The Field of Complex Numbers
... We have been studying the problem of reducibility of polynomials f 2 F [x] where F is a given …eld (or perhaps even just a commutative ring with unity). We have seen examples of polynomials that are not reducible in Z [x] ; Q [x], and R [x] (where R is the …eld of real numbers). One of the very impo ...
... We have been studying the problem of reducibility of polynomials f 2 F [x] where F is a given …eld (or perhaps even just a commutative ring with unity). We have seen examples of polynomials that are not reducible in Z [x] ; Q [x], and R [x] (where R is the …eld of real numbers). One of the very impo ...
Section 5.6 – Complex Zeros: Fundamental Theorem of Algebra
... real coefficients can be uniquely factored into a product of linear and/or irreducible quadratic factors in the real numbers. In this section, we will see that every non-constant polynomial can be factored into a product of linear factors in the complex numbers. We will also find all the complex zer ...
... real coefficients can be uniquely factored into a product of linear and/or irreducible quadratic factors in the real numbers. In this section, we will see that every non-constant polynomial can be factored into a product of linear factors in the complex numbers. We will also find all the complex zer ...
Homework # 7 Solutions
... Proof. (contrapositive) Suppose that a is an odd integer. Then a = 2k + 1 for some integer k. So (a + 1)2 − 1 = (2k + 2)2 − 1 = 4k 2 + 8k + 3 = 4k 2 + 8k + 2 + 1 = 2(2k 2 + 4k + 1) + 1. Since 2k 2 + 4k + 1 is an integer, (a + 1)2 − 1 is odd. 8. Let a, b ∈ Z. If a ≥ 2, then either a - b or a - (b + 1 ...
... Proof. (contrapositive) Suppose that a is an odd integer. Then a = 2k + 1 for some integer k. So (a + 1)2 − 1 = (2k + 2)2 − 1 = 4k 2 + 8k + 3 = 4k 2 + 8k + 2 + 1 = 2(2k 2 + 4k + 1) + 1. Since 2k 2 + 4k + 1 is an integer, (a + 1)2 − 1 is odd. 8. Let a, b ∈ Z. If a ≥ 2, then either a - b or a - (b + 1 ...
Holt Algebra 1 11-EXT
... There are inverse operations for other powers as well. For example 3 represents a cube root, and it is the inverse of cubing a number. To find 3 , look for three equal factors whose product is 8. Since 2 • 2 • 2 = 8. ...
... There are inverse operations for other powers as well. For example 3 represents a cube root, and it is the inverse of cubing a number. To find 3 , look for three equal factors whose product is 8. Since 2 • 2 • 2 = 8. ...
solutions to HW#3
... least common multiple of these numbers, so |σ| = 30. 1.3.7 Write out the cycle decomposition of each element of order 2 in S4 . The six transpositions all have order 2. They are (12), (13), (14), (23), (24), and (34). Products of pairs of disjoint 2-cycles also have order 2. They are (12)(34), (13)( ...
... least common multiple of these numbers, so |σ| = 30. 1.3.7 Write out the cycle decomposition of each element of order 2 in S4 . The six transpositions all have order 2. They are (12), (13), (14), (23), (24), and (34). Products of pairs of disjoint 2-cycles also have order 2. They are (12)(34), (13)( ...