Editing Multiplicative inverse (section)

==Further remarks==
If the multiplication is associative, an element ''x'' with a multiplicative inverse cannot be a [[zero divisor]] (''x'' is a zero divisor if some nonzero ''y'', {{nowrap|1=''xy'' = 0}}). To see this, it is sufficient to multiply the equation {{nowrap|1=''xy'' = 0}} by the inverse of ''x'' (on the left), and then simplify using associativity. In the absence of associativity, the [[sedenion]]s 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 set|finite]], however, then all elements ''a'' which are not zero divisors do have a (left and right) inverse. For, first observe that the map {{nowrap|1=''f''(''x'') = ''ax''}} must be [[injective]]: {{nowrap|1=''f''(''x'') = ''f''(''y'')}} implies {{nowrap|1=''x'' = ''y''}}:
:<math>\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}</math>
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, {{nowrap|1=''ax'' = 1}}, so that ''x'' is an inverse for ''a''.