vector space prove!

Just to verify, vector addition is defined as: $xp(x) + yp(y) = (x+y)p(x+y)$ ?

Let's go through some of the axioms and see if you can do the rest.

1. $\bold{u}, \bold{v} \in S \: \Rightarrow \: \bold{u} + \bold{v} \in S$

Let's choose two arbitrary vectors, ap(a) and bp(b). Then, $ap(a) + bp(b) = (a+b)p(a+b) \in S$ as it is in the form xp(x) (x = a + b) in this case.

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2. Given scalar k, $\bold{u} \in S$ , then $k\bold{u} \in S$ as well.

Again, pick an arbitrary vector ap(a). Then $k \cdot ap(a) = kap(ka) \in S$ (imagine x = ka).

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3. $\bold{u} + \bold{v} = \bold{v} + \bold{u}$

Pick two arbitrary vectors ap(a) and bp(b). Then:
$ap(a) + bp(b) = (a+b)p(a+b) = (b+a)p(b+a) = bp(b) + ap(a)$

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4. $(c+k)\bold{u} = c\bold{u} + k\bold{u}$

Pick arbitrary vector ap(a). Then, $(c+k)ap(a) = (c+k)a p\left[(c+k)a\right] \in S$ (imagine x = (c+k)a)

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5. There exists a zero vector, $\bold{0}$ , such that $\bold{u} + \bold{0} = \bold{0} + \bold{u} = \bold{u}$

Again, choose arbitrary vector ap(a). Then:
$ap(a) + 0p(0) = (a+0)p(a+0) = ap(a)$ (just going by definition)
$0p(0) + ap(a) = ...$

etc. etc.

As you can see, verifying some of these are pretty trivial. Now, it looks like the set is indeed a vector space but I may be wrong. Go through all the axioms and see if they are all satisfied.