
Question 1: Use Euclid`s division algorithm to find the HCF of
... a = 6q + rfor some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 6. Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5 Also, 6q + 1 = 2 × 3q + 1 = 2k1 + 1, where k1 is a positive integer 6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k2 + 1, where k2 is an integer 6q + 5 = (6q ...
... a = 6q + rfor some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 6. Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5 Also, 6q + 1 = 2 × 3q + 1 = 2k1 + 1, where k1 is a positive integer 6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k2 + 1, where k2 is an integer 6q + 5 = (6q ...
2013
... where {x} denotes the fractional part of x, 0 ≤ {x} < 1. Our result must be an integer, so {2013 log 2}+ {2013 log 5} must also be an integer. But since it is equal to the sum of two fractional parts, each less than 1, the result can only be 0 or 1. However, if the sum were 0, both of {log 22013 } a ...
... where {x} denotes the fractional part of x, 0 ≤ {x} < 1. Our result must be an integer, so {2013 log 2}+ {2013 log 5} must also be an integer. But since it is equal to the sum of two fractional parts, each less than 1, the result can only be 0 or 1. However, if the sum were 0, both of {log 22013 } a ...
Final Exam Review Summer 08
... Find all zeros and write h(x) in factored form. You must verify at least two zeros using synthetic division. State all zeros in exact form. Do not use decimal approximations. h( x ) x 4 4 x 3 53 x 2 196 x 196 ...
... Find all zeros and write h(x) in factored form. You must verify at least two zeros using synthetic division. State all zeros in exact form. Do not use decimal approximations. h( x ) x 4 4 x 3 53 x 2 196 x 196 ...
supplemental sheet #7
... would say that this quadratic has "no solution". If you do know about complexes, then you would say that this quadratic has "no real solution" or that is has a "complex solution". In either case, this quadratic had no "real" solution. Since solving "(quadratic) = 0" for x is the same as finding the ...
... would say that this quadratic has "no solution". If you do know about complexes, then you would say that this quadratic has "no real solution" or that is has a "complex solution". In either case, this quadratic had no "real" solution. Since solving "(quadratic) = 0" for x is the same as finding the ...
Unit 9 – Roots, Radicals, and Root Functions
... Complex numbers are of the form a + bi, where a and b are real numbers. The number a is called the real part and b is called the imaginary part. If a is 0 they are called pure imaginary numbers. Add and subtract complex numbers We use the commutative, associative, and distributive properties to add ...
... Complex numbers are of the form a + bi, where a and b are real numbers. The number a is called the real part and b is called the imaginary part. If a is 0 they are called pure imaginary numbers. Add and subtract complex numbers We use the commutative, associative, and distributive properties to add ...
Factorization
In mathematics, factorization (also factorisation in some forms of British English) or factoring is the decomposition of an object (for example, a number, a polynomial, or a matrix) into a product of other objects, or factors, which when multiplied together give the original. For example, the number 15 factors into primes as 3 × 5, and the polynomial x2 − 4 factors as (x − 2)(x + 2). In all cases, a product of simpler objects is obtained.The aim of factoring is usually to reduce something to “basic building blocks”, such as numbers to prime numbers, or polynomials to irreducible polynomials. Factoring integers is covered by the fundamental theorem of arithmetic and factoring polynomials by the fundamental theorem of algebra. Viète's formulas relate the coefficients of a polynomial to its roots.The opposite of polynomial factorization is expansion, the multiplying together of polynomial factors to an “expanded” polynomial, written as just a sum of terms.Integer factorization for large integers appears to be a difficult problem. There is no known method to carry it out quickly. Its complexity is the basis of the assumed security of some public key cryptography algorithms, such as RSA.A matrix can also be factorized into a product of matrices of special types, for an application in which that form is convenient. One major example of this uses an orthogonal or unitary matrix, and a triangular matrix. There are different types: QR decomposition, LQ, QL, RQ, RZ.Another example is the factorization of a function as the composition of other functions having certain properties; for example, every function can be viewed as the composition of a surjective function with an injective function. This situation is generalized by factorization systems.