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#19
Feb404, 11:37 AM

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Matt grime,
It almost seems that you're not listening to me. But first, I will assume that I am very confused about mod2 (being not a math person). First of all, I thought that mod2 could only deal with two distinct discrete objects, for instance 0 and 1. Then, the combination rule is defined such that 1 + 1 = 0, so that you always stay in the group. I didn't know that multiplication, exponentiation, imaginary, or irational numbers were allowed in this scheme. I read through your bit about what i is in mod2 over and over again, and I still don't get it. If i = 1, then why write i? Does it only equal 1 when it is considered by itself, but it equals √1 when it is in a product or exponent? It seems like there is 0 and there is 1, and then there is an infinitude of other "values" (not mod2) that you either have to identify with 0 or with 1. Why not say that i = 0? I never asked about n. I think you may have seen the html "pi" in my post and interpretted it as an "n." Anyway, I'm curious what you would call pi in mod2. And now I realize that I forgot to ask about e. Both of these are irrational. That seems like it would be a problem for any discrete number system. Tell me where I have gone way off. You said something about a ring. What is that? edit: Well, I don't know why I screwed this whole thing up so bad. For some reason, I thought we were talking about Euler's identity. But you were just talking about x = x this whole time (I just read through the thread again). No wonder you think I'm whacko. I am TERRIBLY sorry about that. I would still like to hear what you have to say about the questions/concerns that I posed above, though. 


#20
Feb404, 12:18 PM

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for turin
indeed i was only talking about x=x, nothing to do with e or pi. Firstly, F_2 is the set {0,1} as you know, with the addition and mult. as you stated. Now, this is all I was using. Now, i is defined as the square root of 1 in C, here there is a analogous element in the field because 1=1 has a square root. Notice however that there is a polynomial x^2+x+1 that has no root in F_2. 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. In all these fields 1=1. In the same way as there is a surjection from Z to F_2, there is a way of relating algebraic integers to "an" algebraic closure of F_2. Let (2) be the ideal generated by 2 inside the ring of algebraic integers (that is the set of solutions of all monic polynomials with integer coeffs), the ideal is just the set of all things 2x, where x is an algebraic integer. Just like mod 2 arithmetc, we can declare two elments in the alg, integers to be the same if their difference lies in (2) (actually we require a maximal ideal containing (2) but that's a technicality). Thus it is possible to define the image of any algebraic integer in a field of characteristic 2. However, neither e nor pi are alg. integers so i can't define their images. To explain the high faluting maximal thing  2 must get sent to zero, and so sqrt(2) must also get sent to zero if this were to make any sense, but sqrt(2) is not in the ideal (2). The maximal thing there corrects that problem and makes sure that the quotient is a field. the vast majority of algebraic integers are irrational, in fact the only rational ones are the integers. it's a useful exercise to prove that interestingly, the golden ratio, a root of x^2+x+1 must get sent to A, or 1+A in this scheme. note that A has no value as a real number! just as i doesn't, it's just a symbol we manipulate according to the rule A^2=A+1 Matt 


#21
Feb404, 12:30 PM

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Matt, I understood just about none of that termonology, but thanks for explicating. I suppose I can inch through it and eventually understand it.



#22
Feb404, 05:08 PM

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Here it is in simpler terms.
In the world of complex numbers, a perfectly valid definition of [itex]i[/itex] is: [itex]i[/itex] is a root of [itex]x^2 + 1[/itex]. All of the arithmetic we do with [itex]i[/itex] can be done algebraically using this fact; for instance: [tex] \begin{equation*}\begin{split} (a + bi) (c + di) &= ac + adi + bic + bidi \\ &= ac + (ad + bc) i + bd (i^2) \\ &= ac + (ad + bc) i + bd (i^2 + 1  1) \\ &= ac + (ad + bc) i + bd (0  1) \\ &= (ac  bd) + (ad + bc) i \end{split}\end{equation*} [/tex] So, 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) Why stop there? We could instead declare [itex]\alpha[/itex] is a root of [itex]x^2  2x + 2[/itex]! (1) This is also a perfectly valid way to create the complex numbers; declare them to be all numbers of the form [itex]p + q \alpha[/itex]. (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]?) It turns out in this case that (assuming my hasty algebra was correct) if we make the substitution [itex]\alpha \rightarrow i + 1[/itex] that we can convert from this new definition of the complex numbers to the normal definition. 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) 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]!!! 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]. It turns out that when doing this type of construction over a finite field (such as the integers mod a prime), things are much more interesting than the boring world of the real and complex numbers. (that's boring algebraically; they're still very interesting because of their topology) (1) This sentence was originally incorrectly written as "Why stop there? We could instead declare [itex]\alpha[/itex] is a root of [itex]\alpha^2  2\alpha + 2[/itex]!" 


#23
Feb404, 06:14 PM

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I'm not sure I can countenance using x to be a root of x^22x+2...



#24
Feb404, 06:47 PM

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While my typo was still technically correct, I will admit it was extraordinarily confusing. Fixed.



#25
Feb404, 08:01 PM

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#26
Feb404, 08:17 PM

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It would probably help if you knew what a ring, field etc were,
But ignore that bit. 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. Npw, we could equally have well defined j to be the root of the poly x^2+5, and algebraically,R(i) and R(j) are indistinguishable, that is one is a subfield of the other. So all though you say, well, adjoin other roots, the thing is they are already there. Now, let's take polys with coeffs on F_2 or Z_2 depending on your preference. the equation x^2+1 has both its roots in F_2 already. There is however no root of X^2+x+1 in F_2. Traditionally, that is the first one you look at. Now it as a theorem that is not important (but whose proof is, bizarrely*) that no finite field (find a definition on Wolfram) has roots to every polynomial in it. Look up field, field extension and algebraic closure. * to explain that assertion, for instance, it is not very important that sqrt(2) is irrational, but that there is a proof (not the one you know at a guess) that implies that no integer has a rational square root unless it is a square of an integer  look at prime factors. 


#27
Feb404, 08:41 PM

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These are things that one might worry about when trying to extend a number system. However, there is a nice theorem: Let [itex]F[/itex] be a field, and let [itex]p(x)[/itex] be an irreducible polynomial[itex]^1[/itex] of degree [itex]n[/itex] whose coefficients are in [itex]F[/itex]. Then, there exists a field [itex]F(\alpha)[/itex] that consists of all numbers of the form: [tex]a_0 + a_1 \alpha + a_2 \alpha^2 + ... + a_{n1} \alpha^{n1}[/tex] where all of the [itex]a_i[/itex] are elements of [itex]F[/itex], and we have that [itex]p(\alpha) = 0[/itex]. So the mathematics behind the curtain say we're rigorously justified to do this sort of thing. The mathematics behind the curtain also say there's exactly one extension of the real numbers that you can make in this way; the complex numbers. 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]. Then we consider numbers of the form [itex]a + bh[/itex] where [itex]a, b \in \mathbb{R}[/itex]. (that is, a and b are real numbers) The problem with this system is that division isn't always possible. For instance, what is [itex](1  h)^{1}[/itex]? If we try a generic possibility [itex](1  h)^{1} = (a + bh)[/itex] and multiply, we find: [tex] \begin{equation*}\begin{split} (a + bh) (1  h) &= (a + bh  ah  bh^2) \\ &= (a  b) + (b  a) h \end{split}\end{equation*} [/tex] So no matter what we choose for [itex]a[/itex] and [itex]b[/itex], this cannot equal 1, thus [itex]1  h[/itex] doesn't have a multiplicative inverse; i.e. we cannot divide by [itex]1  h[/itex]. So we have to be careful when we decree things like this; the result might not be quite what we expect! [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]. 1: An irreducible polynomial is one whose only factors are multiples of itself and multiples of 1. The field in question is important; e.g. [itex]2x^2  4[/itex] is irreducible over the rational numbers, but factors as [itex](2x  2\sqrt{2})(x + \sqrt{2})[/itex] over the real numbers. 


#28
Feb404, 10:33 PM

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Thanks Matt and Hurkyl. I think I very well may have learned more math today than in all the 9251 other days of my life combined.
One question: When you say, "it is alg closed," (I'm assuming "alg" means "algebraically") "what" is alg closed? The polynomial? The field? x/y = z. let y = 0 (as far as I understand, this is a perfectly valid member of C). z = ?. edit: I just realized, I said "the" root. What I should have said was "a" root. (An advantage of specificity that the Polish language lacks) So, actually, this renders your example without a problem, because h = 1. Then, (a + bh) = (a  b) = 1/2. Leaving the expression generic for the time being, we must find the two values a & b. We have the equations: a  b = 1/2 (a  b) + (b  a)(1) = 1 Which become: 2a  2b = 1 2a  2b = 1 This system of equations is perfectly selfconsistent. 


#29
Feb404, 11:04 PM

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Ah, but how do you prove that [itex]h = 1[/itex]? You know that 1 is a root of [itex]x^2  1[/itex]... 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. In this new structure (called the hyperbolic numbers), isn't always allowed and nth degree polynomials can have more than n roots! 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. (mainly this is just a warmup for doing the same kind of thing over finite fields; I don't know of any practical use of this exercise other than simply as a good exercise) 


#30
Feb504, 06:34 AM

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Yes, alg means algebraically. Algebraic closure refers to the field. A field is algebraically closed when and only when every polynomial with coeffecients in that field has all its roots in that field.
And x/0 is not allowed in the complexes. The reason so many people think this is ok is that there is a well known and often misused object called the Riemann Sphere which extends the complex plane to include a point at infinity, but this is basic algebra here so we don't touch it. The 'problems' Hurkyl alludes to might include lots of zero divisor issues, and there is alot of the theory of ideals in there. Here are two examples  take the integers mod 8, that is the numbers 0,1,2,3,4,5,6,7 with addition and multiplication taken remainder 8. here 4*2=0 and we say 4 and 2 are divisors of zero. This is not a good place to do arithmetic  we can't divide. Also note that x^2=1 has 4 solutions, 1,3,5,7. All this behind the scenes stuff makes sure that there are no zero divisors in C and degree n polys have n roots. The second. To give you a flavour of what the smoke and mirrors are: Take R[x] to be the ring of all polys in on variable x with real coeffs. Let p=x^2+1, then {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). Now p is irreducible  that is it has no real roots (that's not the general definition but for quadratics it is equivalent). 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 rs, or equivalently there is another poly h with r = s+h*p  that we get a field. we denote this R[x] / {p} 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. 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. 


#31
Feb504, 11:48 AM

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so that is why it is noteworthy that you can do this for algebraic equations, like [itex]x^2=1[/itex] 


#32
Feb504, 11:50 AM

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#33
Feb504, 12:32 PM

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#34
Feb504, 03:06 PM

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I think I'll start a new thread for this, but to answer some specific questions.
1. 'omthe' is a typo that should read 'other' 2. No 0 is defined in C already, you just can't divide by it  check the defintion of a field, F is a field if it is an abelian group, with identity 0 under operation +, and F, omitting 0, is also an abelian group under the operation *. 3. In some structure ( ring usually) we say x divides y (is a divisor of) if there some other z with x*z=y. So in the ring of integers mod 8, 2 divides 0 in a nontrivial way (obviously x.0=0 is a trivial statement), that is what we mean by zerodivisors (the nontrivial is implicit). 4. When I say it is not a good place to do arithmetic, I mean things like finding roots of ax+b=0 is not as easy as it ought to be, because usually we would say x= b/a. However, when nontrivial zero divisors exist this isn't true, as we can no longer divide by a. I mean the multiplicative inverse for 2 does not exist in mod 8 arithmetic. To convince yourself of what's going on, lets do mod 3 arithemetic, what is 1/2? It is by definition the thing that when multiplied by two gives 1, agreed? So we are seeking a y such that 2y=1 (mod 3). By inspection 2*2=4=1 mod 3, so 1/2 = 2! Really we ought not to write 1/2 as it is too suggestive, but instead write 2^{1} In cases where x*y=0 for nonzer x and y we cannot say 0/x =y and vice versa  or at least whilst you may write it, it is not valid as a mathematical statement. To see why, consider mod 16 arithmetic  4*4=0 and 4*8=0, so you cannot define 0/4  there are two possibitlities. LOok out for a new posting. 


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