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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.
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 ...
SRS can be solved in polynomial time in the Real RAM model. [3] However, its run-time complexity in the Turing machine model is open, as of 1997. [1] The main difficulty is that, in order to solve the problem, the square-roots should be computed to a high accuracy, which may require a large number of bits.
A square root of a number x is a number r which, when squared, becomes x: =. Every positive real number has two square roots, one positive and one negative. For example, the two square roots of 25 are 5 and −5. The positive square root is also known as the principal square root, and is denoted with a radical sign:
The principal square root of a real positive semidefinite matrix is real. [3] The principal square root of a positive definite matrix is positive definite; more generally, the rank of the principal square root of A is the same as the rank of A. [3] The operation of taking the principal square root is continuous on this set of matrices. [4]
For example, if one computes the integer square root of 2000000 using the algorithm above, one obtains the sequence In total 13 iteration steps are needed. Although Heron's method converges quadratically close to the solution, less than one bit precision per iteration is gained at the beginning.
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.
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.
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