Recent content by Benn

http://mathematics.gulfcoast.edu/mathprojects/andrew5.jpg We can't forget this one

1: Let's look at the case when a=7 and b=2. Then a+xb = 7+2x. Keeping in mind that x is an integer, we see that there's no x that makes 7+2x = 0. So, in this case, 0 is not in S. Ask yourself when 0 will be in S. ---------- 2: Still with a=7 and b=2, the division algorithm gives us 7= (3)2...

The bit in the parentheses is establishing that the the set S is nonempty. This needs to be known so that the well ordering principle can give us an element r to work with. We know that r = a-qb since r is an element of S. You may be thinking "doesn't S have elements that look like a+xb?" It...

I think you might mean { x }^{ 2 }=4\\ \sqrt{{ x }^{ 2 }}=\sqrt { 4 } \\ |x| = 2 \\ x=\pm 2

If ##p(x) = c_{n}x^{n} + ... + c_{1}x + c_{0}## where ##p## is defined from ##\mathbb{R}## to ##\mathbb{R}## and ##c_{i} \in \mathbb{R}##, then ##p(A): \{ \text{m x m matrices} \} \rightarrow \{ \text{m x m matrices} \}## is defined by ##p(A) = c_{n}A^{n} + ... + c_{1}A + c_{0}I##. ... I'm...

Yes, thank you. I understand that. But I wasn't sure if it was just a coincidence that the theorem seemed so obvious when we considered the characteristic polynomial to be det (A-xI), or if there was some way to make rigorous the idea of plugging in A.

In my linear algebra course, we just finished proving the cayley hamilton theorem (if p(x) = det (A - xI), then p(A) = 0). The theorem seems obvious: if you plug in A into p, you get det (A-AI) = det (0) = 0. But, of course, you can't do that (this is especially clear when you consider what...

Your first link is dead. Here's what is should be: http://www.millersville.edu/~bikenaga/math-proof/math-proof-notes.html

This works for R2... I think it's a little more intuitive than the other proofs I've seen. Let a, b be two vectors. Then a / ||a||, b / ||b|| are to unit vectors. We can let a / ||a|| = <cosm, sinm> and b / ||b|| = <cosn, sinn>. then (a / ||a||) * (b / ||b||) = (cosm)(cosn)+(sinm)(sinn) =...

The group is abelian if the each element a in the group satisfies a2=1 (prove this). The more structure you give something, the more you know about it. Commutativity,closure, etc are examples of structure.

Is the group abelian?

MIT does not offer video lectures of algebra; however, Harvard offers video lectures (using the same book.) Here's a link: http://www.extension.harvard.edu/open-learning-initiative/abstract-algebra

Here's an exposition of a neat way to do it: http://www.millersville.edu/~bikenaga/number-theory/exteuc/exteuc.html He gives a proof of the method, and then gives an example of it in action.

There is such a system in place now. The character "2343678" uniquely determines that number, and, by a clever method, we know exactly where that number is in relation to all the other natural numbers. The rationals have a similar system, but the representation isn't unique (i.e. 1/1 = 2/2 =...

I find that having a different mindset helps. Don't study for the test: study until you understand the material.