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In modular arithmetic, a number g is a primitive root modulo n if every number a coprime to n is congruent to a power of g modulo n. That is, g is a primitive root modulo n if for every integer a coprime to n, there is some integer k for which g k ≡ a (mod n). Such a value k is called the index or discrete logarithm of a to the base g modulo n.
A square root of a number x is a number r which, when squared, becomes x: =. Every positive real number has two square roots, one positive and one negative. For example, the two square roots of 25 are 5 and −5. The positive square root is also known as the principal square root, and is denoted with a radical sign:
The fourth digit of the answer is and carry to the next digit. Continue with the same method to obtain the remaining digits. 2 Finger method. Trachtenberg called this the 2 Finger Method. The calculations for finding the fourth digit from the example above are illustrated at right.
Given real numbers x and y, integers m and n and the set of integers, floor and ceiling may be defined by the equations ⌊ ⌋ = {}, ⌈ ⌉ = {}. Since there is exactly one integer in a half-open interval of length one, for any real number x, there are unique integers m and n satisfying the equation
A method analogous to piece-wise linear approximation but using only arithmetic instead of algebraic equations, uses the multiplication tables in reverse: the square root of a number between 1 and 100 is between 1 and 10, so if we know 25 is a perfect square (5 × 5), and 36 is a perfect square (6 × 6), then the square root of a number greater than or equal to 25 but less than 36, begins with ...
In these examples, the (negative) least absolute remainder is obtained from the least positive remainder by subtracting 5, which is d. This holds in general. When dividing by d, either both remainders are positive and therefore equal, or they have opposite signs. If the positive remainder is r 1, and the negative one is r 2, then r 1 = r 2 + d.
A quadratic with two real roots, for example, will have exactly two angles that satisfy the above conditions. For complex roots, one must also find a series of similar triangles, but with the vertices of the root path displaced from the polynomial path by a distance equal to the imaginary part of the root. In this case, the root path will not ...
After each step, be sure the remainder for that step is less than the divisor. If it is not, there are three possible problems: the multiplication is wrong, the subtraction is wrong, or a greater quotient is needed. In the end, the remainder, r, is added to the growing quotient as a fraction, r ⁄ m.