uncool

But it is possible to get a rod C such that A^4 + B^4 = C^4. Here is how: Draw a line l, and choose any point O on it. Find the point M such that the length of OM is A. Find the point N such that the length of ON is B. Draw another line m through ON, and find the point OP such that the length of OP is B. Draw the line between M and P. Find the line parallel to MP that goes through N, and find the point R where the line intersects M. Then we know that OR has length B^2/A, by similar triangles. Now construct a right triangle with side lengths equal to A and to B^2/A (we can do the latter because we have constructed it). Then the length of the hypotenuse is sqrt(A^2 + B^4/A^2) = sqrt(A^4/A^2 + B^4/A^2) = sqrt(C^4/A^2) = C^2/A. We now have a segment of length C^2/A. Construct a line, and choose a point E. Find F such that EF = A, and G on the other side of E such that EG = C^2/A. Draw the circle such that FG is a diameter. Draw the perpendicular to FG that goes through E, and find the point on the circle H where the line intersects. We know that FHG is a right triangle, so we can get by similar triangles that EH has length equal to the geometric mean of the lengths of EG and EF. That is, the geometric mean of C^2/A and A. But that is just C - the desired length. =Uncool-

I doubt that it is either, because with Watts you can make it happen for as short as you want, while with Joules you can extend the power out over a lifetime. I'm sure that in your lifetime you easily absorb more than a Joule in electric power. =Uncool-

Define "diverging to infinity". Then try using that definition. =Uncool-

Remember, it's not how to find epsilon, n, and m just like that. You have to find an epsilon such that for any N, you can find an n and m such that [stuff]. =Uncool-

How are you getting that c^n - (c - 1)^n < sqrt(c^n)? Because that is pretty much never true. =Uncool-

Everything is good here. Thanks! =Uncool-

It appears to not remember the PM in my inbox, but it's still in notifications apparently. =Uncool-

Will do. Think you can get rid of it? It still tells me that I have a notification every time I log in... =Uncool-

Nope. Says absolutely nothing other than the date, which means I can't click on it to read it. Also, if it was from deleting spam messages, I got one which was never deleted. I deleted it, though, so... =Uncool-

I recently have been told that I got a new message from here, but when I look, all of them are read, except for one with no title that says it was sent at January 1, 1970. I'm guessing that that is a bug. =Uncool-

It's also a joke. 2 + 2 = 5, for very large values of 2... =Uncool-

Ahhh. You probably should expand the proof that C - B and P have no common factors. Proof: (C^n - B^n)/(C - B) = sum from 1 to n of B^(n - i)C^(i - 1). If we look modulo C - B, the latter is equivalent to sum from 1 to n of B^n, or n B^(n - 1). Therefore, the greatest common factor of C - B and P is the same as the greatest common factor of C - B and nB^n. Then since n does not divide C - B and n is prime, n and C - B are relatively prime, so any common factor of C - B and nB^n is a common factor of C - B and B^n. This is then a common factor of B^n and C^n - which can only be 1. =Uncool- Merged post follows: Consecutive posts mergedVictor, you have an arithmetic error when going from step 10 to 11. You go from (C-B)[T+(C-A)x]+(C-A)[T+(C-B)x] to [(C-B)+(C-A)](T+x) when the correct outcome is [(C-B)+(C-A)]T + 2(C-A)(C-B)x. =Uncool-

Why does C - B = a^n, or C - A = b^n? And shouldn't your first a, b, and c be A, B, and C? =Uncool-

You are talking about a pion, not an electron. An electron is truly an elementary particle - a lepton - while a pion is a kind of meson. A pion can decay into an electron and a neutrino, but the biggest difference between the two is that the electron is a fermion, while a pion is a boson.=Uncool-

So being unable to have sex would mean you can't get married? Really? =Uncool-

You claimed that for any sequence that didn't go to infinity, that is, any bounded sequence, that sequence does not converge to any finite limit. the tree showed that there was a sequence that didn't go to infinity that did converge to a finite limit, thereby proving your statement wrong. =Uncool-

I got it from my electrodynamics classes, and from reading on this exact subject. Look up Gauss's law for magnetism. Do you mind showing me where it is used as such in any equations? I can positively say that I have never seen such a definition. You'll notice that Maxwell's equations for magnetism say specifically that the magnetic field has no divergence. That is the exact statement that magnetic monopoles do not exist. =Uncool-

Sha31: Consider the magnetic field [MATH]\overline{B} = x \hat{j}[/MATH]. This is a normal field. A pole is any place where the divergence of the field is nonzero. Where is that here? If we consider the wire, we still get that the divergence of the field is 0 - and therefore, that there are no poles. There are plenty of fields with zero divergence. Consider any vector function on [MATH]\mathbb{R}^3[/MATH] A. Then if we let B be the curl of A, that is a divergenceless field. A is known as the vector potential. Then B has no poles. =Uncool-

Consider the sequence: 1/n if n is even, 1 - 1/n if n is odd. This doesn't exactly alternate, but it does have two limiting values. You can have arbitrarily many limiting values. =Uncool-

18.25^3 = 6 078.390625 =Uncool-

Um. a^3 + b^3 - c^3 is clearly not 0, but 8.20. =Uncool-

It is possible to figure out more about a face than a picture has, but that is more from knowing about faces than using math. That is, you cannot simply "enhance" a jpeg past its original resolution using math, but if you have a good picture, you can rotate it so that it's in a certain direction, and use certain facts about faces to figure out a little more about this particular face. This is extremely difficult, but possible. =Uncool-

If you have a nearly-reversible battery (usually known as a rechargeable battery), then this circuit should light the bulb for about as long as 3 separate batteries do. However, after that, it will stop - there is no more energy coming out of them. You can define a quasi-invariant here: the total energy left in the three batteries. To recharge one battery, you have to lose the same amount of energy in another battery. As such, the total energy keeps decreasing by a constant rate - the rate needed to power the bulb. =Uncool-

His congressional career is a joke. A cruel joke on the idiots he takes down. =Uncool-

Do you actually know what "quantized" means? =Uncool-