Rules for Multiplication

Multiplication has a problem of its own: when you multiply two negative numbers, you get a positive. Why? It is not arbitrary. The rule is forced on us by insisting that the distributive law $a(b+c) = ab + ac$ keep working when negatives are allowed. Once we accept that the distributive law holds, every sign rule below is unavoidable.

The four laws — plus one

Multiplication keeps three of the four laws of addition, then adds the bridge that ties multiplication and addition together.

Commutative. $a \cdot b = b \cdot a$. Order does not matter.

Associative. $(a \cdot b) \cdot c = a \cdot (b \cdot c)$. Grouping does not matter.

Identity. $a \cdot 1 = a$. One is the do-nothing element for multiplication, just as zero is for addition.

Distributive. $a(b + c) = ab + ac$. Multiplication distributes over addition. This is the bridge between the two operations — everything that follows depends on it.

Sign rules — forced, not invented

The product of two integers obeys this table:

Same sign → positive. Different signs → negative. The interesting case is the last one. Here is the proof, two lines long. Start with $(-1)(1 + (-1)) = (-1)(0) = 0$. Distribute: $(-1)(1) + (-1)(-1) = 0$. The first term is $-1$, so $-1 + (-1)(-1) = 0$, hence $(-1)(-1) = 1$. The positive answer was forced by the distributive law and the additive inverse — not chosen.

The number line says it too

If you think of multiplication by $-1$ as a reflection across zero, then multiplying by $-1$ twice reflects you twice — back where you started. Two flips of the number line equal no flip at all. Click the $(-)(-)$ cell in the widget below to walk through the distributive-law proof; click the other three cells to see the area model for each sign rule.

Zero kills everything

For every integer $a$, $a \cdot 0 = 0$. Proof: $a \cdot 0 = a(0 + 0) = a \cdot 0 + a \cdot 0$. Subtract $a \cdot 0$ from both sides and you get $0 = a \cdot 0$. The same distributive law that forces $(-)(-) = (+)$ also forces zero to annihilate every product.

Try it

• Compute $(-3)(-4)(2)(-1)$. Count the negative factors first; the sign of the answer is the parity of that count.
• Expand $-5(2x - 3)$ in one step using the distributive law. What goes wrong if you forget that the $-5$ multiplies the $-3$?
• Factor $6x + 15$ by pulling out the greatest common factor of $6$ and $15$.
• If $ab = 0$ and $a \neq 0$, what must $b$ be? (This is the "zero product property" you will need for quadratics.)

Try it in code

A trap to watch for

Beginners often write $-3^2 = 9$. This is wrong. By the standard order of operations, exponents bind tighter than the unary minus, so $-3^2$ means $-(3^2) = -9$. If you want the $-3$ squared, you must write $(-3)^2 = 9$ explicitly. The fix: when you mean "the square of a negative number," write the parentheses every single time, until your eye learns to flag the missing ones.

What you now know

Multiplication has its own four laws plus the distributive bridge that links it to addition. The sign rules — including the troublesome $(-)(-) = (+)$ — are not arbitrary conventions but consequences of the distributive law. The next section uses these laws to classify integers as even or odd, and to talk about divisibility — the first hint of the deep number theory waiting beyond this textbook.

Quick check

References

• Lang, S. (1971). Basic Mathematics. Springer. Chapter 1, §3 — the canonical distributive-law-forces-everything proof of the sign rules.
• Hardy, G. H. and Wright, E. M. (2008). An Introduction to the Theory of Numbers, 6th ed. Oxford. Chapter 1 — the integers and their ring structure.
• Devlin, K. (1994). Mathematics: The Science of Patterns. Scientific American Library. Chapter 2 explains why $(-1)(-1) = 1$ is a consequence of insisting that the familiar laws keep working.