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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 ...
For example, to factor =, the first try for a is the square root of 5959 rounded up to the next integer, which is 78. Then b 2 = 78 2 − 5959 = 125 {\displaystyle b^{2}=78^{2}-5959=125} . Since 125 is not a square, a second try is made by increasing the value of a by 1.
This corresponds to a set of y values whose product is a square number, i.e. one whose factorization has only even exponents. The products of x and y values together form a congruence of squares. This is a classic system of linear equations problem, and can be efficiently solved using Gaussian elimination as soon as the number of rows exceeds ...
The square root of a positive integer is the product of the roots of its prime factors, because the square root of a product is the product of the square roots of the factors. Since p 2 k = p k , {\textstyle {\sqrt {p^{2k}}}=p^{k},} only roots of those primes having an odd power in the factorization are necessary.
In mathematics, a proof by infinite descent, also known as Fermat's method of descent, is a particular kind of proof by contradiction [1] used to show that a statement cannot possibly hold for any number, by showing that if the statement were to hold for a number, then the same would be true for a smaller number, leading to an infinite descent and ultimately a contradiction. [2]
In number theory, the integer square root (isqrt) of a non-negative integer n is the non-negative integer m which is the greatest integer less than or equal to the square root of n, = ⌊ ⌋. For example, isqrt ( 27 ) = ⌊ 27 ⌋ = ⌊ 5.19615242270663... ⌋ = 5. {\displaystyle \operatorname {isqrt} (27)=\lfloor {\sqrt {27}}\rfloor ...
The modular square root of can be taken this way. Having solved the associated quadratic equation we now have the variables w {\displaystyle w} and set v {\displaystyle v} = r {\displaystyle r} (if C {\displaystyle C} in the quadratic is a natural square).
By performing this iteration, it is possible to evaluate a square root to any desired accuracy by only using the basic arithmetic operations. The following three tables show examples of the result of this computation for finding the square root of 612, with the iteration initialized at the values of 1, 10, and −20.