Multiplicative inverse

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The notation x−1 is sometimes also used for the inverse function, which usually is not equal to the multiplicative inverse. For example, 1/sin x = (sin x)−1 is very different from the inverse of sin x, denoted sin−1 x or arcsin x. Only for linear maps are they strongly related (see below). The terminology difference reciprocal versus inverse is not sufficient to make this distinction, since many authors prefer the opposite naming convention, probably for historical reasons (for example in French, the inverse function is preferably called application réciproque).

Examples and counterexamples[edit]







Complex numbers[edit]

As mentioned above, the reciprocal of every nonzero complex number z = a + bi is complex. It can be found by multiplying both top and bottom of 1/z by its complex conjugate \bar z = a - bi and using the property that z\bar z = \|z\|^2, the absolute value of z squared, which is the real number a2 + b2:

\frac{1}{z} = \frac{\bar z}{z \bar z} = \frac{\bar z}{\|z\|^2} = \frac{a - bi}{a^2 + b^2} = \frac{a}{a^2 + b^2} - \frac{b}{a^2+b^2}i.

In particular, if ||z||=1 (z has unit magnitude), then 1/z = \bar z. Consequently, the imaginary units, ±i, have additive inverse equal to multiplicative inverse, and are the only complex numbers with this property. For example, additive and multiplicative inverses of i are −(i) = −i and 1/i = −i, respectively.


\frac{1}{z} = \frac{1}{r}\left(\cos(-\varphi) + i \sin(-\varphi)\right).



 \frac{d}{dx} x^{-1} = (-1)x^{(-1)-1} = -x^{-2} = -\frac{1}{x^2}.


\int \frac{1}{x}\,dx = \frac{x^0}{0}\ + C

Instead the integral is given by:

\int_1^a \frac{1}{x}\,dx = \ln a,
\int \frac{1}{x}\,dx = \ln x + C.

where ln is the natural logarithm. To show this, note that \frac{d}{dx} e^x = e^x, so if y = e^x and x = \ln y, we have:[2]

\frac{dy}{dx} = y\quad \Rightarrow \quad \frac{dy}{y} = dx \quad\Rightarrow\quad \int \frac{1}{y}\,dy = \int 1\,dx \quad\Rightarrow\quad \int \frac{1}{y}\,dy = x + C = \ln y + C.


Computing the reciprocal is important in many division algorithms, since the quotient a/b can be computed by first computing 1/b and then multiplying it by a. Noting that f(x) = 1/x - b has a zero at x = 1/b, Newton's method can find that zero, starting with a guess x_0 and iterating using the rule:

x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} = x_n - \frac{1/x_n - b}{-1/x_n^2} = 2x_n - bx_n^2 = x_n(2 - bx_n).


x1 = 0.1(2 - 17 × 0.1) = 0.03
x2 = 0.03(2 - 17 × 0.03) = 0.0447
x3 = 0.0447(2 - 17 × 0.0447) ≈ 0.0554
x4 = 0.0554(2 - 17 × 0.0554) ≈ 0.0586
x5 = 0.0586(2 - 17 × 0.0586) ≈ 0.0588





Reciprocals of irrational numbers[edit]

Every number excluding zero has a reciprocal, and reciprocals of certain irrational numbers can have important special properties. Examples include the reciprocal of e (≈ 0.367879)and the golden ratio's reciprocal (≈ 0.618034). The first reciprocal is special because no other positive number can produce a lower number when put to the power of itself; f(1/e) is the global minimum of f(x)=x^x. The second number is the only positive number that is equal to its reciprocal plus one:\phi = 1/\phi + 1. Its additive inverse is the only negative number that is equal to its reciprocal minus one:-\phi = -1/\phi - 1.

The function f(n)=n+\sqrt {(n^2+1)}, n \in N, n>0 gives an infinite number of irrational numbers that differ with their reciprocal by an integer. For example, f(2) is the irrational 2+\sqrt 5. Its reciprocal 1 / (2 + \sqrt 5) is -2 + \sqrt 5, exactly 4 less. Such irrational numbers share a curious property: they have the same fractional part as their reciprocal.

Further remarks[edit]


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:

 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.




See also[edit]



