# Show that $x$ is commutative and $+$ is associative for these binary operations on $\mathbb{R}^2$

\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 $+$.

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Please post your current working so that people can give you some useful help. – David Feb 11 '14 at 1:16
I have no current work, I have no idea where to start. :( – Karl Feb 11 '14 at 1:18
The problem with questions such as this is that they are messy to type up and show and they garner no rep points usually. – Vladhagen Feb 11 '14 at 1:18
@Karl, do you know what "commutative" means? – David Feb 11 '14 at 1:24
@Vladhagen Yes, a 0. – Karl Feb 11 '14 at 1:29

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 $(a,b)$

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End of answer b should be associativity not communtivity, no? – Karl Feb 11 '14 at 1:39
How come in the third line of answer b the term after (ad+bc)f there isn't an (ad+bc)e as well since there's a bde and a bdf? – Karl Feb 11 '14 at 1:52
If you examine the formula you gave for + carefully, we do first entry of first pair time last entry of last pair, then add that to second entry of first pair times first entry of second pair. That should give us what we have above. – Vladhagen Feb 11 '14 at 2:34
When I do the calculations in answer c I'm getting (4,10) for both equations. How did you carry out the operations? – Karl Feb 11 '14 at 2:46
Notice that the plus rule you gave is not just direct addition. In my comment above I detail how the addition is to be done. You are just adding the two points directly, which is not the rule you gave for addition in your question. Again, you cannot just directly add the points. Follow your rules exactly. – Vladhagen Feb 11 '14 at 3:11