Mathematician Keith Devlin has written several articles about the relationship between multiplication and addition (1, 2, 3, 4, 5). Drawing on a variety of sources, including pure mathematics and empirical research about learning, Devlin argues that since multiplication isn't repeated addition, students shouldn't be taught that it is. I think that's basically right, though in practice it is complicated, as Devlin acknowledges.

In this sidebar, I would like to mathematize the basic question. "Multiplication is repeated addition" might initially sound like a mathematical claim, something straightforwardly either true or false, but as it stands, this statement is too vague to admit of either proof or disproof. Some precise definitions are necessary before we can consider the question from the standpoint of mathematics.

Everybody agrees that repeated addition is useless as a way to understand multiplication of real numbers, as in cases like π ×

*e*. Where I think I differ from most people is that I do not even grant that multiplication reduces to repeated addition for the case of whole numbers. But that claim is also vague, so let's get started. You will let me know if anything below is incorrect or needs fixing.**Definition 1**. A

*whole-number operation*is a function that maps every pair of non-negative integers to some non-negative integer, which is called the

*result*of the operation, or the resulting

*value*. Since we only consider whole numbers, for brevity we simply call a whole-number operation an

*operation*. The result of an operation B on two numbers

*m*and

*n*can be denoted

*m*B

*n*.

Examples:

1) Addition (+) and multiplication (×) are operations.

2) Let F be defined by

*m*F*n*= 7 ×*n*. Then F is an operation, in particular a noncommutative operation. For example, 4F6 = 42, but 6F4 = 28.
3) Let G be defined by

*m*G*n*= |*m*−*n*|. Then G is an operation, in particular a nonassociative operation. For example, 1G(1G2) = 0, but (1G1)G2 = 2.**Definition 2**. An

*addition expression*is a string of symbols consisting of

*k*≥ 2 whole numbers separated from one another by

*k*− 1 addition symbols.

Examples.

4) "2 + 3" is an addition expression.

5) "2 + 3 + 9" is an addition expression.

6) " + 1" is not an addition expression.

7) "1" is not an addition expression.

8) "6 × 7" is not an addition expression.

Remark 1. "1" and "6 × 7" are algebraic expressions, but not addition expressions.

Remark 2. An addition expression can be said to have a value in the obvious way, e.g., the value of "2 + 3 + 9" is 14.

**Definition 3**. An operation C is

*completely reducible to addition*if, for all whole numbers

*m*and

*n*, there exists an addition expression that equals

*m*C

*n*and contains no numbers besides

*m*or

*n*.

Examples.

9) The operation

*m*F*n*= 7 ×*n*is completely reducible to addition. Here is a particular instance: the addition expression 4 + 4 + 4 + 4 + 4 + 4 + 4 equals 5F4 and contains no numbers besides 5 or 4.
Remark 3. One might have thought that only commutative operations are completely reducible to addition; Example 9 shows that this is not the case.

10) The operation

*m*G*n*= |*m*−*n*| is not completely reducible to addition. To see this, consider that 1G1 = 0, whereas it follows immediately from the definition of an addition expression that an addition expression containing only 1s has value at least 2.**Theorem**. Multiplication is not completely reducible to addition.

Proof. As noted in Example 10, an addition expression containing only 1s has value at least 2. So an addition expression containing only 1s cannot equal 1 × 1 in value. ∎

The theorem states the obvious: 1 × 1 isn't equal in value to any well formed expression consisting of 1s and addition symbols. According to the sense of Definition 3, this means that multiplication is not completely reducible to addition.

This being obvious, it must be that when people talk of reducing whole-number multiplication to addition, they have something else in mind besides Definition 3.

Wikipedia provides the following quasi-definition of multiplication as repeated addition:

*m*×

*n*=

*n*+

*n*+ ... +

*n*(

*m*addends).

I'm not sure the Wikipedia authors have thought about what happens to this prescription when

*m*=*n*= 1. In this case, it says
1 × 1 = 1 (1 addend).

The equation is true, of course, but to our point today I can't help noticing that there's no addition in it! Addition is a binary operation. If you are truly adding, then there's never just 1 addend. Whatever the Wikipedia formula reduces multiplication to, it isn't addition.

Maybe the Wikipedia definition ought to say something like this instead?

*m*×

*n*= 0 +

*n*+

*n*+ ... +

*n*(

*m*instances of

*n*)

For

*m*=*n*= 1, this becomes
1 × 1 = 0 + 1 (1 instance of 1)

which is a true equation, and, well, if it contains a number other than 1, then at least it is an addition expression. But now what happens when we take

*m*= 0? Since, according to the prescription, we mustn't write any instances of 5, the result is an ill-formed expression:
0 × 5 = 0 +

Maybe we could patch things up by agreeing that the phrase "no instances of 5" means "insert the number 0." But wait—isn't that just multiplying?

I'll leave off trying to patch up Wikipedia's formula (feel free to take up that project here) and just list some representative cases that prevent us from interpreting multiplication as repeated addition:

1 × 1

1/5 × 3/4

−1/5 × −3/4

π ×

*e**xy*

Hmm. Kind of a lot. It makes me wonder what the value is of defending the idea.

In my own work, when I have needed to describe the relationship between multiplication and addition, I have pointed first to the distributive property:

*a*× (

*b*+

*c*) =

*a*×

*b*+

*a*×

*c*.

Observe that both operations appear on both sides of this identity—it is not a prescription for eliminating one operation in favor of another. Nothing in the axioms is inviting us to undertake that project.

Indeed the distributive property is the only field axiom that refers to both operations, which means that it captures the entirety of the relationship between the two operations. For that reason, to describe the relationship between multiplication and addition using formulations other than the distributive property itself is to invite error. But is there any true and valuable statement that juxtaposes multiplication with repeated addition? I'll try one:

Repeated addition is a special-case algorithm for calculating whole-number products,m×n, whenm> 1 orn> 1.

Proving that the algorithm works involves the distributive property, of course:

Proof: By assumption, either

*m*> 1 or*n*> 1. First suppose*n*> 1. Then, and only then, can*n*be written as a sum of 1s:*n*= 1 + 1 + ... + 1 (*n*addends). So by the distributive property,*m*×*n*=*m*× (1 + 1 + ... + 1) =*m*× 1 +*m*× 1 + ... +*m*× 1. Since 1 is the multiplicative identity, we have*m*×*n*=*m*+*m*+ ... +*m*(*n*addends). Likewise, if (and only if)*m*> 1,*m*can be written as a sum of 1s:*m*= 1 + 1 + ... + 1 (*m*addends). So*m*×*n*= (1 + 1 + ... + 1) ×*n*= 1 ×*n*+ 1 ×*n*+ ... + 1 ×*n*=*n*+*n*+ ... +*n*(*m*addends). Thus, if (and only if)*n*> 1, the product*m*×*n*may be calculated by adding*n*addends*m*, and if (and only if)*m*> 1, then the product*m*×*n*may be calculated by adding*m*addends*n*. ∎
Repeated addition is an algorithm, not an operation. This is one reason among many why multiplication and repeated addition are best thought of as two different things, even for whole numbers.