Recent content by BSMSMSTMSPHD

Proof that (-1)(-1) = 1 Let a be any real number. Then, a = (a + 1) - 1, and so, a2 = [(a + 1) - 1]2. Now, just expand both sides and simplify: a2 = (a + 1)2 - 2(a + 1) + (-1)(-1) a2 = a2 + 2a + 1 - 2a - 2 + (-1)(-1) 0 = -1 + (-1)(-1) 1 = (-1)(-1).

Use a double angle formula to change the cosine term into a sine term. Then you'll have solvable equation.

You lost the 7 in the original numerator. The correct values are A = -3, B = 1, C = 9.

It's not "more correct" (not even really sure what that means!) but I personally find it more concise to write answers with the minimum number of symbols, including negative signs. Your answer is fine.

Not sure where to put a question about topology, but I'll try here. I'm trying to show that a certain topology for the Real line is not normal. The topology in question has no disjoint open sets (they are all nested) and therefore, no disjoint closed sets. If a topology has no disjoint...

cos(x) has no limit at infinity, because the function oscillates. But, it's bounded, and the denominator (x) is growing without bound. What will happen to the fraction as x gets larger? As for the first one, try replacing 1/x with a new variable, say w. Then, the limit as z approaches...

Homework Statement Define a quasimetric on the Sorgenfrey Line.Homework Equations I know how to show the distance function is always nonnegative, equal to zero if evaluating the distance of a point from itself, and the triangle inequality. I'm having trouble coming up with the function.The...

I'm trying to prove the following Theorem. Suppose T1 and T2 are topologies for X. The following are equivalent: 1. T1 is a subset of T2; 2. if F is closed in (X, T1), then F is closed in (X, T2); 3. if p is a limit point of A in (X, T2), then p is a limit point of A in (X, T1)...

Man, it's always something really obvious. Thanks.

Homework Statement Let f(z) = \frac{1-iz}{1+iz} and let \mathbb{D} = \{z : |z| < 1 \} . Prove that f is a one-to-one function and f(\mathbb{D}) = \{w : Re(w) > 0 \} . 2. The attempt at a solution I've already shown the first part: Assume f(z_1) = f(z_2) for some z_1, z_2 \in...

Sorry, that's unfair - it's your first post. Can you show us the work you have so far? Do you understand what absolute value means?

Yes I can. Thanks for inquiring!

Which it is because \mathbb{Q}(\sqrt3, \sqrt7) is degree 2 over \mathbb{Q}(\sqrt7) which is degree 2 over \mathbb{Q}. The minimal polynomials are x^2-3 and x^2-7 respectively.

Yep, I made the fix. I do understand why this shows that the splitting field for \sqrt3 + \sqrt7 includes the elements \sqrt3 and \sqrt7. Does this lead to the conclusion that the order of the extension is 4?

\frac{1}{\sqrt3 + \sqrt7} = \frac{\sqrt7 - \sqrt3}{4} So, if I do (\sqrt3 + \sqrt7) + 4(\sqrt3 + \sqrt7)^{-1} = 2\sqrt7 that shows that \sqrt7 is in the splitting field?