My question has to do with the exercise above. I am to prove that the following set is a vector space. When I come across Axiom 3, the associative property, a necessary property of the set to be a vector space, I basically arrive at something like this:
I'm going to define $3$ polynomials $u, v,$ and $w$ such that:
$$u = a_0 + a_1x + a_2x^2$$$$v= b_{0} + b_{1}x + b_{2}x^2$$
$$w= c_{0}+c_{1}x+c_{2}x^2$$
For $P_2$ to be a vector space, the following must be true:
$$u + (v+w) = (u+v)+w$$
However.. the only way I can prove it is by plugging in my expressions for $u, v$ and $w$ and just stating that the following are equivalent because.. they are. That's just how Real numbers behave. Is there something I can do to prove it other than just saying, well, $P_2$ is a subset of real numbers and due to this have this associative property? Is there anything else I can do, or leave it at that?
This will show you how to write a proof of something seemingly obvious. You just need to be a little careful that you do not make loose statements. One such made in your question was "$P_2$ is a subset of the real numbers", which is definitely not true.
1 : Note that $$(u+v)+w = ((a_0 + b_0) + c_0) + ((a_1 + b_1) + c_1)x + ((a_2 + b_2) + c_2)x^2 \tag{def'n}$$ by definition of how we add polynomials.
2 : By associativity of addition of real numbers (we do not need to prove this), we see that:$$ (a_i + b_i) + c_i = a_i + (b_i + c_i) \quad ;\quad i = 0,1,2 \tag{assoc} $$
3: Substituting $(\mathrm {assoc})$ in $(\mathrm{def'n})$ gives us: $$ ((a_0 + b_0) + c_0) + ((a_1 + b_1) + c_1)x + ((a_2 + b_2) + c_2)x^2 \\ =(a_0 +( b_0 + c_0)) + (a_1 + (b_1 + c_1))x + (a_2 + (b_2 + c_2))x^2 \tag{3} $$
4 : Finally, from definition, we observe that: $$ u + (v + w) = (a_0 +( b_0 + c_0)) + (a_1 + (b_1 + c_1))x + (a_2 + (b_2 + c_2))x^2 =^{3}(u + v) + w $$
which completes the proof of associativity of addition of polynomials.