Written By Madhurima Das
Last Modified 25-01-2023

Commutative Property Formula: Definition, Facts, and Examples

Commutative Property Formula: Math is all about numbers. In everything, we deal with numbers in daily life. These numbers can be whole numbers, natural numbers, integers, fractions, and many more. We apply basic operations on numbers such as addition, subtraction, multiplication, and division. Sometimes the calculations can be a bit lengthy. Now, to reduce these complexities, some properties have been introduced.

There are three main properties for the algebra of numbers: associative property, distributive property, and commutative property, with which we interact again and again while learning the properties of numbers. In this article, we will cover about commutative property formula.

What is Commutative Property?

The sum or the product of two numbers is said to be commutative for addition or multiplication if their sum or the product remains the same even if the order of the addition or multiplication is changed, respectively. In other words, if changing the operands’ position does not change the result of the arithmetic operation such as addition and multiplication, then that particular arithmetic operation is commutative.

This property can not be applied for subtraction and division. Let us discuss here the commutative property over addition and multiplication in detail.

Commutative Property Over Addition

Commutative property over addition says the sum of two numbers is said to be commutative for addition if their sum remains the same even if the order of the addition is changed.
Consider the following table where \(A\) and \(B\) are natural numbers,

\(A\)	\(B\)	\(A+B\)	\(B+A\)	\(A+B=B+A\)
\(5\)	\(3\)	\(5+3=8\)	\(3+5=8\)	\(5+3=3+5\)
\(12\)	\(13\)	\(12+13=25\)	\(13+12=25\)	\(12+13=13+12\)

Consider the following table where \(A\) and \(B\) are whole numbers,

\(A\)	\(B\)	\(A+B\)	\(B+A\)	\(A+B=B+A\)
\(5\)	\(0\)	\(5+0=5\)	\(0+5=5\)	\(5+0=0+5\)
\(10\)	\(13\)	\(10+13=23\)	\(13+10=23\)	\(10+13=13+10\)

Similarly, consider the following table where \(A\) and \(B\) are integers,

\(A\)	\(B\)	\(A+B\)	\(B+A\)	\(A+B=B+A\)
\(2\)	\(-3\)	\(2+(-3)=-1\)	\((-3)+2=-1\)	\(2+(-3)=(-3)+2\)
\(-2\)	\(-3\)	\((-2)+(-3)=-5\)	\((-3)+(-2)=-5\)	\((-2)+(-3)=(-3)+(-2)\)

Let us take an example where \(A\) and \(B\) are rational numbers,

\(A\)	\(B\)	\(A+B\)	\(B+A\)	\(A+B=B+A\)
\(\frac{1}{2}\)	\(\frac{2}{3}\)	\(\frac{1}{2} + \frac{2}{3} = \frac{7}{6}\)	\(\frac{2}{3} + \frac{1}{2} = \frac{7}{6}\)	\(\frac{1}{2} + \frac{2}{3} = \frac{2}{3} + \frac{1}{2}\)
\(\frac{4}{5}\)	\(\frac{2}{5}\)	\(\frac{4}{5} + \frac{2}{5} = \frac{6}{5}\)	\(\frac{2}{5} + \frac{4}{5} = \frac{6}{5}\)	\(\frac{4}{5} + \frac{2}{5} = \frac{2}{5} + \frac{4}{5}\)

Therefore, we can say the natural numbers, whole numbers, integers, and rational numbers are commutative over addition.

Commutative Property Over Addition Formula

The commutative property says if \(A\) and \(B\) be the operands, then changing its position does not change the result of the addition.
Therefore, \(A+B=B+A\).

Commutative Property Over Multiplication

Commutative property over multiplication says if the product two whole numbers or integers is said to be commutative for multiplication if their product remains the same even if the order of the multiplication is changed.
Consider the following table where \(A\) and \(B\) are natural numbers,

\(A\)	\(B\)	\(A×B\)	\(B×A\)	\(A×B=B×A\)
\(6\)	\(3\)	\(6×3=18\)	\(3×6=18\)	\(2×6=6×2\)
\(4\)	\(5\)	\(4×5=20\)	\(5×4=20\)	\(4×5=5×4\)

Consider the following table where \(A\) and \(B\) are whole numbers,

\(A\)	\(B\)	\(A×B\)	\(B×A\)	\(A×B=B×A\)
\(2\)	\(6\)	\(2×6=12\)	\(6×2=12\)	\(2×6=6×2\)
\(0\)	\(10\)	\(0×10=0\)	\(10×0=0\)	\(0×10=10×0\)

Similarly, consider the following table where \(A\) and \(B\) are integers,

\(A\)	\(B\)	\(A×B\)	\(B×A\)	\(A×B=B×A\)
\(4\)	\(-3\)	\(4×(-3)=-12\)	\((-3)×4=-12\)	\(4×(-3)=(-3)×4\)
\(-1\)	\(6\)	\((-1)×6=-6\)	\((6)×(-1)=-6\)	\(\left( { – 1} \right) \times \left( 6 \right) = \left( 6 \right) \times \left( { – 1} \right)\)

Let us take an example where \(A\) and \(B\) are rational numbers,

\(A\)	\(B\)	\(A×B\)	\(B×A\)	\(A×B=B×A\)
\(\frac{1}{2}\)	\(\frac{2}{5}\)	\(\frac{1}{2} \times \frac{2}{5} = \frac{1}{5}\)	\(\frac{2}{5} \times \frac{1}{2} = \frac{1}{5}\)	\(\frac{1}{2} \times \frac{2}{5} = \frac{2}{5} \times \frac{1}{2}\)
\(\frac{2}{3}\)	\(\frac{3}{7}\)	\(\frac{2}{3} \times \frac{3}{7} = \frac{2}{7}\)	\(\frac{3}{7} \times \frac{2}{3} = \frac{2}{7}\)	\(\frac{2}{3} \times \frac{3}{7} = \frac{3}{7} \times \frac{2}{3}\)

Therefore, we can say the natural numbers, whole numbers, integers, and rational numbers are commutative over multiplication also.

Commutative Property Over Multiplication Formula

The commutative property says if \(A\) and \(B\) be the operands, then changing its position does not change the result of the multiplication.
Therefore, \(A×B=B×A\).

Commutative Property Over Subtraction

Let us see whether the commutative property is applicable to the subtraction of numbers or not.

Let’s take two integers \(-2\) and \(3\).
\(-2-(3)=-5\) and \(3-(-2)=5\).

Here, \(-5≠5\)
If \(A\) and \(B\) are integers, then \(A-B≠B-A\)
Therefore, the difference between the two integers is not the same if their places are interchanged.

Thus, the integers are not commutative under subtraction.

Let’s take two rational numbers \(\frac{1}{2}\) and \(3\).
\(\frac{1}{2} – (3) = \frac{{ – 5}}{2}\) and \(3 – \left( {\frac{1}{2}} \right) = \frac{{ 5}}{2}.\)
Here, \( – \frac{{ – 5}}{2} \ne \frac{5}{2}\)
If \(A\) and \(B\) are rational numbers, then \(A – B \ne B – A\)

Therefore, the difference between the two rational numbers is not the same if their places are interchanged.

Thus, the rational numbers are not commutative under subtraction.

As the integers and rational numbers are not commutative under subtraction, the natural numbers and the whole numbers are also not commutative under subtraction

Commutative Property Over Division

Let us see whether the commutative property is applicable to the division of numbers or not. Let’s take two integers \(-2\) and \(4\).

Now, \(\frac{{ – 2}}{4} = \frac{{ – 1}}{2}\) and \(\frac{4}{{ – 2}} = – 2\)
Here, \(\frac{{ – 1}}{2} \ne – 2\)
If \(A\) and \(B\) are integers, then \(\frac{A}{B} \ne \frac{B}{A}\).
Therefore, the result of the division of two integers is not the same if their places are interchanged.

Thus, the integers are not commutative under division.

Let’s take two rational numbers, \(5\) and \(15\).
\((5\) and \(15\) are rational numbers as they can be written as \(\frac{5}{1}\) and \(\frac{{15}}{1}\) respectively)
Now, \(\frac{5}{{15}} = \frac{1}{3}\) and \(\frac{{15}}{5} = 3\)
Here, \(\frac{5}{{15}} \div \frac{{15}}{5}\)

If \(A\) and \(B\) are rational numbers, then \(\frac{A}{B} \ne \frac{B}{A}.\).
Therefore, the result of the division of the two rational numbers is not the same if their places are interchanged.

Thus, the rational numbers are not commutative under division.

As the integers and rational numbers are not commutative under the division then, the natural numbers and the whole numbers are not commutative under division.

Solved Examples – Commutative Property Formula

Q.1. Prove that the addition of \(20\) and \(30\) comes under the commutative property.
Ans: The commutative property says if \(A\) and \(B\) be the operands, then changing its position does not change the result of the addition.
Therefore, \(A+B=B+A\).
Now, \(20+30=50\), and if we change its order, we have \(30+20=50\).
Therefore, \(20+30=30+20\)
Hence, the addition of \(20\) and \(30\) comes under the commutative property.

Q.2. Check whether the subtraction of \(10\) and \(15\) comes under commutative property or not.
Ans:
\(10-15=-5\) and \(15-10=5\).
Here, \(-5≠5\)
If \(A\) and \(B\) are two numbers, then \(A-B≠B-A\).
Therefore, the difference between the two numbers is not the same if their places are interchanged.
Thus, the given numbers are not commutative under subtraction.

Q.3. Prove that the division of \(40\) and \(4\) does not satisfy the commutative property.
Ans:
Now, \(\frac{{40}}{4} = 10\) and \(\frac{4}{{40}} = \frac{1}{{10}}\)
Here, \(10 \ne \frac{1}{{10}}\)
If \(A\) and \(B\) are the numbers, then \(\frac{A}{B} \ne \frac{B}{A}.\).
Therefore, the result of the division of two numbers is not the same if their places are interchanged.
Thus, the numbers are not commutative under division.

Q.4. Verify that \(A×B=B×A\), if \(A=6\) and \(B=7\).
Ans: Given that, \(A=6\) and \(B=7\).
LHS \(=A×B=6×7=42…(i)\)
RHS \(=B×A=7×6=42…(ii)\)
By equations (i) and (ii) we have,
LHS = RHS
Hence, proved.

Q.5. Verify that \(A÷BB÷A\) if \(A = \frac{2}{5}\) and \(B = \frac{3}{{10}}\).
Ans:
Given that, \(A = \frac{2}{5}\) and \(B = \frac{3}{{10}}\)
LHS \( = A \div B = \frac{2}{5} \div \frac{3}{{10}} = \frac{2}{5} \times \frac{{10}}{3} = \frac{4}{3} \ldots (i)\)
RHS \( = B \div A = \frac{3}{{10}} \div \frac{2}{5} = \frac{3}{{10}} \times \frac{5}{2} = \frac{3}{4} \ldots (ii)\)
By equations (i) and (ii) we have,
LHS RHS
Hence, proved.

Summary

In this article, we have learned about one of the properties of numbers, namely commutative property and its formula and the constraints of this property. We have discussed how we can apply it to numbers such as natural numbers, whole numbers, integers, and rational numbers.

Frequently Asked Questions – Distributive Property Formula

Q.1. What is commutative property? Explain it by an example.
Ans: The sum or the product of two numbers is said to be commutative for addition or multiplication if their sum or the product remains the same even if the order of the addition or multiplication is changed.
For example,
Let’s take two integers \(-2\) and \(3\).
\(-2+(3)=1\) and \(3+(-2)=1\).
Here, \(1=1\)
If \(A\) and \(B\) are integers, then \(A+B=B+A\).
Now, \(-2×3=-6\) and \(3×-2=-6\).
Here, \(-6=-6\)
If \(A\) and \(B\) are integers, then \(A \times B = B \times A.\)

Q.2.What is the formula for commutative and associative property?
Ans: The formula for the commutative property over addition is \(A+B=B+A\) when \(A\) and \(B\) are operands.
The formula for the commutative property over multiplication is \(A×B=B×A\) when \(A\) and \(B\) are operands.
The formula for the associative property over addition is \(A + \left( {B + C} \right) = \left( {A + B} \right) + C\) when \(A, B\) and \(C\) are operands.
The formula for the associative property over multiplication is \(A \times \left( {B \times C} \right) = \left( {A \times B} \right) \times C\) when \(A, B\) and \(C\) are operands.

Q.3. What are \(2\) examples of commutative property?
Ans: The commutative property can be applied only over the addition and multiplication of any two numbers.
Let’s take two numbers \(2\) and \(3\).
\(2+(3)=5\) and \(3+(2)=5\).
Here, \(5=5\)
If \(A\) and \(B\) are the numbers, then \(A+B=B+A\).
Now, \(2×3=6\) and \(3×2=6\).
Here, \(6=6\)
If \(A\) and \(B\) are the numbers, then \(A×B=B×A\).

Q.4.What is the commutative property of addition?
Ans: According to the commutative property, if the sum of two numbers is said to be commutative for addition if their sum remains the same even if the order of the addition is changed.
If \(A\) and \(B\) are two numbers, then \(A+B=B+A\).

Q.5.What is the commutative property of subtraction?
Ans: The commutative property can not be applied over the subtraction of two numbers.
For example, let’s take two integers \(-2\) and \(3\).
\(-2-(3)=-5\) and \(3-(-2)=5\).
Here,\( -5≠5\)
If \(A\) and \(B\) are integers, then \(A-B≠B-A\).
Therefore, the difference between the two integers is not the same if their places are interchanged.
Thus, the integers are not commutative under subtraction.

Q.6.Can a commutative property have \(3\) numbers?
Ans: Yes, the commutative property can have three numbers. This property states if we change the order of the numbers in an arithmetic operation, then the result of the arithmetic operation will be the same. This property is only applicable to addition and multiplication.
For example, \(1+2+3=1+2+3=2+1+3=2+3+1=3+1+2=3+2+1=6\)

We hope you find this article on commutative property helpful. In case of any queries, you can reach back to us in the comments section, and we will try to solve them. Happy learning!