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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 ...
A necessary (but not sufficient) condition for solvability is that n is not divisible by 4 or by a prime of form 4k + 3. [note 3] Thus, for example, x 2 − 3 y 2 = −1 is never solvable, but x 2 − 5 y 2 = −1 may be. [27] The first few numbers n for which x 2 − n y 2 = −1 is solvable are with only one trivial solution: 1
Lighting and reflection calculations, as in the video game OpenArena, use the fast inverse square root code to compute angles of incidence and reflection.. Fast inverse square root, sometimes referred to as Fast InvSqrt() or by the hexadecimal constant 0x5F3759DF, is an algorithm that estimates , the reciprocal (or multiplicative inverse) of the square root of a 32-bit floating-point number in ...
Notation for the (principal) square root of x. For example, √ 25 = 5, since 25 = 5 ⋅ 5, or 5 2 (5 squared). In mathematics, a square root of a number x is a number y such that =; in other words, a number y whose square (the result of multiplying the number by itself, or ) is x. [1]
A root of degree 2 is called a square root and a root of degree 3, a cube root. Roots of higher degree are referred by using ordinal numbers, as in fourth root, twentieth root, etc. The computation of an n th root is a root extraction. For example, 3 is a square root of 9, since 3 2 = 9, and −3 is also a square root of 9, since (−3) 2 = 9.
In the case of two nested square roots, the following theorem completely solves the problem of denesting. [2]If a and c are rational numbers and c is not the square of a rational number, there are two rational numbers x and y such that + = if and only if is the square of a rational number d.
Thought of quotitively, a division problem can be solved by repeatedly subtracting groups of the size of the divisor. [1] For instance, suppose each egg carton fits 12 eggs, and the problem is to find how many cartons are needed to fit 36 eggs in total. Groups of 12 eggs at a time can be separated from the main pile until none are left, 3 groups:
The fourth quotient being 1, we say 333 times 1 is 333, and this plus 22, the numerator of the fraction preceding, is 355; similarly, 106 times 1 is 106, and this plus 7 is 113. In this manner, by employing the four quotients [3;7,15,1], we obtain the four fractions: 3 / 1 , 22 / 7 , 333 / 106 , 355 / 113 , ....