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For example, taking the statement x + 1 = 0, if x is substituted with 1, this implies 1 + 1 = 2 = 0, which is false, which implies that if x + 1 = 0 then x cannot be 1. If x and y are integers, rationals, or real numbers, then xy = 0 implies x = 0 or y = 0. Consider abc = 0. Then, substituting a for x and bc for y, we learn a = 0 or bc = 0.
The equations 3x + 2y = 6 and 3x + 2y = 12 are inconsistent. A linear system is inconsistent if it has no solution, and otherwise, it is said to be consistent. [7] When the system is inconsistent, it is possible to derive a contradiction from the equations, that may always be rewritten as the statement 0 = 1. For example, the equations
An example of using Newton–Raphson method to solve numerically the equation f(x) = 0. In mathematics, to solve an equation is to find its solutions, which are the values (numbers, functions, sets, etc.) that fulfill the condition stated by the equation, consisting generally of two expressions related by an equals sign.
Vertical line of equation x = a Horizontal line of equation y = b. Each solution (x, y) of a linear equation + + = may be viewed as the Cartesian coordinates of a point in the Euclidean plane. With this interpretation, all solutions of the equation form a line, provided that a and b are not both zero. Conversely, every line is the set of all ...
An example of an equation involving x and y as unknowns and the parameter R is x 2 + y 2 = R 2 . {\displaystyle x^{2}+y^{2}=R^{2}.} When R is chosen to have the value of 2 ( R = 2), this equation would be recognized in Cartesian coordinates as the equation for the circle of radius of 2 around the origin.
Then, f(x)g(x) = 4x 2 + 4x + 1 = 1. Thus deg(f⋅g) = 0 which is not greater than the degrees of f and g (which each had degree 1). Since the norm function is not defined for the zero element of the ring, we consider the degree of the polynomial f(x) = 0 to also be undefined so that it follows the rules of a norm in a Euclidean domain.
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When X 0 (n) has genus one, it will itself be isomorphic to an elliptic curve, which will have the same j-invariant. For instance, X 0 (11) has j-invariant −2 12 11 −5 31 3, and is isomorphic to the curve y 2 + y = x 3 − x 2 − 10x − 20. If we substitute this value of j for y in X 0 (5), we obtain two rational roots and a factor of ...