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In the factor ring /, the equivalence class of 3 is nilpotent because 3 2 is congruent to 0 modulo 9. Assume that two elements a {\displaystyle a} and b {\displaystyle b} in a ring R {\displaystyle R} satisfy a b = 0 {\displaystyle ab=0} .
More generally, every integral domain is a reduced ring since a nilpotent element is a fortiori a zero-divisor. On the other hand, not every reduced ring is an integral domain; for example, the ring Z [ x , y ]/( xy ) contains x + ( xy ) and y + ( xy ) as zero-divisors, but no non-zero nilpotent elements.
For example, any nonzero 2 × 2 nilpotent matrix is similar to the matrix []. That is, if is any nonzero 2 × 2 nilpotent matrix, then there exists a basis b 1, b 2 such that Nb 1 = 0 and Nb 2 = b 1. This classification theorem holds for matrices over any field. (It is not necessary for the field to be algebraically closed.)
In mathematics, specifically in ring theory, a nilpotent algebra over a commutative ring is an algebra over a commutative ring, in which for some positive integer n every product containing at least n elements of the algebra is zero.
A simple and sufficient test for the absence of a dependence is the greatest common divisor (GCD) test. It is based on the observation that if a loop carried dependency exists between X[a*i + b] and X[c*i + d] (where X is the array; a, b, c and d are integers, and i is the loop variable), then GCD (c, a) must divide (d – b).
Thus, the second partial derivative test indicates that f(x, y) has saddle points at (0, −1) and (1, −1) and has a local maximum at (,) since = <. At the remaining critical point (0, 0) the second derivative test is insufficient, and one must use higher order tests or other tools to determine the behavior of the function at this point.
Take each digit of the number (371) in reverse order (173), multiplying them successively by the digits 1, 3, 2, 6, 4, 5, repeating with this sequence of multipliers as long as necessary (1, 3, 2, 6, 4, 5, 1, 3, 2, 6, 4, 5, ...), and adding the products (1×1 + 7×3 + 3×2 = 1 + 21 + 6 = 28). The original number is divisible by 7 if and only if ...
Divide the first term of the dividend by the highest term of the divisor (x 3 ÷ x = x 2). Place the result below the bar. x 3 has been divided leaving no remainder, and can therefore be marked as used by crossing it out. The result x 2 is then multiplied by the second term in the divisor −3 = −3x 2.