Multiplicative Inverses

Subtraction was solved by inventing additive inverses. Division is solved by inventing multiplicative inverses. In $\mathbb{Z}$, you can subtract anything because every integer $a$ has a partner $-a$ with $a + (-a) = 0$. The same idea lifts to multiplication once we move into $\mathbb{Q}$: every nonzero rational has a partner that multiplies with it to give $1$.

Defining the reciprocal

The multiplicative inverse (or reciprocal) of a nonzero number $a$ is the unique number $a^{-1}$ such that $a \cdot a^{-1} = 1$. Notation: we also write $\dfrac{1}{a}$. The two notations mean the same thing.

For an integer $a \neq 0$: $a^{-1} = \dfrac{1}{a}$. For a fraction: flip it. $\left(\dfrac{a}{b}\right)^{-1} = \dfrac{b}{a}$. Check: $\dfrac{a}{b} \cdot \dfrac{b}{a} = \dfrac{ab}{ba} = 1$.

The widget makes the flip visual

(Set A = 3/4 in the widget. The readout shows the reciprocal 4/3, and the product A × reciprocal(A) = 1. Now try A = 7/2: the reciprocal becomes 2/7. The relationship is purely a swap of numerator and denominator.)

Why zero is excluded

The rule "every nonzero number has a reciprocal" cannot extend to zero. Suppose $0^{-1} = x$ existed for some number $x$. Then by definition $0 \cdot x = 1$. But the distributive law forces $0 \cdot x = 0$ for every $x$ (we proved this in §1.3). So $0 = 1$, which is false. Hence no such $x$ exists. Zero is the one and only number in $\mathbb{Q}$ without a multiplicative inverse, and this single exception is the entire reason division by zero is undefined.

Division IS multiplication by the inverse

The deepest consequence: once we have inverses, division is no longer a fundamental operation. It is just multiplication in disguise:

This is why "divide by a fraction = multiply by its reciprocal" works. The rule looks like a memorisation hack; it is actually the definition of division unpacked.

Solving simple equations

To solve $\dfrac{2}{3} x = 8$, multiply both sides by the inverse $\dfrac{3}{2}$:

$\dfrac{3}{2} \cdot \dfrac{2}{3} x = \dfrac{3}{2} \cdot 8 \quad\Rightarrow\quad x = 12$.

The left side collapsed because $\dfrac{3}{2}$ and $\dfrac{2}{3}$ are reciprocals. This is the engine for every linear equation you will solve from now on.

Try it

• Predict first: what is the reciprocal of $5/8$? Set $A = 5/8$ in the widget to verify, then multiply to confirm the product is $1$.
• Find the reciprocal of $-\dfrac{3}{7}$. Watch the sign carefully — reciprocals preserve sign.
• Solve $\dfrac{3}{4} x = 9$ by multiplying both sides by the reciprocal of $\dfrac{3}{4}$.
• What is $\left( a^{-1} \right)^{-1}$ in general? (Take the reciprocal twice.)

A trap to watch for

The reciprocal of a sum is not the sum of the reciprocals. Concretely, $\dfrac{1}{a + b} \neq \dfrac{1}{a} + \dfrac{1}{b}$. Try $a = 2, b = 3$: the left side is $\dfrac{1}{5}$, the right side is $\dfrac{1}{2} + \dfrac{1}{3} = \dfrac{5}{6}$. The two are wildly different. The fix: reciprocals distribute over multiplication but not addition. Always rewrite $(a+b)^{-1}$ as $\dfrac{1}{a + b}$ first and resist the urge to split the denominator.

What you now know

You can now construct the reciprocal of any nonzero rational, and you understand why zero is excluded. Division is multiplication by the inverse, and that single insight will run every linear equation, every fraction simplification, and every cancellation you do for the rest of this book. Chapter 2 builds on this: with addition, multiplication, and inverses all settled, we can finally solve systems of linear equations.

Quick check

References

• Lang, S. (1971). Basic Mathematics. Springer. Chapter 1, §6 — reciprocals defined and the field axioms completed.
• Stewart, I. and Tall, D. (2015). The Foundations of Mathematics, 2nd ed. Oxford. Chapter 5 — rationals as the field of fractions of $\mathbb{Z}$.
• Spivak, M. (2008). Calculus, 4th ed. Publish or Perish. Chapter 1 — the field axioms presented as the rules of algebra that we never bother to think about.