  # HalfWit

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• ### hypervalent_iodine

1. The mathematical abstraction that describes distances is called a metric space. In a metric space, distance is required to be a nonnegative real number. There's no mathematical theory of negative lengths. That's not to say that someday someone won't come up with something like that, but it's not currently available. The field's wide open to you. http://en.wikipedia.org/wiki/Metric_space
2. I hear the blacksmiths have been put out of business by these newfangled automobiles.
3. Pi is finite in every base. It's between 3 and 4 no matter how you represent it. It can't possibly have a terminating or repeating expansion in any integer or rational base (however you define rational bases, it's tricky) because pi is an irrational number.
4. Young man, in mathematics you don't understand things. You just get used to them. -- John von Neumann http://en.wikiquote.org/wiki/John_von_Neumann
7. Where do you live? If you are in the SF Bay area you can get a job easily. There's a huge tech boom right now. If you live in a small town in a place with no tech, you will find it much more difficult. Perhaps you might consider moving to where the jobs are.
8. How do you figure that? If a/i = ai and you multiply both sides by i you get a = -a. Surely you don't believe that every number is equal to its negative.
9. Yes. 1, x, x^2, ... are a basis for P(x). There are infinitely many vectors in the basis. And each element of P(x) can be written as a finite linear combination of basis vectors. It doesn't make a whole lot of sense to point out that x^2 = 0 + 0*x + 1*x^2 + 0*x^3 ... It's true, but so what? In any vector space V, any vector v whatsoever can be written v = v + 0*x1 + 0*x2 + ... where the xi's are all the other vectors in the entire vector space. But so what? What's the significance of this to you? That would be true about any basis vector in any vector space. In the Cartesian plan with standard basis {(1,0), (0,1)} you could certainly make the point that (1,0) = (1,0) + 0 * (0,1) but what would be the point of saying that?
10. No, because "root" is ambiguous. Every nonzero complex number has two distinct square roots. And there is no canonical way to distinguish them. So you have to say which of the two roots you mean in the equation above. All the problems in this thread come down to choosing one square root on the left side and the other square root on the right. This is different from defining sqrt(2) as the positive of the two real numbers whose square is 2. In the real numbers, we can define a privileged subset of positive numbers. In the real numbers, 2 can be algebraically distinguished from -2. But in the complex numbers, there are no positive numbers so that we can not distinguish between the two square roots of a number without explicitly saying which square root we're choosing. There's a somewhat heavygoing discussion of this subject here ... http://en.wikipedia.org/wiki/Branch_point Basically they teach you this stuff in a math major class on complex analysis. The key point is that in the reals, sqrt can be defined unambiguously as the positive one. In the complex numbers, sqrt can not be defined unambiguously except by explicit saying which square root you mean.
11. No, -1^3 = 1(1^3) since exponentiation has precedence over negation. One of the many persistent errors in this thread. Hello, people, two things please: 1) (-2)^2 = 2^2 does not imply that -2 = 2. 2) -1^48 = -1. (-1)^48 = 1.
12. I haven't been following this (silly IMO) thread in any detail. But I'm not aware of the above either. Please explain.
13. If you find a polynomial with rational coefficients that has e as a zero, then e would not be transcendental. That's the definition.
14. How do you figure that? Multiplying both sides by i gives a = -a, which is your mistake.
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