Multiplicative Inverse - Further Remarks

Further Remarks

If the multiplication is associative, an element x with a multiplicative inverse cannot be a zero divisor (meaning for some y, xy = 0 with neither x nor y equal to zero). To see this, it is sufficient to multiply the equation xy = 0 by the inverse of x (on the left), and then simplify using associativity. In the absence of associativity, the sedenions provide a counterexample.

The converse does not hold: an element which is not a zero divisor is not guaranteed to have a multiplicative inverse. Within Z, all integers except −1, 0, 1 provide examples; they are not zero divisors nor do they have inverses in Z. If the ring or algebra is finite, however, then all elements a which are not zero divisors do have a (left and right) inverse. For, first observe that the map ƒ(x) = ax must be injective: ƒ(x) = ƒ(y) implies x = y:

\begin{align} ax &= ay &\quad \rArr & \quad ax-ay = 0 \\ & &\quad \rArr &\quad a(x-y) = 0 \\ & &\quad \rArr &\quad x-y = 0 \\ & &\quad \rArr &\quad x = y. \end{align}

Distinct elements map to distinct elements, so the image consists of the same finite number of elements, and the map is necessarily surjective. Specifically, ƒ (namely multiplication by a) must map some element x to 1, ax = 1, so that x is an inverse for a.

The multiplicative inverse of a fraction is simply

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