Multiplicative Inverses

Multiplicative Inverses

Additive inverses let us undo addition: to cancel a summand, we add its inverse. Multiplicative inverses play a similar role for multiplication. The field axioms establish that every nonzero real number has a unique multiplicative inverse. Here we develop the notation for that inverse, establish how it behaves in products and quotients, and use it to solve simple equations.

The multiplicative identity \(1\) is the number that leaves every real number unchanged under multiplication. A multiplicative inverse is therefore the number that combines with \(x\) to produce that identity. As with additive inverses, uniqueness matters: if two real numbers both multiply by \(x\) to give \(1\), they must be equal. We use this uniqueness, established from the field axioms in Field Axioms, to identify inverse expressions.

The condition \(x\ne0\) is essential. Multiplication by zero always gives zero, so there is no real number \(y\) for which \(0y=1\). Consequently, zero has no multiplicative inverse. This is why the notation \(\frac{1}{x}\) is defined only when \(x\ne0\), and why division by zero is not defined.

For \(x\ne0\), the notation \(x^{-1}\) names a number; it does not mean that \(x\) is being raised to a power that must be evaluated by some separate process. Its defining property is the equation \(xx^{-1}=1\). Likewise, the fraction notation \(\frac{1}{x}\) denotes this same number. The field axioms guarantee its existence and uniqueness.

Basic Identities for Multiplicative Inverses

Several useful identities follow directly from the defining equation and uniqueness. The inverse of \(1\) is \(1\), since \(1\cdot1=1\). The inverse of an inverse returns the original number: if \(x\ne0\), then \(x^{-1}\ne0\), and \(x^{-1}x=1\) shows that \(x\) is the multiplicative inverse of \(x^{-1}\). By uniqueness, \((x^{-1})^{-1}=x\).

Proof. Since \(1\cdot1=1\), the number \(1\) is the multiplicative inverse of \(1\), and uniqueness gives \(1^{-1}=1\). Next, if \(x^{-1}=0\), then the defining equation \(xx^{-1}=1\) would give \(x\cdot0=1\). But multiplication by zero gives \(x\cdot0=0\), so this would imply \(0=1\), contrary to the field axioms. Hence \(x^{-1}\ne0\). Finally, \(x^{-1}x=1\), so \(x\) is a number whose product with \(x^{-1}\) is \(1\). By uniqueness of the multiplicative inverse of \(x^{-1}\), \((x^{-1})^{-1}=x\). \(\square\)

The Inverse of a Product

When two nonzero numbers are multiplied, their product is also nonzero. Indeed, if \(x\ne0\) and \(y\ne0\), then \(xy\ne0\) by the zero-product property from Field Axioms. Therefore \(xy\) has a multiplicative inverse. The inverse of the product is obtained by multiplying the individual inverses.

Proof. Since \(x\ne0\) and \(y\ne0\), both \(x^{-1}\) and \(y^{-1}\) exist, and \(xy\ne0\). By associativity and commutativity of multiplication,

Thus \(x^{-1}y^{-1}\) is a multiplicative inverse of \(xy\). By uniqueness of that inverse, \((xy)^{-1}=x^{-1}y^{-1}\). \(\square\)

The order of the factors does not cause a problem in the real numbers, because multiplication is commutative. The same formula can be applied repeatedly: the inverse of a product of finitely many nonzero real numbers is the product of their inverses. The requirement that every factor be nonzero ensures each individual inverse exists.

Reciprocals and Quotients

Division is defined using multiplicative inverses, just as subtraction is defined using additive inverses. If \(y\ne0\), the quotient \(x/y\) means the product of \(x\) and the inverse of \(y\). This definition keeps the restriction visible: the denominator must be nonzero.

When both \(x\) and \(y\) are nonzero, the quotient is also nonzero, and its inverse has a useful form. The numerator and denominator exchange places. This follows from the product formula, together with the fact that the inverse of an inverse is the original number.

Proof. Since \(x\ne0\) and \(y\ne0\), both quotients in the claim are defined, and \(xy^{-1}\ne0\). By the inverse-of-a-product theorem and the basic identities,

By commutativity of multiplication, \(x^{-1}y=yx^{-1}\). By the definition of quotient, \(yx^{-1}=y/x\). Therefore \(\left(\frac{x}{y}\right)^{-1}=y/x\). \(\square\)

The Reciprocal Map

The operation of taking a reciprocal can also be viewed as a function. Its domain must exclude zero, because zero has no multiplicative inverse. Its outputs are also nonzero, so the nonzero reals are a natural codomain. The identity \((x^{-1})^{-1}=x\) shows that applying this function twice recovers the input.

Proof. For every nonzero \(x\), the basic inverse identities give

Hence \(R\circ R=\operatorname{id}_{\mathbb{R}\setminus\{0\}}\). To prove injectivity, suppose \(R(x)=R(y)\). Apply \(R\) to both sides. Then

so \(x=y\). To prove surjectivity, let \(z\in\mathbb{R}\setminus\{0\}\). Choose \(x=z^{-1}\), which is also nonzero. Then

Thus every element of the codomain is an output, and \(R\) is surjective. It is both injective and surjective, hence bijective. Since applying \(R\) twice is the identity, \(R\) is its own inverse function: \(R^{-1}=R\). \(\square\)

This theorem parallels the earlier result that negation is a bijection and its own inverse. The domain restriction is different: negation is defined for every real number, whereas taking a multiplicative inverse requires a nonzero input. On its proper domain, taking reciprocals loses no information.

Using Inverses to Solve Equations

If \(a\ne0\), multiplication by \(a\) can be undone by multiplication by \(a^{-1}\). This gives a direct way to solve an equation of the form \(ax=b\). The earlier result that multiplication by a nonzero real number is bijective expresses this reversibility as a function; the calculation below shows how the inverse acts on a particular equation.

Why the Restrictions Matter

The inverse-of-a-product formula is sometimes remembered as “take the reciprocal of each factor.” The hypotheses are part of the rule: each factor must be nonzero. If even one factor is zero, the product is zero, and the product has no multiplicative inverse. Similarly, the inverse-of-a-quotient formula requires both numerator and denominator to be nonzero. A quotient with zero numerator equals zero, which has no reciprocal.

One should also distinguish a reciprocal from an additive inverse. The additive inverse \(-x\) satisfies \(x+(-x)=0\), while the multiplicative inverse \(x^{-1}\) satisfies \(xx^{-1}=1\), when \(x\ne0\). They serve different operations and generally name different numbers. For example, the additive inverse of \(4\) is \(-4\), whereas its multiplicative inverse is \(\frac14\).

A dependable way to check a proposed multiplicative inverse is to multiply it by the original number and verify that the product is \(1\). This check also exposes missing conditions: if the original number is zero, no candidate can pass, since multiplying any real number by zero gives zero rather than one.