# What is multiplication?

Ordinarily, multiplication is interpreted as repeated addition:

$8\times 5 = 8+8+8+8+8 = 40$.

But what does it mean to multiply two arbitrary real numbers, such as $(8.304\ldots)\times(-5.692\ldots)$?

Well, if the second number is a positive integer like $5$, our interpretation still applies no matter what the first number is: just add together five copies of it. So we can do multiplication of the form $n\times 5$, or really $n \times$(any positive integer). All that’s left is to decide what to do when instead of $5$ we have something like $-5.692\ldots$, with a possible minus sign and maybe digits after the decimal point.

For decimals, we can figure out what multiplying by $0.6$ should mean: since $0.6\times 10 = 6$, we know that whatever $n$ is, the product $n\times 0.6$ should have the property that

$(n \times 0.6) \times 10 = n\times (0.6\times 10) = n\times 6$.

(The validity of shifting parentheses around is called associativity.) It turns out that solving a problem of the form “What times 10 is a given number?” is easy: the unique solution is found by moving the decimal point in the given number one place to the left, and this is called dividing by 10. So we can calculate $n\times 0.6$ by first computing $n\times 6$ (which we know how to do: add $n$ together six times) and then divide the result by 10.

Similarly, since $0.09 = \frac{9}{100}$, we can calculate $n\times 0.09$ by first calculating $n\times 9$ and then dividing by 10 twice, and so on. This lets us calculate all the individual pieces $n\times 5$, $n\times 0.6$, $n\times 0.09$, $n\times 0.002$, etc., and we can add them up to multiply $n$ by a decimal with any finite number of digits:

\begin{aligned} n\times 5.6 &= n\times (5 + 0.6) \\&= (n\times 5) + (n\times 0.6),\\ n\times 5.69 &= n\times (5 + 0.6 + 0.09) \\&= (n\times 5) + (n\times 0.6) + (n\times 0.09),\\ n\times 5.692 &= n\times (5 + 0.6 + 0.09 + 0.002) \\ &= (n\times 5) + (n\times 0.6) + (n\times 0.09) + (n\times 0.002). \end{aligned}

Here we’re distributing the $n\times$ to each term in the finite sums using the law of distributivity. But distributivity by itself isn’t enough to handle an infinite number of digits after the decimal point; to pin down the value of $n\times (5.692\ldots)$ we need something like continuity:

Since $5.692\ldots$ is the limit of the sequence $(5, 5.6, 5.69, 5.692, \ldots)$, the product $n\times (5.692\ldots)$ is the limit of the sequence $(n\times 5, n\times 5.6, n\times 5.69, n\times 5.692, \ldots)$.

Or we could use the ordering of the real numbers:

Since $5.692\ldots$ is between $5$ and $6$, and between $5.6$ and $5.7$, and between $5.69$ and $5.70$, etc., we can deduce that $n\times (5.692\ldots)$ is

• between $n\times 5$ and $n\times 6$, and
• between $n\times 5.6$ and $n\times 5.7$, and
• between $n \times 5.69$ and $n\times 5.70$, and

There’s only one number between all these pairs of products, so that’s the number we should call $n\times (5.692\ldots)$.

Either way, we now know what it means to multiply any real number by any positive real number. To handle the rest of the cases, we just have to know how to multiply by zero or by negative numbers, and for that we use distributivity again:

Whatever $n\times 0$ is, it has the property that $(n\times 0) + (n\times 0) = n\times (0+0) = n\times 0$. Subtracting one copy of $n\times 0$ from each side, we get $n\times 0 = 0$.

(A more modern viewpoint is that the rule $n\times 0 = 0$ is really a special case of distributivity: just like $5+3$ is a sum of two numbers and $4+6+21+93+12$ is a sum of five numbers, we regard $0$ as the result when we add up no numbers, the “empty sum.” So if we start with $n\times 0$, think of it as $n$ times the empty sum, and then distribute the $n\times$ across the (no) terms, we end up with another empty sum, so the result is again $0$. In general, the “empty” version of a binary operation is the identity element for that operation, so the “empty sum” is the additive identity, $0$, while the “empty product” is the multiplicative identity, $1$.)

And for negative numbers:

If $-m$ is a negative number and we want to calculate $n\times (-m)$, start by multiplying $n$ by the positive number $m$. Whatever $n\times (-m)$ is, its sum with $n\times m$ must be zero:

$n\times (-m) + n\times m = n\times (-m+m) = n\times 0 = 0$.

Therefore $n\times (-m)$ and $n\times m$ are additive inverses, so $n\times -m = -(n\times m)$.

And there we go! To multiply $(8.304\ldots)\times(-5.692\ldots)$, we first:

• Compute $(8.304\ldots)\times 5$$(8.304\ldots)\times 0.6$$(8.304\ldots)\times 0.09\ldots)$$(8.304\ldots)\times0.002$, etc.
• Add up that infinite series to get $(8.304\ldots)\times(5.692\ldots)$.
• And then take the negative of that to get $(8.304\ldots)\times(-5.692\ldots)$.

Now we can multiply any two real numbers!

To summarize the process we used to extend multiplication to all real numbers:

(This is how we found the values for things like $n\times 0.6$ or $n$ times a negative number: if associativity is going to keep holding, $n\times 0.6$ needs to equal one tenth of $n\times 6$, and distributivity forces $n\times (-m)$ to be the negative of $n\times m$. In general this tells us what $n\times$ any rational number must equal.)
(We used continuity/order properties to extend from $n\times$ any rational number to $n\times$ any real number.)