Recent content by Chaos2009

Yea, pretty much. I also believe you may have seen somewhere that a gcd is the only common divisor which can be written as a linear combination. And the monic condition means it is the unique monic gcd. Then you've just shown that making the field bigger doesn't change the gcd.

Well, suppose |X| is strictly less than |Y| and both finite. Then f wouldn't be bijective since it cannot be onto. Similarly g couldn't be bijective since it cannot be one-to-one. However, if we restrict our attention to only the subset of |Y| that is hit by f (say Z := f(X)), then f and g could...

Okay, well the proof that \left\{ \left( x, y \right) \in \mathbb{R}^{2} : x^{2} + y^{2} > 1 \right\} should be similar. Perhaps if you consider that \left| \left( x, y \right) \right| < \left| \left( x, y \right) - \left( a, b \right) + \left( a, b \right) \right|. It's similar to what you...

I believe you are on the right track, however, I believe you have made an assumption. Sylow's second theorem says that all the Sylow 5-subgroups are conjugate to each other. This does imply normality of H iff H is the only Sylow 5-subgroup of G. I think you can use Sylow's third theorem to show...

Hmm, what if you let f_{n} \left( x \right) = \left\{ \begin{array}{rl} f \left( x \right) &, x \in I_{n} \\ 0 &, x \not \in I_{n} \end{array} \right.. I'm not sure if f_{n} \in \mathcal{L}^{1} \left( \mathbb{R}^{k} \right) but it is an increasing sequence of functions which converges point-wise...

That radius r is good. You may need to also prove that such an r > 0 exists for each (x, y) in the set, but that shouldn't be too bad. Also, what method did you use to show that x^2 + y^2 < 7 is open?

Hmm, well, you the monic assumption is usually used to prove the uniqueness of the gcd. In this case, I'd say the monic assumption is useful since if you do not define the gcd to be monic, then h \left( x \right) a gcd of f \left( x \right) and g \left( x \right) means that r h \left( x \right)...

Sometimes a picture helps me a lot with these kinds of problems, so let's try and see what this set looks like. Well, we know that x^2 + y^2 is the points distance from the origin squared, so it should look something like a donut with outer radius \sqrt{7} and inner radius 1. Also, since...

Hmm, I'm afraid that doesn't quite work. Say A = \left( \frac{1}{2}, \frac{3}{4} \right) \cup \left( \frac{3}{4}, 1 \right), then what would I_{2} be? I think it would be better to use the fact that the rational numbers are dense within the real numbers. You should be able to show that each C...

For the last part, one idea that comes to mind is that b_1 \in W and b_2 \in W^{\perp} means that b_1 = a \begin{pmatrix} 1 \\ 2 \\ 3 \\ 4 \end{pmatrix} + b \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix} for some a,b \in \mathbb{R} and similarly b_2 would be a linear combination of the basis you...

Alright, well, I think it might be easier to define a mapping between these C \left( x \right) and a subset of the rational numbers. Can you think of a way to assign a unique r \in \mathbb{Q} to each U \in W?

Have you covered in your class that the elements of H / N and G / N are the cosets N in H and G? I might try this problem by taking an arbitrary element h' \in H / N and g' \in G / N and showing that g' h' g'^{-1} \in H / N. Since the elements chosen were arbitrary, H / N \triangleleft G / N.

This may just be me, but I don't see the question in this.

I believe what this is saying is that you first select \alpha \leq \pi and then form D := \left\{ z : \left| \mathit{arg} \ z \right| < \alpha \right\}. This would not be the entire complex plane. I believe for say \alpha = \frac{\pi}{2} would look like D = \left\{ z : \mathit{real} \ z > 0...

I'm sorry, but what is C \left( x \right)?