Recent content by wsalem

I would choose the exponential form which makes it a matter of algebraic manipulation, but using trigonometric identities is equality valid. sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i} and cos(\theta) = = \frac{e^{i\theta} + e^{-i\theta}}{2} substituting in the original equation, you'll...

Keeping score! Huh. Just because I misread your proof and replied with incorrect analysis doesn't make your proof more likely true! Heck, suppose I was a complete idiot, that wouldn't make your proof correct either. Also, you're misquoting me, in an unethical way to be sure! Well, I actually...

Reread my post, yes T=c must be prohibited, as well as T=1 (You can't write equation 4 without explicitly assuming this). And no matter what you do, if any step after (4) proved or assumed that T=c or T=1 then you should know that you're in deep trouble!

You reached a very obvious contradiction. In 8 and 9, you showed that c = c^2. This implies c = 1. But then c = a^x + b^y and by assumption a,b are positive integers, a contradiction. You can be assured that you did something wrong! Assuming that T=c, and dividing by \frac{ln(T)}{ln(c)}-1...

Wouldn't this make f a little bit trivial. What am I missing here? It doesn't look very interesting! Which object are you trying to study? The group G, the subset H, or the map f? Also why G is a group and not just a set, what advantage does this give with f only a map and not a homomorphism?

mnb96, I believe dropping the homomorphism requirement is not the right direction, you'll lose plenty of information then. Here's a couple of question for you. Let G be a group, f an endomorphism, H the subgroup of all elements satisfying f(x) = x in G. Is the subset J where f(x) doesn't equal...

The analog of invariant subspaces for groups is the the characteristic subgroup which is a subgroup H of a group G that is invariant under ALL automorphisms of G. If we let f be an arbitrary automorphism then for H to be a characteristic subgroup, f(x) is in H for every x in H. But it doesn't...

I actually never read about it, so I can't make any claims about it's significance. Your last phrasing of the problem I believe doesn't help much. You are given a group G, and a non-trivial subset H, finding an f that is the identity when restricted to H is trivial. Just let f be the identity...

I'm not familiar with the name but this is actually a property shared with all endomorphisms. Since if you consider the trivial subgroup (of only the identity element), any endomorphism must preserve the identity. Hence e \in H = \{e\} \Rightarrow f(e)=e.

in general, \mathbb{Z}[\frac{1}{3}] is the smallest sub-ring of \mathbb{Q} which contains both \frac{1}{3} and \mathbb{Z}. From the viewpoint of localization, we consider \mathbb{Z} as our domain. Choose 3 in \mathbb{Z}, and let the multiplicative set S = \{3^n | n \in \mathbb{N}\} (Note: 3^0=1...

Not at all. If S is a subring, then S contains the additive identity 0, but then the localization will be the zero ring, in fact the localization S^{-1} Z = 0 if and only if 0 \in S. (This should be intuitive, recall that when Q was constructed, we required the denominator to never be zero, in...

You should along with Z, choose the multiplicative subset you want to do localization with. Perhaps an intuitive example is the construction of the quotient field Q. First, we know that \frac{8}{2}= \frac{4}{1}, i.e the two are equivalent since 8 * 1 - 4*2 = 0. We also need to be careful...

I'm not sure what "proof of existence" do you seek here, would you agree that the real numbers exist? I'll restrict myself to the mathematics though, leaving existence and usefulness in waves, quantum mechanics, ... to physicists! Given any field R, there exist a field C which contains R as a...

if you insist on that example, if each kid need 5 candy and we have 3 such confused kids, then they actually have 15 candies. Here a "confused kid" is a kid who thinks he need but actually have a given number of candies, and vice verse!

Algebraicly speaking, there's a structure called wheels, which does something similar to that. Not sure how useful it is though! See http://dx.doi.org/10.1017/S0960129503004110 Division by zero is still not defined in the extended real line, since you would get both +\infty and -\infty...