\begin{align*} + : \mathbb{R}^2\times\mathbb{R}^2 &\rightarrow \mathbb{R}^2\\ ((a,b),(c,d))&\mapsto(ad+bc,bd)\\[1cm] \times: \mathbb{R}^2\times \mathbb{R}^2 &\rightarrow \mathbb{R}^2\\ ((a,b),(c,d))&\mapsto(ac,bd) \end{align*} Questions:
a) Show that $\times$ is commutative.
b) Show that $+$ is associative.
c) Show that $\times$ is not distributive over $+$.
d) Show that there is a neutral element for $+$.
a. Assuming that we commute in $\mathbb{R}$, $(a,b) \times (c,d) = (ac,bd) = (ca,db) = (c,d)\times(a,b)$
b. $$((a,b) + (c,d)) + (e,f) $$$$= (ad+bc,bd) + (e,f) $$$$= ((ad+bc)f+bde, bdf)$$ and we also have $$(a,b) + ((c,d) + (e,f)) $$$$= (a,b) + (cf+de,df) $$ $$= (adf + b(cf+de), bdf)$$ $$=((ad+bc)f+bde, bdf)$$
So we have associativity.
c. Here is a good counter example. We are going to go step by step on this first one now: $(1,2) \times ((3,4) + (1,1)) =(1,2) \times(3\cdot 1+4\cdot 1, 4 \cdot 1) =(1,2) \times (7,4) = (1\cdot 7, 2\cdot 4)$ $$ = (7,8)$$ $(1,2) \times (3,4) +(1,2) \times (1,1) = (3,8) + (1,2) $ $$= (14, 16)$$
So we have shown that distributive law does not work.
d. (0,1) is neutral. Notice that $(a,b) + (0,1) = (a\cdot1+ b\cdot 0 , b\cdot 1) = (a,b)$ for any $(a,b)$