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Guns n'Roses changing it's name to attract the Math, Logician crowd: Guns if and only if Roses.

In France, Marseille 1, nothing Tolouse.

I'm still unable to drink fluids to take care of a cold. Liquids, yes. Gases, not quite yet.

Same with me pointing out in Bruno Mars' video " It will Rain", that morphine is used for physical, not emotional, pain. No wonder I'm writing this comment from my home on Friday night , while all others are out in bats, clubs.

Now no one will be able to steal the wall:

Recent causes

High-level discussion on the web, as usual

Well, part of the issue is that exp(z) plays a role when dealing with Complexes, e.g., Polar representation, but it's not 1-1, and as such has no global inverse log. This requires you to patch together local inverses. A mess. Edit: Here the square root too, may involve arguments, branches, and...

$$ _always_ works. you just need enough of it ;).

Sure, https://en.wikipedia.org/wiki/Compactness_theorem Through the concept of Elementary Equivalence , aka, the Transfer Principle, 1st -order properties are preserved "upwardly", from lower to higher cardinality, between models of different cardinality. Edit: Is that what you were asking?

You only need to check through ## \aleph_0## models of the Reals , per Lowenheim -Skolem. Outside of the standard Reals, you lose " standard " metrizability, since metric are Real-valued.

Figure it out and you'll get a Figure one out and go collect your Fields/Abel medal.

You aint nothing but a Groundhog.

No German porn, please.

Yes, kind of a nightmare dealing with the branching alone.

Or, to be pedantic, use that the Complexes can be seen as a field extension by ## x^2+ 1 ##. Then, in the extension ## \mathbb R[x]/(x^2 +1) ; x^2 +1 =0 ## You designate this new element by ##i ##..

Every tab everything working except YouTube. I was searching for SysInternals. Which I never had??

If at first you don't succeed, don't try skydiving.

Eigenvalues ##\lambda## for a matrix ##A##are defined to satisfy ##Det(A-\lambda I)=0##. This comes from ##Ax=\lambda x ##, so that ##(A-\lambda I )x=0 ##.

But it was an _open_ exam. ;).

One result I remember is that given ##G## a Lie Group; here ## \mathbb T^2##, a subset is a Lie Group if it's both a subgroup and topologically closed in ##G##.

Nor would it be onto, if we consider ##\mathbb N =\{1,2,..\} ##, as nothing would hit ##1##.

Worse comes to worse, time is running out on the exam, cube both sides. Only one term will remain to a non-unity power p/q. Leave it alone in one side and raise both sides to the qth power. Hardly elegant, but at least you'll get some credit. Edit: In this exercise, you first cube both sides...

Had to correct my friend . I told him to put a potato in his bathing suit at the beach to help him impress women. But In the front, not in the back, as he did last time.

Winner gets a Lapp dance.

You mean as in using Order Statistics? Say horse number five finished the race before horse number 1( of course, numbers were given before the race)?

That's music to Meir.

Weird exchange: I have something in my ear Well, it can't be Golda, she's been dead for a while.

So, the expression 5 < 1 by itself does not provide enough information to determine what this '<' means, nor how it works by itself.

Still, to keep some standards, we need to deal with Mathematical issues: What requirements must an ordering satisfy?

This works, but needs rescaling to fit the OP: Assume there are numbers S between 0 and 1. Then since the set is bounded, it has a least upper bound L, so that: 0<L<1 Now multiply thtough by L: 0< L^2<L Then L^2 is a bound lower than L. Contradiction.

Not too confident about an organization called the " Feil" organization.

A club sandwich?

I think Turkey in Turkish is 'Gindi'

Abraham includes Pastrami

Geography, religion and meats: Turkey: Ottoman Empire, Ham: Muslims Bacon: Britain ( Roger). Rotisserie: 15:3.