Is zero considered a number in mathematics?

[lethe]

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?

try to declare j a root of e^x=0 over the reals. you won't get a field.

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

 

[lethe]

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 is this field F? i am unfamiliar with the algebraic closure of a finite field. does it have a name? does it depend on which finite field we start with? (i am pretty sure that the answer to this is "yes") is there a nice way to write a generic element?

 

[turin]

Originally posted by Hurkyl
The proof that x^2 - 1 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, h, we are no longer working over a field, so this proof fails.

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
Basically, I'm suggesting you do the same thing with these numbers. Just like you use the fact i^2 = -1 to simplify the expressions for ordinary complex numbers, you can use the fact \alpha^2 = 2\alpha - 2 to simplify expressions for complex numbers written in this new way.

I get it. That's kind of interesting.

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.

Does one need the Riemann sphere to define 0 as a member of C?

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.

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
... {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).

What is a "omthe poly?"

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.

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
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.

This looks like Chinese to me (and I am not oriental).

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.

Thanks for the reference. I will try to motivate myself through it (first I have to motivate myself to the library).

Originally posted by lethe
try to declare j a root of e^x=0 over the reals. you won't get a field.

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.

 

[matt grime]

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 non-trivial way (obviously x.0=0 is a trivial statement), that is what we mean by zero-divisors (the non-trivial 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 non-trivial 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, let's 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 non-zer 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.





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