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The reciprocal function: y = 1/x.For every x except 0, y represents its multiplicative inverse. The graph forms a rectangular hyperbola.. In mathematics, a multiplicative inverse or reciprocal for a number x, denoted by 1/x or x −1, is a number which when multiplied by x yields the multiplicative identity, 1.
t 2 = 6 is the modular multiplicative inverse of 5 × 11 (mod 7) and t 3 = 6 is the modular multiplicative inverse of 5 × 7 (mod 11). Thus, X = 3 × (7 × 11) × 4 + 6 × (5 × 11) × 4 + 6 × (5 × 7) × 6 = 3504. and in its unique reduced form X ≡ 3504 ≡ 39 (mod 385) since 385 is the LCM of 5,7 and 11. Also, the modular multiplicative ...
Once we have defined multiplication for formal power series, we can define multiplicative inverses as follows. The multiplicative inverse of a formal power series A is a formal power series C such that AC = 1, provided that such a formal power series exists. It turns out that if A has a multiplicative inverse, it is unique, and we denote it by ...
In mathematics, the concept of an inverse element generalises the concepts of opposite (−x) and reciprocal (1/x) of numbers. Given an operation denoted here ∗, and an identity element denoted e, if x ∗ y = e, one says that x is a left inverse of y, and that y is a right inverse of x.
The multiplicative inverse of an element may be computed by using the extended Euclidean algorithm (see Extended Euclidean algorithm § Modular integers). Let F {\displaystyle F} be a finite field. For any element x {\displaystyle x} in F {\displaystyle F} and any integer n {\displaystyle n} , denote by n ⋅ x {\displaystyle n\cdot x} the sum ...
The multiplicative identity 1 and its additive inverse −1 are always units. More generally, any root of unity in a ring R is a unit: if r n = 1, then r n−1 is a multiplicative inverse of r. In a nonzero ring, the element 0 is not a unit, so R × is not closed under addition.
Finally, given a, the multiplicative inverse of a modulo n is an integer x satisfying ax ≡ 1 (mod n). It exists precisely when a is coprime to n , because in that case gcd( a , n ) = 1 and by Bézout's lemma there are integers x and y satisfying ax + ny = 1 .
For example, given b = 5, e = 3 and m = 13, dividing 5 3 = 125 by 13 leaves a remainder of c = 8. Modular exponentiation can be performed with a negative exponent e by finding the modular multiplicative inverse d of b modulo m using the extended Euclidean algorithm. That is: c = b e mod m = d −e mod m, where e < 0 and b ⋅ d ≡ 1 (mod m).