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a simple question
what is zero "0" does it count as a number?
any answers
what is zero "0" does it count as a number?
any answers
That should beOriginally posted by suyver
Zero is the only solution of the equation [tex] x = -x[/tex].
Also, zero is both the limit of the largest negative numbers and the limit of the smallest positive numbers...
Naturally, even more important (due to Euler):
[tex] 0 \; = \; e^{i\pi}-1[/tex]
Had enough?
Originally posted by suyver
Zero is the only solution of the equation [tex] x = -x[/tex].
I hang my head in deep shame. You are (of course) very right.Originally posted by kishtik
That should be
[tex] 0 \; = \; e^{i\pi}+1[/tex]
Originally posted by turin
What is i or π in mod2?
Not n, π. That is pi.Originally posted by matt grime
... n is either 0 or 1 depending on n odd or even resp.
Originally posted by quartodeciman[/b]
I like the title of a monograph by nineteenth-century German mathematician Richard Dedekind.
"Was sind und was sollen die Zahlen?"
This can be rendered roughly in English by the following.
What are numbers, and what should they be?"
I don't see how this could be nonsensible. It is a definition, and I don't see how it could be self-contradictory in this case. I'm assuming we already take for granted what ^2, +, 1, and "root" mean? Then, we just introduce another entry into our dictionary: i. What could be wrong with that?Originally posted by Hurkyl
... for any practical algebraic purpose, we can simply say that we've simply declared that [itex]i[/itex] is a root of [itex]x^2 + 1[/itex].
(Behind the curtain, there is quite a bit of mathematics involved to prove that such a declaration can be sensible, but you don't need to know what goes on behind the curtain to use such declarations)
It doesn't seem to be an issue to continue with replacement. Why stop with the definition for α, when we could instead declare β is a root of x2 + x + 1? And then why stop here, when we could instead declare is a root of x2 + x + ? In answer to "why stop there?" I would reply, because it is just as good a place as any (better IMO). Are you saying that this use of α is a better way to define complex numbers?Originally posted by Hurkyl
Why stop there? We could instead declare [itex]\alpha[/itex] is a root of [itex]x^2 - 2x + 2[/itex]
I don't get it. Isn't it already in that form, that is, two factors of that form?Originally posted by Hurkyl
(exercise for those of you at home, and I strongly suggest you do it, things will make more sense!: what is [itex](a + b \alpha) (c + d \alpha)[/itex] written in the form [itex]p + q \alpha[/itex]?)
I think I missed what the idea was. You mean we can define things in mod2? I don't get why that is so special. Are you talking about the analogy of extending the real numbers into the complex numbers? Is this F2 the short way of writing "the field of integers mod 2?"Originally posted by Hurkyl
Anyways, we can apply this same idea to any other field, such as the integers mod 2 ([itex]F_2[/itex]). (a field is essentially just a number system in which you can divide by nonzero things)
I don't understand this. What does "[itex]p, q \in F_2[/itex]" mean? So what if it's not helpful if it's valid. Does this not prove that i is, by definition, 1 in mod 2? (since the definition of i is the root of x2 + 1)Originally posted by Hurkyl
One could try and define a new number system that is all numbers of the form [itex]p + q \alpha[/itex] where [itex]p, q \in F_2[/itex] and [itex]\alpha[/itex] is a root of [itex]x^2 + 1[/itex]... but that isn't helpful because [itex]x^2 + 1[/itex] already has a root in [itex]F_2[/itex]!!!
Is this just a random polynomial that you chose as an example that doesn't have roots in mod 2, or is there something special/conventional about it?Originally posted by Hurkyl
But, as before, we can simply pick another polynomial and declare some new field (let's call it [itex]F_2(\alpha)[/itex]) as all numbers of the form [itex]p + q \alpha[/itex] where [itex]p, q \in F_2[/itex] and [itex]\alpha[/itex] is a root of [itex]x^2 + x + 1[/itex]. This system is interesting because [itex]x^2 + x + 1[/itex] doesn't already have a root in [itex]F_2[/itex].