Let V denote the set of ordered pairs of real numbers. If (a, b) and (c, d) are elements of V and x is an element of R, define:
(a, b) + (c, d) = (a+c, bd)
x(a, b) = (xa, b).
Is V a vector space over R? Justify.
Well, it isn't, but the book claims this is due to the fact (which is referred to by (VS4)), that for each element a in V there exists an element b in V such that
a + b = 0.
However, if c = -a, and d = 0, then (a+c, bd) = (0, 0) always. Am I missing something here?
Firstly, note that a vector space forms an Abelian group under +. The zero is called the identity of the group, and the b such that a+b=0 is called the inverse of a.
You are quite right, every element has another elements such that . However, this element is not unique. Taking to be the identity for the group, then is the additive inverse for the set .
In a group, and so in a vector space whilst under +, the inverse is always unique. If and then . Thus, your set set cannot form an Abelian group under +, and so is not a vector space.
The underlying reason for this not working is the fact that you have no identity. You assumed it was , but actually it should be (and it it, in fact, neither):
Let us examine the case of . Then, , so is clearly not our identity. It is obvious, however, that is the unique identity for this set. Therefore, is the unique identity for the set under +, which is a subset of our set.
However, if this is the case then set has no additive inverse, . Thus, the entire set has no single additive inverse, which contradicts the additive Abelian group axioms.
What you have is a fine example of a semigroup, but I'm afraid it's not a group.
Hope that makes sense!