Section 3 - Divisibility
... Divisibility and Non-divisibility • There is another way to test for divisibility: If d | n, there is integer k with n = dk, then k = (n/d). So, if (n/d) is an integer, then d | n. • This leads to an easy way to test for nondivisibility: If (n/d) is not an integer, then d cannot divide n. • Examples ...
... Divisibility and Non-divisibility • There is another way to test for divisibility: If d | n, there is integer k with n = dk, then k = (n/d). So, if (n/d) is an integer, then d | n. • This leads to an easy way to test for nondivisibility: If (n/d) is not an integer, then d cannot divide n. • Examples ...
Word - University of Georgia
... (when the number is reduced to lowest terms) is odd. This is because if the exponent is a fraction with an even denominator (such as ½), the exponent indicates an even root. An even root of a negative number does not give a real result. If we restrict the domains of x, m, and n to the real numbers, ...
... (when the number is reduced to lowest terms) is odd. This is because if the exponent is a fraction with an even denominator (such as ½), the exponent indicates an even root. An even root of a negative number does not give a real result. If we restrict the domains of x, m, and n to the real numbers, ...
Problem Set 1 - University of Oxford
... Solution: From the theory of cyclic groups, a homomorphism from the abelian group Z to any group is completely determined by the image of 1 ∈ Z: if f : Z → (R, +) is a homomorphism of abelian groups then f (n) = f (1)+f (1)+. . . f (1) (n terms on the righthand side) when n ≥ 0 and f (n) = −f (−n) ...
... Solution: From the theory of cyclic groups, a homomorphism from the abelian group Z to any group is completely determined by the image of 1 ∈ Z: if f : Z → (R, +) is a homomorphism of abelian groups then f (n) = f (1)+f (1)+. . . f (1) (n terms on the righthand side) when n ≥ 0 and f (n) = −f (−n) ...
Complex Numbers
... If b2 – 4ac < 0, then the equation has no real solution. But in the complex number system, this equation will always have solutions, because negative numbers have square roots in this expanded setting. ...
... If b2 – 4ac < 0, then the equation has no real solution. But in the complex number system, this equation will always have solutions, because negative numbers have square roots in this expanded setting. ...
Lecture 3.5
... If b2 – 4ac < 0, then the equation has no real solution. But in the complex number system, this equation will always have solutions, because negative numbers have square roots in this expanded setting. ...
... If b2 – 4ac < 0, then the equation has no real solution. But in the complex number system, this equation will always have solutions, because negative numbers have square roots in this expanded setting. ...
a 2
... When multiplying exponential forms that have the same base, we can add the exponents and keep the same base n a nb = na +b Simplify monomials raised to a power To simplify an exponential form raised to a power, we can multiply the exponents and keep the same base ( n a ) b = nab Multiply polynomials ...
... When multiplying exponential forms that have the same base, we can add the exponents and keep the same base n a nb = na +b Simplify monomials raised to a power To simplify an exponential form raised to a power, we can multiply the exponents and keep the same base ( n a ) b = nab Multiply polynomials ...
