*By William McCallum*

Somewhere back in days of Facebook fury about the Common Core there was a post from an outraged parent whose child had been marked wrong for something like this:

$$

6 \times 3 = 6 + 6 + 6 = 18.

$$

Apparently the child was supposed to do

$$

6 \times 3 = 3 + 3 + 3 + 3 + 3 +3 = 18

$$

because of this standard:

*3.OA.A.1. Interpret products of whole numbers, e.g., interpret $5 \times 7$ as the total number of objects in 5 groups of 7 objects each. *For example, describe a context in which a total number of objects can be expressed as $5 \times 7$.

The parent had every right to be upset: a correct answer is a correct answer. Comments on the post correctly pointed out that, since multiplication is commutative, it shouldn’t matter in what order the calculation interpreted the product. But hang on, I hear you ask, doesn’t that contradict 3.OA.A.1, which clearly states that $6 \times 3$ should be interpreted as 6 groups of 3?

The fundamental problem here is a confusion between ways of thinking and ways of doing. 3.OA.A.1 proposes a way of thinking about $a \times b$, as $a$ groups of $b$. In other words, it proposes a definition of multiplication. It could have proposed the other definition: $a \times b$ is $b$ groups of $a$. The choice is arbitrary, so why make it? Well, there’s an interesting discovery to me made here: the two definitions are equivalent. That’s how you prove that multiplication is indeed commutative. It’s not obvious that $a$ groups of $b$ things each amounts to the same number of things as $b$ groups of $a$ things each. At least, not until you prove it, for example by arranging the things into an array:

You can see this as 3 groups of 6 by looking at the rows,

and as 6 groups of 3 things each by looking at the columns,

Since it’s the same number things no matter how you look at it, and using our definition of multiplication, we see that $3 \times 6 = 6 \times 3$. (We leave it as an exercise to the reader to generalize this proof.)

None of this dictates the way of doing $6 \times 3$, that is, the method of computing it. In fact, it expands the possibilities, including deciding to work with the more efficient $3 \times 6$, as this child did. The way of thinking does not constrain the way of doing. If you want to test whether a child understands 3.OA.A.1, you will have to come up with a different task than computation of a product. There are some good ideas from Student Achievement Partners here.

#####
William McCallum

Bill McCallum, founder of Illustrative Mathematics, is a University Distinguished Professor of Mathematics at the University of Arizona. He has worked in both mathematics research, in the area of number theory and arithmetical algebraic geometry, and mathematics education, writing textbooks and advising researchers and policy makers. He is a founding member of the Harvard Calculus Consortium and lead author of its college algebra and multivariable calculus texts. In 2009–2010 he was one of the lead writers for the Common Core State Standards in Mathematics. He holds a Ph. D. in Mathematics from Harvard University and a B.Sc. from the University of New South Wales.

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