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In modular arithmetic, the modular multiplicative inverse of a is also defined: it is the number x such that ax ≡ 1 (mod n). This multiplicative inverse exists if and only if a and n are coprime. For example, the inverse of 3 modulo 11 is 4 because 4 ⋅ 3 ≡ 1 (mod 11). The extended Euclidean algorithm may be used to compute it.
The set {3,19} generates the group, which means that every element of (/) is of the form 3 a × 19 b (where a is 0, 1, 2, or 3, because the element 3 has order 4, and similarly b is 0 or 1, because the element 19 has order 2).
The sum, the difference and the product are the remainder of the division by of the result of the corresponding integer operation. The multiplicative inverse of an element may be computed by using the extended Euclidean algorithm (see Extended Euclidean algorithm § Modular integers). Let be a finite field.
A modular multiplicative inverse of a modulo m can be found by using the extended Euclidean algorithm. The Euclidean algorithm determines the greatest common divisor (gcd) of two integers, say a and m. If a has a multiplicative inverse modulo m, this gcd must be 1. The last of several equations produced by the algorithm may be solved for this gcd.
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 ...
This integer a −1 is called a modular multiplicative inverse of a modulo m. If a ≡ b (mod m) and a −1 exists, then a −1 ≡ b −1 (mod m) (compatibility with multiplicative inverse, and, if a = b, uniqueness modulo m). If ax ≡ b (mod m) and a is coprime to m, then the solution to this linear congruence is given by x ≡ a −1 b (mod m).
The number 3 is a primitive root modulo 7 [5] because = = = = = = = = = = = = (). Here we see that the period of 3 k modulo 7 is 6. The remainders in the period, which are 3, 2, 6, 4, 5, 1, form a rearrangement of all nonzero remainders modulo 7, implying that 3 is indeed a primitive root modulo 7.
For instance, to solve the inequality 4x < 2x + 1 ≤ 3x + 2, it is not possible to isolate x in any one part of the inequality through addition or subtraction. Instead, the inequalities must be solved independently, yielding x < 1 / 2 and x ≥ −1 respectively, which can be combined into the final solution −1 ≤ x < 1 / 2 .