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Since it is closed under addition, has an additive identity, and has additive inverses for each element, this forms a group. Given a suitable definition for scalar multiplication (where it essentially ignores the 1,2,3 at the beginning), then we can even make this a vector space.
Yes: this is how you show it is unique. Your proof is correct; although it could be made a lot less verbose by just saying i1 =i1 +i2 = i2 +i1 = i2 i 1 = i 1 + i 2 = i 2 + i 1 = i 2. That's a little verbose, but essentially it. Yes, although you can trim it down a bit; for clarity's sake. If in i n is an additive identity, then for any a a in ...
$\begingroup$ What exactly don't you understand: what is meant by expressions "$0+U$" and "$-v+U$", or why it's true that they are the identity and the inverse in the quotient space? Those are two different questions. $\endgroup$
I’m not sure what definition you’re using, but semirings are usually defined to require that the additive identity is absorbing. If you’d like an example of something slightly less than a semiring which has a non absorbing zero, see Examples for almost-semirings without absorbing zero .
$\begingroup$ Generally $1 \neq 0$ is part of the definition of a field. e.g., in Wikipedia, "To exclude the trivial ring, the additive identity and the multiplicative identity are required to be distinct." $\endgroup$ –
By definition of the additive identity, then $0=c*0$. However, I feel like this is wrong, ...
1) If a, b are vectors, a + b is a vector. 2) (a + b) + c = a + (b + c) for vectors a, b, c. 3) a + 0 = 0 + a = a by definition of the zero vector 0 (i.e. the zero vector is defined to be an additive identity as in the vector space axioms). 4) a + (− a) = (− a) + a = 0 , i.e. additive inverses exist. In a group, one can easily show an ...
The additive identity property of a vector space can be broken into two parts. First, the addition operation must allow for an identity to exist. Second, that identity must actually be in the space. The operation dictates what the identity has to be, and the set determines whether that identity is included. There are a few common errors that ...
The subspace {${0}$} is an additive identity for the operation of addition on the subspaces of V. More precisely, if U is a subspace of V, then U + {${0}$} = {${0}$} + U = U. This is quite obvious as subspaces are itself vector spaces and so must have {${0}$}, or even their addition, whether uniquely expressed or not.
The fact that (until we prove otherwise) we suspect that there may be other elements of the vector space with the same property of being an additive identity doesn't change what the axioms have told us about the one whose existence it asserts. $\endgroup$