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hamlet69
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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].
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 x^{2} + x + 1? And then why stop here, when we could instead declare is a root of x^{2} + 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 F_{2} 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 x^{2} + 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].
I don't see how this could be nonsensible.
Are you saying that this use of ? is a better way to define complex numbers?
I don't get it. Isn't it already in that form, that is, two factors of that form?
Are you talking about the analogy of extending the real numbers into the complex numbers?
Is this F_{2} the short way of writing "the field of integers mod 2?"
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?
This is something I never before realized. So, basically, I never realized how phenomenal the field of complex numbers is until today. Again, thank you for this supurb enlightenment.Originally posted by matt grime
The key here is that algebraically the real numbers are not complete - there is a polynomial with real coeffs that has no real root - x^2+1. We can declare that there is some element i, the
at behaves accoridng to the rule i.i=-1, and form C as R(i). Now it is an important result that you've now gone far enough - every polynomial with coeffs in C has roots in C, that is it is alg closed.
Isn't this already a "problem" with complex numbers:Originally posted by Hurkyl
It could lead to a new number system that doesn't have nice properties. (e.g. maybe division isn't well-defined)
Can you give an example of how this contradicts something.Originally posted by Hurkyl
(maybe we have to represent things like [itex]a + bi + ci^2[/itex]) The definition might not be self-contradictory, but it might lead to a contradiction in the future!
But this, to me, seems like a self-contradictory definition. Do we not know that the root of this polynomial is 1? So, h = 1, but then we define h /= 1. So h /= h. How is this not a contradiction.Originally posted by Hurkyl
Here's an example of something that yields a system that has less nice properties (though this system is an interesting one):
Define [itex]h[/itex] to be a root of [itex]x^2 - 1[/itex], but also require that [itex]h \neq 1[/itex].
ABSOLUTELY! And again, thank you for posing it. I feel like my level of understanding has sky-rocketed.Originally posted by Hurkyl
No, I just wanted to make another example before leaving the comfortable world of the real and complex numbers. Besides, IMHO, playing around with at least one alternate definition of [itex]\mathbb{C}[/itex] is a good exercise for understanding things.
Are you demonstrating that it cannot be put into the desired form, or that puting this into the desired form is not trivial? Is this a demonstrated "future contradiction?"Originally posted by Hurkyl
Well, if we foil, we get:
[tex]
(a + b\alpha) (c + d\alpha) = ac + (ad + bc) \alpha + bd \alpha^2
[/tex]
Which is not of the form [itex]p + q \alpha[/itex].
Isn't this already a "problem" with complex numbers:
x/y = z.
let y = 0 (as far as I understand, this is a perfectly valid member of C).
z = ?.
Can you give an example of how this contradicts something.
But this, to me, seems like a self-contradictory definition. Do we not know that the root of this polynomial is 1? So, h = 1, but then we define h /= 1. So h /= h. How is this not a contradiction.
Are you demonstrating that it cannot be put into the desired form, or that puting this into the desired form is not trivial?
Originally posted by turin
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 matt grime
f Just like in the real/complex case, we simply define A to be a root of this polynomial (the other root is 1+A), then there is an extension F_2[A] which we call F_4, it is the field with 4 elements, it is still not posible to find roots of every polynomial in F_4, so we can extend again and again. Each extension has 2^r elements for some r. The 'limit' of this construction, we'll call F, and it is the algebraic closure of F_2. It is infinite.
What does it mean to be "working over a field?" I thought that the coefficients of the polynomial had to be members of a field.Originally posted by Hurkyl
The proof that [itex]x^2 - 1[/itex] has only two roots (1 and -1) depends on the fact that you're working over a field; that is division always works (for nonzero things). When I introduced this new thing, [itex]h[/itex], we are no longer working over a field, so this proof fails.
I get it. That's kind of interesting.Originally posted by Hurkyl
Basically, I'm suggesting you do the same thing with these numbers. Just like you use the fact [itex]i^2 = -1[/itex] to simplify the expressions for ordinary complex numbers, you can use the fact [itex]\alpha^2 = 2\alpha - 2[/itex] to simplify expressions for complex numbers written in this new way.
Does one need the Riemann sphere to define 0 as a member of C?Originally posted by matt grime
... the Riemann Sphere ... extends the complex plane to include a point at infinity, but this is basic algebra here so we don't touch it.
How can we have divisors without being able to divide? What is this new (to me) definition of divisor? Why do you only list 4 and 2? I thought that 0/anything = 0. Is it that 0/2 = 4 and 0/4 = 2? Is that true? If so, that's wierd.Originally posted by matt grime
... take the integers mod 8, ... here ... we say 4 and 2 are divisors of zero. This is not a good place to do arithmetic - we can't divide.
What is a "omthe poly?"Originally posted by matt grime
... {p} is the set of all multiples of p, that is all polynomials of the form p(x)q(x) for some omthe poly q(x).
When we say "divide," do we always assume that there is no remainder? So, basically, by saying "p divides r-s," that is another way of saying that "p is a factor of r-s?"Originally posted by matt grime
This irreducibility means that when we form the mod p analogue for polynomials - ie two polys r and s are defined to be equivalent if p divides r-s, or equivalently there is another poly h with r = s+h*p -- that we get a field.
This looks like Chinese to me (and I am not oriental).Originally posted by matt grime
notice that we can define a map to C by sending x to i. This is acutally an isomorphism - that is R[x]/{p} and C are actually the same field.
Thanks for the reference. I will try to motivate myself through it (first I have to motivate myself to the library).Originally posted by matt grime
Ian Stewart's book on Galois Thoery is a good place to learn about this - unlike most university level texts this emphasizes the examples and works inside C most of the time.
I realize that this addresses my question with a counterexample, so I do not dispute it. But, for the sake of future discussion, will infinite order polynomials be relevant? If so, then this seems like an example to show that not even the complex field is alg closed.Originally posted by lethe
try to declare j a root of [itex]e^x=0[/itex] over the reals. you won't get a field.
In mathematics, a number is a mathematical object used to count, measure, and label. It can be represented by symbols or words and can be used in mathematical operations.
Yes, zero is considered a number in mathematics. It is a whole number and is used to represent the absence of quantity or value.
Zero is considered a number in mathematics because it follows the same rules and properties as other numbers. It can be added, subtracted, multiplied, and divided like any other number.
In mathematics, zero is a number that represents the absence of value, while null is a placeholder that represents the absence of any value or object. In other words, zero has a specific numerical value, while null does not have a value at all.
Zero is used in mathematics to indicate an empty set, to represent the origin on a number line, to indicate a neutral element in operations, and to represent the absence of value in equations and calculations.