How do you teach students to multiply numbers when one or both of them is negative?  This is a surprisingly challenging question.  In this post I offer some thoughts about how to accomplish this in a way that is mathematically legitimate and yet still easy to grasp.

Notion #1: What is a negative number?

Just what is a “negative number,” anyway?  Just to be quirky I’ll refer to “anti-numbers,” such as 3 and anti-3.  (Thinking of matter and anti-matter.)  The diagram below shows 3 piles of sand and 3 anti-piles, or more simply, 3 holes.


The important thing to know here is that we can squish a pile into a hole, and when we are done, we are left with the same flat sandbox we started with before building any piles or digging any holes.  So here we have a visual representation of the idea that 3 and anti-3 “cancel out.”  In symbols, we write 3 + -3 = 0.

Notion #2: Multiplication as Repeated Addition

When students first encounter multiplication, they learn to compute 5×3 by writing out 5 groups of 3, which gives 15:


Problem #1: What is 5 x -3?

Using Notion #2, we think of this as being 5 groups of anti-3.  The diagram below shows that this is simply anti-15:


Here we are assuming that students know already how to add negative numbers.  To read about how to develop this notion using piles and holes, see here.

Okay, so (positive) x (negative) turned out to be a piece of cake!

Problem #2: What is -5 x 3?

When we try to interpret this using Notion #2, as we did with Problem #1, things get a bit awkward: what is “anti-5 groups” of 3?  It’s not at all clear what this should mean, so we are going to have to appeal to a new technology called…

Notion #3: Multiplication Distributes over Addition

Michael drew 5 groups of 4 dots on a piece of paper, then tore it into 2 pieces as shown below:


On piece A, there are 3 groups of 4 dots.  On piece B, there are 2 groups of 4 dots.  Together, there must be 5 groups of 4 dots.  The diagram below shows how to represent all of this mathematically:


As a check, we can see that 12 dots and 8 dots make 20 dots altogether.

Problem #2 Revisited!

Let’s try to tackle Problem #2 with the help of high-tech Notion #3.  What exactly is anti-5 groups of something?  Well, perhaps it is, in some sense, the opposite of 5 groups of something.  Let’s see where this idea takes us:


I want to know what anti-5 groups of 3 is.  I already know that 5 groups of 3 make 15.  If the distributive property, which holds for positive numbers, is also to be believed for negative numbers, then we can work out the answer to Problem #2: 5 groups plus anti-5 groups must make no groups at all, as the diagram above shows.  But if we also believe that “zero groups of something” makes zero, then the answer is staring at us!  If 15 + ? = 0, then the missing quantity must be anti-15.  Thus we would like to believe that -5 x 3 = -15.  That is, if we believe the distributive property still holds for negative numbers, we are forced to conclude that -5 x 3 = -15.  Problem solved!

Problem #3: What is -5 x -3?

Here we have anti-5 groups of anti-3.  Wow!  This one looks like a toughie.  Invoking Notion #3 again, we get this puzzle:


It looks as though the distributive property is telling us that anti-5 groups of anti-3 is simply the opposite of 5 groups of anti-3, which we know from Problem #2 is anti-15.  We are again faced with zero groups of something, and we are committed to the belief that 0 x -3 = 0, so the solution to -15 + ? = 0 must be anti-anti-15!  Whoa — things are getting a bit crazy — what exactly is anti-anti-15?  Well, since 15 + -15 = 0, it must be the case that anti-anti-15 is just 15!  So once again our prior belief in the distributive property necessitates the conclusion that -5 x -3 = 15.


  1. To compute 5×3, we use the notion of “repeated addition” to get 5×3 = 15.  This means that (pos)(pos) = pos.
  2. To compute 5x-3, we can again use repeated addition, giving 5x-3 = -15.  This shows us that (pos)(neg) = neg.
  3. To compute -5×3, we invoke the distributive rule to see that the answer must be -(5×3) = -15, so that (neg)(pos) = neg.
  4. To compute -5x-3, we also use the distributive rule.  This shows us that the answer must be -(5x-3) = -(-15) = 15.  So (neg)(neg) = pos.

That’s it!  All the sign rules for multiplication can be worked out using straightforward logic, provided we understand 1) what a negative number (or anti-number) is, 2) how to perform addition with both positive and negative numbers, and 3) how the distributive rule works.  (Technically, we also used the fact that (0)(a) = 0 and the fact that additive inverses are unique, but surely these are minor details.)