1st Grade · 1.OA.B.3

The commutative property of addition in 1st grade

The commutative property of addition says that two numbers can be added in either order and the total stays the same. $6 + 3$ and $3 + 6$ both make 9. In 1st grade, your child does not need to know the name. Your child uses the idea as a shortcut, along with other properties of addition, to add and subtract more easily.

What the standard says

The official Common Core wording, quoted exactly. The plain version follows.

1.OA.B.3 Apply properties of operations as strategies to add and subtract.

What it means, in plain words

In class, the commutative property of addition is often called the turn-around rule. If your child knows $7 + 4 = 11$ , then $4 + 7 = 11$ is known too. This cuts the number of facts to learn almost in half. It also lets your child start with the bigger number and count on. For $3 + 8$ , start at 8 and count 9, 10, 11.

Properties of addition are rules that are true for every number. Two more come up in 1st grade. The grouping rule, called the associative property, says that with three numbers you can add any pair first. For $5 + 7 + 3$ , adding $7 + 3$ first makes 10, and $10 + 5 = 15$ . The zero rule says that adding 0 leaves a number the same, so $17 + 0 = 17$ .

The turn-around rule works for addition only. Taking 4 away from 9 leaves 5, but you cannot take 9 apples away from 4 apples. So your child cannot swap the numbers in a subtraction.

Your child should be able to:

Use a known fact to get its turn-around, such as $4 + 7$ from $7 + 4$
Start with the bigger number and count on
Add three numbers by picking an easy pair first, such as two that make 10
Explain that adding 0 leaves a number the same

Example

If $5 + 8 = 13$ , what is $8 + 5$ ?

The two sums use the same numbers in a different order.
Picture 5 red blocks and 8 blue blocks in a row. Turning the row around changes the order, not how many blocks there are.
So $8 + 5 = 13$ too, with no need to count again.

Example

What is $6 + 9 + 1$ ?

Look for two numbers that make 10: 9 and 1.
Add them first: $9 + 1 = 10$ .
Then add the 6: $10 + 6 = 16$ .
The total is 16.

Where children usually get stuck

Your child answers $3 + 9 = 12$ quickly but counts all over again for $9 + 3$ .
To the child these look like two different problems. Nobody has pointed out yet that they use the same two numbers in a different order.
What helps. Line up 3 spoons and 9 forks and count them. Then walk to the other side of the table and count again from there.
For $3 + 8$ , your child starts at 3, counts 8 more on fingers and often lands on 10 or 12.
Counting 8 steps is long and easy to slip on. Starting at 8 and counting 3 steps gives the same total with fewer chances to slip.
What helps. Before your child starts, ask which number is bigger. Then count on from that number together: 9, 10, 11.
Your child says that $8 - 2$ and $2 - 8$ give the same answer, because $8 + 2$ and $2 + 8$ do.
The turn-around rule was learned as a rule for any two numbers, not only for adding them.
What helps. Try it with crackers. Put out 8 and eat 2. Then put out 2 and try to eat 8.

Try four practice questions

Real 1st grade questions from our question bank, easiest first. Answer them right here, check each one, and see a worked solution for any question.

1Warm-up
Maria looked at this equation:
$\displaystyle 1 + 2 = 3$
Maria wrote a new equation:
$\displaystyle 2 + 1 = 3$
Which statement best describes Maria's new equation?
A
Maria did not switch the order. The numbers are the same.
B
Maria switched the order. The sum is different.
C
Maria did not switch the order. The numbers are different.
D
Maria switched the order. The sum is the same.
Show the answer and how to solve it
Maria's new equation is $2 + 1 = 3$ . She switched the order of the numbers $1$ and $2$ , and the sum is still $3$ .
1. 1Look at Maria's two equations.
2. 2Did Maria switch the order? Yes — her new equation is $2 + 1 = 3$ . The two numbers are the same, just in a new order. The sum is still $3$ .
3. 3The answer is the choice that says Maria switched the order and the sum is the same.
Common mistake. A common mistake is thinking that switching the order must change the sum. The two numbers $1$ and $2$ are in a different order, but the sum is still $3$ .
The idea underneath. Switching the order keeps the same two numbers — only their positions change. If a number is replaced with a different number, that is not switching. The sum stays the same when you switch (the commutative property), but it changes when you change a number.
2On level
Aiyana wants to add the numbers below.
$\displaystyle 2 + 9$
She wants to count on from the bigger number. Which addition has the bigger number first?
A
$\displaystyle 9 + {}$
$\displaystyle 2$
B
$\displaystyle 9 - 2$
C
$\displaystyle 9 + {}$
$\displaystyle 3$
D
$\displaystyle 2 + {}$
$\displaystyle 9$
Show the answer and how to solve it
The bigger number in $2 + 9$ is $9$ . Putting $9$ first gives $9 + 2$ . Now Aiyana can count on from $9$ : $10, 11$ .
1. 1Look at the addition $2 + 9$ . The bigger number is $9$ .
2. 2Switch the order so $9$ comes first.
3. 3Now Aiyana can count on from $9$ : $10, 11$ . The answer is $9 + 2$ .
Common mistake. A common mistake is picking $2 + 9$ because the numbers are the same. But to count on from the bigger number, the bigger number $9$ must come first.
The idea underneath. Two numbers added in any order make the same total. Putting the bigger number first means fewer count-on steps.
3On level
Find the total.
$\displaystyle 3 + 7 + 2 = ?$
A
9
B
12
C
10
D
13
Show the answer and how to solve it
The total is 12. First, $3 + 7$ makes 10. Then, $10 + 2 = 12$ .
1. 1Look at the three numbers. Find two that make 10.
2. 2Now add the third number to 10.
3. 3The total is 12.
Common mistake. A common mistake is adding $3 + 7$ to make 10 and stopping there, choosing 10 instead of 12. Remember: after making 10, you still need to add the third number.
The idea underneath. Making 10 first turns a three-number addition into an easy one. Once you have 10, adding any single-digit number is simple: $10 + 2 = 12$ .
4Stretch
Look at Maya's addition.
$\displaystyle \begin{array}{l} 4 + 9 + 6 \\ 4 + 6 = 11 \\ 11 + 9 = 20 \end{array}$
Which sentence is true?
A
Maya only added two of the numbers.
B
Maya counted wrong in the first step.
C
Maya counted wrong in the last step.
D
Maya correctly added all three numbers.
Show the answer and how to solve it
Maya counted wrong in the first step. She wrote $4 + 6 = 11$ instead of $10$ , which made the rest of the work wrong too.
1. 1Look at the 3 lines of Maya's work.
2. 2Check the make-ten step. Maya wrote $4 + 6 = 11$ , but $4 + 6$ is $10$ .
3. 3Maya used the wrong $11$ from above and wrote $11 + 9 = 20$ .
4. 4But starting from $10$ , $10 + 9$ is $19$ , not $20$ .
5. 5So Maya counted wrong in the first step.
Common mistake. A common mistake is picking the last-step error. But $11 + 9 = 20$ follows correctly from Maya's (wrong) $11$ . The error started in the first step, where $4 + 6 = 11$ instead of $10$ .
The idea underneath. When checking make-ten work, both steps need to be right — the make-ten step AND the second step. Check each one separately.