Commutative Property: Definition and Example | EDU.COM
Commutative PropertyCommutative Property: Definition and ExampleTable of Contents
Definition of Commutative Property
The commutative property is a fundamental mathematical concept that states that numbers in an operation can be moved or swapped from their positions without affecting the final result. This property only applies to addition and multiplication operations, but not to subtraction and division. When we rearrange numbers in addition or multiplication, the answer remains the same, which greatly simplifies many mathematical calculations.
The commutative property can be formally expressed in two forms. For addition, it's written as a+b=b+aa + b = b + aa+b=b+a, where aaa and bbb represent any two whole numbers. For multiplication, it's expressed as a×b=b×aa \times b = b \times aa×b=b×a, where aaa and bbb represent any two non-zero whole numbers. This powerful property allows mathematicians and students to rearrange numbers to make calculations easier and more efficient.
Examples of Commutative Property
Example 1: Using Known Products to Find New Ones
Problem:
Use 14×15=21014 \times 15 = 21014×15=210 to find 15×1415 \times 1415×14.
Step-by-step solution:
Step 1, Identify the property we can use. When we're asked to find 15×1415 \times 1415×14 and we already know 14×1514 \times 1514×15, we can apply the commutative property of multiplication.
Step 2, Recall the commutative property of multiplication: a×b=b×aa \times b = b \times aa×b=b×a for any numbers aaa and bbb.
Step 3, Apply this property to our specific numbers: 15×14=14×1515 \times 14 = 14 \times 1515×14=14×15.
Step 4, Therefore, since we know that 14×15=21014 \times 15 = 21014×15=210, we can conclude that 15×14=21015 \times 14 = 21015×14=210 as well.
Example 2: Using Known Sums to Find New Ones
Problem:
Use 827+389=1,216827 + 389 = 1,216827+389=1,216 to find 389+827389 + 827389+827.
Step-by-step solution:
Step 1, Identify the property we can use. When we're asked to find 389+827389 + 827389+827 and we already know 827+389827 + 389827+389, we can apply the commutative property of addition.
Step 2, Recall the commutative property of addition: a+b=b+aa + b = b + aa+b=b+a for any numbers aaa and bbb.
Step 3, Apply this property to our specific numbers: 389+827=827+389389 + 827 = 827 + 389389+827=827+389.
Step 4, Therefore, since we know that 827+389=1,216827 + 389 = 1,216827+389=1,216, we can conclude that 389+827=1,216389 + 827 = 1,216389+827=1,216 as well.
Example 3: Real-life Application with Equal Groups
Problem:
Ben bought 333 packets of 666 pens each. Mia bought 666 packets of 333 pens each. Did they buy an equal number of pens?
Step-by-step solution:
Step 1, Let's calculate how many pens Ben bought. Ben purchased 333 packets with 666 pens in each packet. Total pens for Ben =3×6= 3 × 6=3×6 pens
Step 2, Let's calculate how many pens Mia bought. Mia purchased 666 packets with 333 pens in each packet. Total pens for Mia =6×3= 6 × 3=6×3 pens
Step 3, Let's analyze if these expressions are equivalent. According to the commutative property of multiplication, 3×6=6×33 \times 6 = 6 \times 33×6=6×3.
Step 4, Calculate the total for each person:
Ben's total: 3×6=183 \times 6 = 183×6=18 pens
Mia's total: 6×3=186 \times 3 = 186×3=18 pens
Step 5, Therefore, both Ben and Mia bought exactly the same number of pens (181818 pens each), even though they bought different numbers of packets with different quantities in each packet.
svg]:px-3 dk:w-auto h-10 dk:h-14 rounded-lg dk:rounded-[15px] px-5 dk:px-7.5 text-sm dk:text-xl text-[#3467FF] bg-white hover:bg-white">Load More
© 2025 EDU.COM. All rights reserved.