PM 464
... 1.22 Proposition. If f , g ∈ |[x, y] have no common factors then V ( f , g) = V ( f ) ∩ V (g) is at most a finite set of points. PROOF: Since f and g have no common factor in |[x, y] = |[x][ y], they have no common factors in |(x)[ y]. Therefore gcd( f , g) exists and is 1 in |(x)[ y], so there are ...
... 1.22 Proposition. If f , g ∈ |[x, y] have no common factors then V ( f , g) = V ( f ) ∩ V (g) is at most a finite set of points. PROOF: Since f and g have no common factor in |[x, y] = |[x][ y], they have no common factors in |(x)[ y]. Therefore gcd( f , g) exists and is 1 in |(x)[ y], so there are ...
Algebra 1 ELG HS.A.1: Interpret the structure of expressions.
... o A.SSE.1 Interpret expressions that represent a quantity in terms of its context. ★ o A.SSE.1.a Interpret parts of an expression, such as terms, factors, and coefficients. o A.SSE.1.b Interpret complicated expressions by viewing one or more of their parts as a single entity. For example, interpret ...
... o A.SSE.1 Interpret expressions that represent a quantity in terms of its context. ★ o A.SSE.1.a Interpret parts of an expression, such as terms, factors, and coefficients. o A.SSE.1.b Interpret complicated expressions by viewing one or more of their parts as a single entity. For example, interpret ...
Lesson 2 – The Unit Circle: A Rich Example for
... the triple (3, 4, 5). We will discuss various methods for tackling this problem: a geometric, an analytic, and an algebraic method. The Geometric Method: Parameterization. The geometric solution turns the problem of finding integral solutions of into the equivalent one of finding rational points on ...
... the triple (3, 4, 5). We will discuss various methods for tackling this problem: a geometric, an analytic, and an algebraic method. The Geometric Method: Parameterization. The geometric solution turns the problem of finding integral solutions of into the equivalent one of finding rational points on ...
103B - Homework 1 Solutions - Roman Kitsela Exercise 1. Q6 Proof
... The easiest of these to check is identity, so we need to check whether 0 is in the set {π n : n ∈ Z}? i.e. can find some integer k such that 0 = π k ? Clearly this is impossible and so {π n : n ∈ Z} is not a group under addition. Exercise 2. Q8 Proof. We need to determine whether the n × n real matr ...
... The easiest of these to check is identity, so we need to check whether 0 is in the set {π n : n ∈ Z}? i.e. can find some integer k such that 0 = π k ? Clearly this is impossible and so {π n : n ∈ Z} is not a group under addition. Exercise 2. Q8 Proof. We need to determine whether the n × n real matr ...
No nontrivial Hamel basis is closed under multiplication
... For a field F and a variable x, we let F[x] denote the set of all polynomials in the variable x whose coefficients are in F. So our usual set of polynomials is R[x]. We then denote by F(x) the set of all fractions of elements of F[x] (just don’t divide by zero). Our field of rational functions from ...
... For a field F and a variable x, we let F[x] denote the set of all polynomials in the variable x whose coefficients are in F. So our usual set of polynomials is R[x]. We then denote by F(x) the set of all fractions of elements of F[x] (just don’t divide by zero). Our field of rational functions from ...
The Power of Depth 2 Circuits over Algebras
... straight-line program of length at most 4d using only 3 registers. The following fact can be readily derived from their result (see Appendix): From an arithmetic formula E of depth d and fan-in bounded by 2, we can efficiently compute the expression, P = ∏im=1 ∑nj=0 Aij x j , where m ≤ 4d and Aij ∈ ...
... straight-line program of length at most 4d using only 3 registers. The following fact can be readily derived from their result (see Appendix): From an arithmetic formula E of depth d and fan-in bounded by 2, we can efficiently compute the expression, P = ∏im=1 ∑nj=0 Aij x j , where m ≤ 4d and Aij ∈ ...
CCSS-HSAlgebra - LSU Mathematics
... system. A solution for such a system must satisfy every equation and inequality in the system. An equation can often be solved by successively deducing from it one or more simpler equations. For example, one can add the same constant to both sides without changing the solutions, but squaring both si ...
... system. A solution for such a system must satisfy every equation and inequality in the system. An equation can often be solved by successively deducing from it one or more simpler equations. For example, one can add the same constant to both sides without changing the solutions, but squaring both si ...
SOLUTIONS TO EXERCISES FOR
... In Part (d) of the preceding problem, one cannot find a choice function without assuming something like the Axiom of Choice. The following explanation goes beyond the content of this course but is hopefully illuminating. The first step involves the results from Section I.3 which show that the set of ...
... In Part (d) of the preceding problem, one cannot find a choice function without assuming something like the Axiom of Choice. The following explanation goes beyond the content of this course but is hopefully illuminating. The first step involves the results from Section I.3 which show that the set of ...
Dedekind domains and rings of quotients
... We study the relation of the ideal class group of a Dedekind domain A to that of As, where S is a multiplicatively closed subset of A. We construct examples of (a) a Dedekind domain with no principal prime ideal and (b) a Dedekind domain which is not the integral closure of a principal ideal domain. ...
... We study the relation of the ideal class group of a Dedekind domain A to that of As, where S is a multiplicatively closed subset of A. We construct examples of (a) a Dedekind domain with no principal prime ideal and (b) a Dedekind domain which is not the integral closure of a principal ideal domain. ...
