MATH 160 MIDTERM SOLUTIONS
... constructible, it would follow that e2πi/7 would be an algebraic number of 2-power degree, whereas in fact it has degree 6, being a zero of the irreducible polynomial x6 + x5 + x4 + x3 + x2 + x + 1.) (c) Starting from two points in the plane 1 unit apart, it is possible to construct p√ ...
... constructible, it would follow that e2πi/7 would be an algebraic number of 2-power degree, whereas in fact it has degree 6, being a zero of the irreducible polynomial x6 + x5 + x4 + x3 + x2 + x + 1.) (c) Starting from two points in the plane 1 unit apart, it is possible to construct p√ ...
Lesson Plans 11/24
... and/or the Least Common Multiple (LCM) for sets of monomials. A1.1.1.3.1 Simplify/evaluate expressions involving properties/laws of exponents, roots, and/or absolute values to solve problems. Note: Exponents should be integers from 1 to 10. A1.1.1.4.1 Use estimation to solve problems. A1.1.1.5.1 Add ...
... and/or the Least Common Multiple (LCM) for sets of monomials. A1.1.1.3.1 Simplify/evaluate expressions involving properties/laws of exponents, roots, and/or absolute values to solve problems. Note: Exponents should be integers from 1 to 10. A1.1.1.4.1 Use estimation to solve problems. A1.1.1.5.1 Add ...
Lecture 4 Divide and Conquer Maximum/minimum Median finding
... and is known as a Vandermonde matrix. The determinant is known to have the closed form ∏1≤i< j≤N+1 (xi − x j ) 6= 0. The proof of this determinant equality is by induction on N, using elementary row operations. We leave it as an exercise (or you can search for a proof online). Thus we will employ a ...
... and is known as a Vandermonde matrix. The determinant is known to have the closed form ∏1≤i< j≤N+1 (xi − x j ) 6= 0. The proof of this determinant equality is by induction on N, using elementary row operations. We leave it as an exercise (or you can search for a proof online). Thus we will employ a ...
Math 113 Final Exam Solutions
... 2. a) (6 points) Let H = h(2, 4)i be a subgroup of Z × Z. Show that the cosets of H in Z × Z are precisely those of the form (0, n)H and (1, n)H, where n can be any integer. Suppose (r, n)H = (s, m)H where s, r = 0 or 1. Then we have (r − s, n − m) ∈ H. Note that |r − s| = 0 or 1. At the same time, ...
... 2. a) (6 points) Let H = h(2, 4)i be a subgroup of Z × Z. Show that the cosets of H in Z × Z are precisely those of the form (0, n)H and (1, n)H, where n can be any integer. Suppose (r, n)H = (s, m)H where s, r = 0 or 1. Then we have (r − s, n − m) ∈ H. Note that |r − s| = 0 or 1. At the same time, ...
Finally, we need to prove that HomR(M,R∧ ∼ = HomZ(M,Q/Z) To do
... Proof. It is easy to see that the first two conditions are equivalent. Suppose that x ∈ D and n ≥ 0. Then, A = nZ is a subgroup of the cyclic group B = Z and f : nZ → D can be given by sending the generator n to x. The homomorphism f : nZ → D can be extended to Z if and only if D is divisible. Thus ...
... Proof. It is easy to see that the first two conditions are equivalent. Suppose that x ∈ D and n ≥ 0. Then, A = nZ is a subgroup of the cyclic group B = Z and f : nZ → D can be given by sending the generator n to x. The homomorphism f : nZ → D can be extended to Z if and only if D is divisible. Thus ...
