}

is commutative. So write the binary operations and use the commutative property. You will get the answer to try it.

So in question, it is mentioned that we have to check whether

^{'} o^{'}

is commutative.
It is given in the question that

a o b = \frac{a b}{2}

.
A binary operation is simply a rule for combining two values to create a new value. The most widely known binary operations are those learned in elementary school: addition, subtraction, multiplication, and division on various sets of numbers.
It is possible to define "new" binary operations.
Sometimes, a binary operation on a finite set (a set with a limited number of elements) is displayed in a table that shows how the operation is to be performed. Binary operations on a set are calculations that combine two elements of the set (called operands) to produce another element of the same set.
The binary operations on a non-empty set A are functions from

A \times A

to

A

. The binary operation:

A \times A

to A. It is an operation of two elements of the set whose domains and co-domain are in the same set.
An operation of arity (arity is the number of arguments or operands taken by a function or operator.) two that involve several sets are sometimes also called a binary operation. For example, scalar multiplication of vector spaces takes a scalar and a vector to produce a vector, and the scalar product takes two vectors to produce a scalar. Such binary operations may be called simply binary functions.
So commutative property is the one that refers to moving stuff around. In addition, the rule is

a + b = b + a

; in numbers, this means

2 + 3 = 3 + 2

.
For multiplication, the rule is

a b = b a

; in numbers, this means

2 \times 3 = 3 \times 2

. Any time they refer to the Commutative Property, they want you to move stuff around; any time a computation depends on moving stuff around, they want you to say that the computation uses the Commutative Property.
So

a o b = \frac{a b}{2}

So let us check for

b o a = \frac{b a}{2}

So we get,

a o b = b o a

So we can see that commutative property is proved.

Note: Read the question carefully. Don’t jumble yourself with the operations. Also, you should know the commutative and associative properties. Don’t get confused about the concepts of properties. You should be familiar with binary operations.

Let o be a binary operation on the set Q0 of all non-zero rational numbers defined by aob=ab2, for all a,b∈Q0. Show that operation ′o′ is commutative.

Let o be a binary operation on the set $Q_{0}$ of all non-zero rational numbers defined by $a o b = \frac{a b}{2}$ , for all $a, b \in Q_{0}$ . Show that operation $^{'} o^{'}$ is commutative.