Recent content by Hotsuma

Thanks for the disappearing act; I was about to go to sleep.

C=k\cdot\frac{\epsilon_0A}{d}?

Homework Statement A wind power plant will not produce as much energy when the wind slows down. In order to provide power during these time periods it is proposed to store some power when it is operating in high winds in a capacitor. If it is desired to store the energy to power a town at 20...

Thanks, it is late. Woo silly mistakes. How do I prove this result, or is that sufficient? I mean, should I just plug in the typical supremum proof for this result? Thanks by the way

Homework Statement Find the infimum and supremum of each of the following sets; state whether the infimum and supremum belong to the set E. \item 1. ~~~~E={p/q \in \mathbf{Q} | p^2 < 5q^2 \mbox{ and } p,q >0}. \mbox{ Prove your result. } \item 2. ~~~~E={2-(-1)^n/n^2|n \in \mathbf{N}...

Alright. How does that look?

\mbox{Pf: Assume p is prime. Then} (x+y)^p= \left(\begin{array}{l c} p\\ 0\\ \end{array}\right) x^p+ \left(\begin{array}{l c} p\\ 1\\ \end{array}\right) x^{p-1}y+ \left(\begin{array}{l c} p\\ 2\\ \end{array}\right) x^{p-2}y^2+ ... + \left(\begin{array}{c c} p\\ p-1\\ \end{array}\right)...

I discovered my error. Give me a minute.

Homework Statement \mbox{Prove or give a counterexample: If p is a prime integer, then for all integers x and y, } (x+p)^p \equiv_p x^p+y^p. Homework Equations \equiv_p \mbox{just means (mod p). Can you please check and see if this proof is well-formed?} The Attempt at a Solution...

Oh right, lol, I started with the wrong assumption! Yeah, so, is it really that simple? He he, awesome.

Doing this simplistic proof for one of my computer math classes. I've already taken Abstract Algebra and I'm having trouble with this lol. Actually, I just need someone to verify this is correct for me. It seems far too simple. Homework Statement Prove the following n^3 > n^2...

Yeah, that was my fear. I e-mailed the professor at like 4PM but he has not yet responded (nor do I expect him to).

And I'd love to prove that the two are tautologies, but since I am not sure about the former, that doesn't really help me much :(. (p \rightarrow q) \wedge (q \rightarrow r) \equiv p \rightarrow q \rightarrow r

Dear lanedance, I considered this but wasn't sure if that was correct or not. What you described to me in your past post could be written as "if p then q" and "if q then r," which I don't think is the same thing, or rather, I'm not sure whether it is or is not. Thanks for the suggestion...

Dear All, Having trouble with a seemingly simple logic truth table. Are these answers correct? \begin{tabular}{| c | c | c | c | c | c |} \hline p & q & r & (p \vee q)\wedge(q \vee r) & (\neg p \wedge q) \vee ( p \wedge \neg r) & p \rightarrow q \rightarrow r\\ \hline T & T & T...