- #1
1MileCrash
- 1,342
- 41
Is -(x^-1) = (-x)^-1 true for all nonzero x in any ring, where x^-1 denotes the multiplicative inverse of x?
1MileCrash said:I was only able to prove it for a commutative ring.
micromass said:Please post your proof. Where do you use commutativity?
1MileCrash said:-x • -(x)^-1 = -1 • x • -1 • (x)^-1 = -1 • -1 • x • (x)^-1 [with commutativity] = 1 • 1 = 1
Is the proof I did.
micromass said:OK, so the step where you used commutativity is
[tex](-1) * x = x* (-1)[/tex]
But in fact, this is true in any ring, even noncommutative. So your goal should now be to prove this without using commutativity.
1MileCrash said:Yeah, I haven't responded to this because I am just so embarrassed. I am so careful not to assume what I'm not given in proofs regarding algebraic structures that I often make mistakes like this. Thanks again.
WWGD said:AFAIK, the unity satisfies 1r=r1 for all r in the ring, by definition.
1MileCrash said:That's right, I just somehow chose not to see that this implied the same for -1.
mm, I just adjoined this as a lemma:
-1 • x = x • -1
Pf:
x + (-1 • x) = (1 • x) + (-1 • x) = (1-1)x = 0x = 0
x + (x • -1) = (x • 1) + (x • -1) = x(1•1 + 1•-1) = x0 = 0
Additive inverses are unique, so -1•x = x•-1
micromass said:Seems good. In fact, if ##n\in \mathbb{Z}## (interpreted suitably), you can prove that ##nx = xn## for all ##x\in R##.
1MileCrash said:The proof is quick (I didn't write one, but it is obvious that (nx = (1+1+..+1)x = x+x+...x = (x•1+x•1+..+x•1) = x(1+1+..+1) = xn) - this seems even easier than the specific -1 case, but let me make sure I intuitively understand this..
If we have nx - even if these integers aren't in the ring R at all, this still makes sense because this can be considered to be "x to the (additive) power of n," x operated additively on itself n times, and that idea is commutative. It can't be considered a binary operation on members of R (because n is not in R here) but it is still well defined for any ring. Like, if I talk about the ring of mxm matrices in the usual way, and A is an mxm matrix, nA is just A added to itself n times, it doesn't matter that n isn't in the ring, nA makes sense.
If R DOES contain these integers n, then the above still applies, but we can additionally consider it to be exactly the same as a multiplicative operation on two members of the ring, n and x, which will be commutative since it is the same idea as the "x to the nth power" concept above.
Is that roughly correct, or am I spewing nonsense?
1MileCrash said:Of course, this is all beyond my current studies, but I do plan to learn more about this eventually. Very cool stuff. So it appears by what you are saying is that by "a vector space being an abelian group with scalar multiplication" that this is actually not so different from a normal abelian group, it's not that scalar multiplication is a completely new operation, it is that scalar multiplication allows for all elements from an arbitrary field to act just as the integers already do for any abelian group. So the only thing keeping all abelian groups from being vector spaces is the fact that Z is not a field. That is quite an eye opener to me.
1MileCrash said:Now hold on, what about vector spaces whose underlying field F do not contain integers? Since they are abelian groups with additional features, scalar multiplication by integers in a vector space must always be possible, and so the set of scalars for the vector space would actually be F union Z, which could never be a field.
So what's the deal with that? I can't think of an example off hand, but I have no immediate objections to a field that does not contain integers.
A basic ring is a mathematical structure that consists of a set of elements and two operations, addition and multiplication, that follow certain rules.
A basic ring must have closure, associativity, commutativity, identity elements, and inverse elements for addition and multiplication.
A basic ring does not necessarily have a multiplicative identity element, unlike other types of rings such as commutative rings or integral domains.
No, a basic ring can only have one identity element for both addition and multiplication.
Basic rings are used in various scientific fields, such as physics and chemistry, to model and solve problems involving quantities and operations that follow the rules of a basic ring.