<s>[INST] Let $A = \mathbb{R} \times \mathbb{R}$, and the operation $* : A \times A \rightarrow A$ on $A$. Check if associative, commutative, identity, inverse. I have this exercise:
on the set $A = \mathbb{R} \times \mathbb{R}$ is defined the operation $* : A \times A \rightarrow A$ such that
$\forall (a,x), (b,y) \in A \quad (a,x)*(b,y) = \left ( \frac{2}{5}ab , \frac{3}{4} + y + x \right )$  


*

*Check if the operation is associative;  

*Check if the operation is commutative;  

*Check if exists the identity element of the algebric structure $(A, *)$;  

*Check if exists the inverse of $(2,-3)$ in $(A,*)$.


This is what I have done for the first two points, and I think that is right:  


*

*Associativity:
I have to check if the following equation holds
$\left [ (a,x) * (b,y) \right ] * (c,z) \overset{?}{=} (a,x) * \left [ (b,y) * (c,z) \right ]$  


hence,
LHS:
in the following I have assigned $m = \frac{2}{5}ab, n = \frac{3}{4}+y+x$
$\begin{array}{lcl}\left [ (a,x) * (b,y) \right ] * (c,z) & = & \left ( \frac{2}{5} ab, \frac{3}{4} +y+x\right ) * (c,z) \\ & = & (m,n)*(c,z) \\ & = & \left [ \frac{2}{5}mc, \frac{3}{4} + z +n \right ] \\ & = & \left [ \frac{2}{5} \left ( \frac{2}{5}ab \right )c, \frac{3}{4} + z + \left ( \frac{3}{4} +y +x \right )\right ] \\ & = & \left [ \frac{4}{25}abc, \frac{3}{2} +z+y+x\right ] \end{array}$
RHS:
in the following I have assigned $g = \frac{2}{5}bc, h = \frac{3}{4}+z+y$
$\begin{array}{lcl}(a*x)*\left [ (b,y) * (c,z) \right ] & = & (a,x) * \left [ \frac{2}{5}bc, \frac{3}{4}+z+y \right ] \\ & = & (a,x) * (g,h) \\ & = & \left [ \frac{2}{5} ag, \frac{3}{4} + h + x \right ] \\ & = & \left [ \frac{2}{5}a \left ( \frac{2}{5}bc \right ), \frac{3}{4}+\left ( \frac{3}{4} + z + y \right ) +x\right ] \\ & = & \left [ \frac{4}{25}abc, \frac{3}{2} +z+y+x \right ] \end{array}$
the equation holds and the operation $*$ is associative.


*Commutative element:
I have to check if the following equation holds
$(a,x) * (b,y) \overset{?}{=} (b,y) * (a,x)$  


LHS:
$(a,x) * (b,y) = \left ( \frac{2}{5}ab, \frac{3}{4} + y + x \right )$
RHS:
$(b,y) * (a,x) = \left ( \frac{2}{5}ba, \frac{3}{4}+x+y \right )$
the equation holds and the operation $*$ is commutative.


*Identity element:
here, I have some problems, I am not sure on what to do, I have tried this but without success:


maybe, we have to check the following equation?
$(a,x)*(I_b, I_y) \overset{?}{=} (a,x)$
hence,
$\begin{array}{lcl}(a,x)*(I_b, I_y) & = & \left [ \frac{2}{5}aI_b, \frac{3}{4} + I_y + x \right ] \end{array}$
i.e. we should find a value to $(I_b,I_y)$ such that $\left [ \frac{2}{5}aI_b, \frac{3}{4} + I_y + x \right ] = (a,x)$ 
and therefore I have problems in point 4.
I don't know!
Please, can you help me?
Many thanks really!
 [/INST] $(I_b, I_y)=(\frac{5}{2},\frac{-3}{4})$ and inverse of $(-3,2)$ is $(\frac{-25}{12},\frac{-14}{4})$ because $(\frac{-6}{5}b,\frac{3}4+2+y)=(\frac{5}{2},\frac{-3}{4})\Rightarrow (b,y)=(\frac{-25}{12},\frac{-14}{4})$
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