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(mod m1) mod m2 ? I've never seen this done as an algebra topic. Creating equivalence classes on equivalence classes loses a lot of information. I'm not even sure if its a well-defined operation. If you're just looking for the remainder and m1 <= m2, then the answer doesn't change, i.e. y=z...

You can get theta with the dot product. u (dot) v = |u|*|v|* cos(theta)

The direction of the cross product of two linearly independent vectors in R^2 will be either "out of the page" or "into the page", not given by theta or anything. However the Mangnitude of the cross product can be calculated as u x v = |u|*|v|*|sin(theta)| where theta is the angle bewteen them.

The cross product is certainly defined on R2, just as it is defined on every other plane. Given any two linearly independent vectors, there is a single plane which is spanned by them, so R2 is simply a special case spanned by the two standard coordinate vectors (1,0) and (0,1). Just "pretend"...

If I am interpreting this correctly, I would start with partitioning the real line by the zeros of the function stuck into the absolute values: -1/2 and 2 x <= -1/2 You get: -(2x+1)-(x-2)=a -1/2 < x <= 2 You get: (2x+1)-(x-2)=a x > 2 You get: (2x+1)+(x-2)=a In the first case: -3x+1=a...

edit:nevermind I'm not so sure this is any different than concluding P(black)=1/2 implies one is black :?

Re: Jumping from what today's "Calculus" is and abstract analysis can be frustrating, so let us try a more concrete example. Suppose that x_n = b+2^{-n} (n may vary through all the real numbers if you wish as x_n > b for any real number n). Then it is clear that \lim_{n \to \infty}x_n =...

Yes, in fact the proof would have been the same even if restricted to rational numbers or any dense set. If it helps, let a_n be a sequence which tends to b from the right (monotonically). Then it is the same result to show: if a \le a_n for all n, then a \le b. Showing these two statements...

Yep, first thing to prove is the easier result if a is a non-negative real number such that a \le \epsilon for every positive number \epsilon, then a=0 - read: "a non-negative number smaller than every positive number must be zero". Make sense? Note this is a special case of the kaelbu's...

You are expecting to get a numerical answer, so your outermost integral must only contain numbers as its bounds. Now assuming you are working in the first octant, draw the region of intersection of your plane with the xy plane. (set z=0) This is how you get the relation x=3-y/2. Now, as x is...

Here's a direct proof using the "epsilon argument" (which i believe was proved prior to a similar question in bartle's baby analysis). In the proof an element of contradiction is hidden, so this is essentially the same as the above. Assume a \neq b. Let \epsilon > 0. Then by assumption we have...

I would take a gander at the fundamental theorem of finite abelian groups. Use the fact that Zn + Zm is not cyclic if gcd(m,n) > 1

Okay, so F(a) is a subfield of K. What is [F(a):F]? There is only one step to do for the induction... a^(p^(n+1)) = (a^p^n)^p = ... Plus you know that if m divides n then Fp^m is a subfield of Fp^n. So the automorphism x -> x^p^n fixes Fp^m. That gives you your base case.

If the field is Z_7, we have 0,1,2,3,4,5,6 The product of the nonzero elements is 1*2*3*4*5*6, or we may reorder them as 1*6*(2*4)(3*5) = 1*(-1)*(1)*(1) = -1. Get it now? I don't know of any easier method. You know that x(x-1) is a factor though.

No, g(f(l))=g(0), since f(l)=0. But, g is an automorphism, and so fixes 0. For the other one, any map of finite dimensional vector spaces of the same dimension satisfy "injective iff surjective". You can show the kernel contains only 0.

1) what is N? B is an element of K? 2) yes, so plug in l. as its a root, you get f(l)=0. Then apply g(x) = x^q to the value f(l). It is an automorphism, so... 3) this is actually an if-and-only-if. x^p^n - x is a separable polynomial and has splitting field Z_p^n over Z_p[x] (it actually IS that...

1) The => direction is simple: Let A be the prime subfield. Then [F:K][K:A] = [F:A] = n. For the other direction, if m | n then x^m-1 divides x^n-1. Though theres more to do after this. 2) Finite products in a commutative group may be reordered. 3) Hmm... Trial and error? What divides 4? 1,2,4...

Your question hasn't been completed. And what is that jumbled mess? f_a=p_a f : Y->X_a?? Is that composition? Where is uniformly continuous coming into this?

There is a theorem that states the intersection of a nested sequence of compact sets in R^n is non-empty. In some cases, this arises. For instance the intersection of all sets [-1/n,1/n] is [-0,0], or just [0,0]. I see no reason in having "-0" but some classes which rely strictly on pure rigor...

Here \frac{\sqrt{n}}{\sqrt{3n-2}} = \sqrt{\frac{n}{3n-2}} = \sqrt{\frac{1}{3-\frac{2}{n}}} \to ???