Recent content by Tac-Tics

For sure, it is tedious having to go through all 8 axioms. In any class beyond an intro to linear algebra class, you could simply say "it's easy to see that W is a subspace". You can also phrase things in a slightly clever way (again, if your audience has the background): * W is the image of a...

Having an intuition about linear algebra will help enormously if you take abstract algebra. Vectorspaces are just modules over the real (or the complex) numbers. You can immediately relate group and ring homomorphisms to linear maps. Kernels are just the nullspace, the dimension is closely...

@phoenixthoth - Mathematical objects are primitive notions of whatever formal system you happen to be using. Most mathematicians ostensibly use something like set theory. To them, a mathematical object is anything that can be encoded as a set. (This includes pairs, relations, functions...

It's *not* just a sum. It's conceptually very sum-like, though. As you have probably noticed, you cannot sum an infinitely many things. It takes too long (because it takes literally forever). The best you can do is approximate. You will notice that for *certain* series, they 'tend...

Yes. I meant "abelian", but omitted it to introduce some confusion :)

Ah. Thank you. Now that I think about it, you can' "choose" what group you want the coefficients to be in if your generating your groups freely anyway. (I'm guessing that would be some quotient of the free group, determined by the type of coefficient you're interested in, but I'll worry about...

Sounds about right. Note that some sets in the subspace are open sets even if they aren't open in the larger space. For instance, [0, 1] is closed in R. But when we consider [0, 1] as a subspace, it's open (because the entire topological space is required to be open in any topology)...

I've been poking around, learning a little about homology theory. I had a question about the boundary operator. Namely, how it's defined. There's two definitions I've seen floating around. The first is at: http://en.wikipedia.org/wiki/Simplicial_homology The second, at...

Like Halls said above, this is nonsense. Then again, the value 1∞ is too. It's like asking "what is 2 times a walrus". A walrus isn't a real number, and neither is ∞. When people SAY 1∞, they MEAN some kind of limit towards infinity.

Also, I'm not sure "intuition" is what you should be looking for here. What is the intuition that x^(yz) = (x^y)^z? If you can do that one, perhaps it isn't so hard to understand why (g ∘ f)' = (g' ∘ f) f'

Let: h(x) = e^(2x) f(x) = e^x g(x) = x^2 Our the function we're interested in (g ∘ f)(x) = g(f(x)) = (f(x))^2 = ((e^x)^2) = e^(2x) = h(x) You know how to use the chain rule here. What you're asking is why can't we write it: h(x) = e^(2x) = e^(x2) = (e^2)^x = ... and go on from there to...

The answer is vacuous. P -> Q is *defined* to be that way. One way to think of it, though, is to think of what P → Q means under the Curry-Howard Correspondence. Under Curry-Howard, you can think of P → Q as "the set of functions from P to Q". Or maybe if you're a computer scientist: "the...

The expression you're devoting your attention is important. You can also use this to define the notion of constancy of one variable with respect to another. In the above case, e is constant with respect to x. The reasoning would be something like "restricting your attention to the scope of the...

The "official" definition for constant is probably a distraction from the topic at hand, but if anyone is interested: Variables don't exist on their own. Every variable you use in an expression must first be declared. We call expressions that declare variables "binding forms". Here are some...

I could see how the word closure is still useful there. Just to be clear, though, in the definition of a group on its own, it is strictly redundant: The fact that the domain is G x G and the codomain of * is G implies that * is closed. For a subgroup (U, **), you have to show that ** a...