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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 ...
Two to the power of n, written as 2 n, is the number of values in which the bits in a binary word of length n can be set, where each bit is either of two values. A word, interpreted as representing an integer in a range starting at zero, referred to as an "unsigned integer", can represent values from 0 (000...000 2) to 2 n − 1 (111...111 2) inclusively.
An exact number has an infinite number of significant figures. If the number of apples in a bag is 4 (exact number), then this number is 4.0000... (with infinite trailing zeros to the right of the decimal point). As a result, 4 does not impact the number of significant figures or digits in the result of calculations with it.
One half is the rational number that lies midway between 0 and 1 on the number line. Multiplication by one half is equivalent to division by two, or "halving"; conversely, division by one half is equivalent to multiplication by two, or "doubling". A square of side length one, here dissected into rectangles whose areas are successive powers of ...
Given two real numbers x and y, find exponents m and n such that = by solving = for integers m and n to whatever degree of precision you want by expanding the continued fraction of = to whatever convergent is as close as you want.
Assuming the mass of ordinary matter is about 1.45 × 10 53 kg as discussed above, and assuming all atoms are hydrogen atoms (which are about 74% of all atoms in the Milky Way by mass), the estimated total number of atoms in the observable universe is obtained by dividing the mass of ordinary matter by the mass of a hydrogen atom.
Let be the number of possible values of a hash function, with =. With a birthday attack, it is possible to find a collision of a hash function with 50 % {\textstyle 50\%} chance in 2 l = 2 l / 2 , {\textstyle {\sqrt {2^{l}}}=2^{l/2},} where l {\textstyle l} is the bit length of the hash output, [ 1 ] [ 2 ] and with 2 l − 1 {\textstyle 2^{l-1 ...