Consider the function represented by the equation 6q = 3s - 9. Write the equation in function notation, where q is the independent variable.

 

 

 

 

Let alpha = Squareroot (2 + Squareroot 2)(3 + Squareroot 3) (positive real squareroot s for concreteness) and consider the extension E = Q(alpha). Show that a = (2 + Squareroot 2)(3 + Squareroot 3) is not a square in F = Q(Squareroot 2, Squareroot 3). [If a = c^2, c F, then a phi (a) = (2 + Squareroot 2)^2 (6) = (c phi c)^2 for the automorphism phi Gal(F/Q) fixing Q(Squareroot 2). Since c phi c = N_F/Q (Squareroot 2) (c) Q(Squareroot 2) conclude that this implies Squareroot 6 Q (Squareroot 2), a contradiction.] Conclude from (a) that [E: Q] = 8. Prove that the roots of the minimal polynomial over Q for alpha are the 8 elements plusminus Squareroot 2 plusminus Squareroot 2) (3 plusminus Squareroot 3). Let beta = Squareroot (2 - Squareroot 2)(3 + Squareroot 3). Show that alpha beta = Squareroot 2(3 + Squareroot 3) F so that beta E. Show similarly that the other roots are also elements of E so that E is a Galois extension of Q. Show that the elements of the Galois group are precisely the maps determined by mapping alpha to one of the eight elements in (b). Let sigma Gal(E/Q) be the automorphism which maps alpha to beta. Show that since sigma(alpha^2) = beta^2 that sigma(Squareroot 2) = - Squareroot 2 and sigma(Squareroot 3) = Squareroot 3. From alpha beta = Squareroot 2(3 + Squareroot 3) conclude that sigma(alpha beta) = - alpha beta and hence sigma(beta) = -alpha. Show that sigma is an element of order 4 in Gal(E/Q).

Let r in and u =(1 1 0), v = (1 4 1), w = (r2 1 r2), b = (3+2r 5+12r 2r). For which values r in is the set {u. v, w} linearly independent? For which values r in is the vector b a linear combination of u. v and w? For which of these values of r can b be written as a linear combination of u. v and w in more than one way? The set 3 of all column vectors of length three, with real entries, is a vector space. Is the subset B = {(x y z) in 3 | xy + yz = 0} a subspace of 3? Justify your answer. Show that the set of all twice differentiable functions f : satisfying the differential equation sin(x) f"(x) + x2f(x) = 0 is a vector space with respect to the usual operations of addition of functions and multiplication by scalars. Here, f" denotes the second derivative of f. Let S be the following subset of the vector space 3 of all real polynomials p of degree at most 3: S = {p in 3 | p(l) = 0, p'(l) = 0}, where p' is the derivative of p. Determine whether S is a subspace of 3. Determine whether the polynomial q(x) = x - 2x2 + x3 is an element of S. u = (1 1 0), v = (1 4 1), w = (r2 1 r2) if {u, v, w} is linearly independent set 1(4r2 - 1)-1(r2 - r2) ne 0 4r2 ne 1 r ne plusmn1/2 this set is linearly independent set for values of e ne plusmn1/2 - - - - - - - (3+2r 5+12r 2r) = a1 (1 1 0) + a2 (1 4 1) + a3 (r2 1 r2) this forms system of equations in a1, a2, a3 if 4r2 - 1 ne 0 then system has unique solution if 4r2 - 1 = 0 and -2 - 4r = 0 rArr r = -1/2, the system has more than one solution if 4r2 - 1 ne 0 b can be written as linear combination of u,v,w as uniquely if r = -1/2 then b can be written as linear combination of u,v,w more than one way - - - - - - - - - (1 1 -1) in B and (1 0 1) in B (1 1 -1) + (1 0 1) = (2 1 0) not in B since 2times1+1times0 ne 0 so addition of vector is failed, this set is not forms subspace consider the following differential equation (sinx)f"(x) + x2f (x) = 0 each solution is continuous function the set of all solutions is subset of space of set of all continuous functions n from R to R let f,g are two solutions (sinx)f"(x) + x2f (x) = 0 (sinx)g"(x)+x2g(x) = 0 ccheck whether af+bg is solution or not (sinx)f"(x) + x2f(x) = 0 rArr (sinx) af"(x) + x2?f (x) = 0 (sinx)g"(x) + x2g (x) = 0 rArr (sinx)bg" (x) + x2bg (x) = 0 (sinx)(af"(x)+bg"(x)) +x2 (af (x) + bg(x)) = 0 af+bg is also solution to differential equation the set of all solutions to differential equation forms subspace - - - - - - - - - - consider the following subset of P3 {p in P3/.p(l)=0, p'(l) = 0} let p, q in S rArr p(l)= 0, p'(l) = 0 and q(l) = 0, q'(l) = 0 (ap + bq) (l) = ap (l) +b q(l) = a (0) + b (0) = 0 (ap + bq)'(1) = ap'(l) +bq'(1) = a (0) + b(0) = 0 p, q in S rArr ap +bq in S then S is subspace q(x) = x - 2x2 + x3 q(1) = 1 - 2 + 1 = 0 q'(x)=1 - 4x + 3x2 q'(l) = 1 - 4 + 3 = 0 q is element of S