Let $A = \mathbb{R} \times \mathbb{R}$, and the operation $* : A \times A \rightarrow A$ on $A$. Check if associative, commutative, identity, inverse.

Question

I have this exercise:

on the set $A = \mathbb{R} \times \mathbb{R}$ is defined the operation $* : A \times A \rightarrow A$ such that

$\forall (a,x), (b,y) \in A \quad (a,x)*(b,y) = \left ( \frac{2}{5}ab , \frac{3}{4} + y + x \right )$

1. Check if the operation is associative;
2. Check if the operation is commutative;
3. Check if exists the identity element of the algebric structure $(A, *)$;
4. 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:

1. 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.

2. 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.

3. 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!

Answer 1

$(I_b, I_y)=(\frac{5}{2},\frac{-3}{4})$ and inverse of $(-3,2)$ is $(\frac{-25}{12},\frac{-14}{4})$ because $(\frac{-6}{5}b,\frac{3}4+2+y)=(\frac{5}{2},\frac{-3}{4})\Rightarrow (b,y)=(\frac{-25}{12},\frac{-14}{4})$

Answer 2

Supposing $(a,x)$ is an identity element immediately tells us that $\left(\frac{2}{5}(1)(a), \frac{3}{4} + x + 1\right) =(1,1) \ast (a,x) = (1,1)$. Now you can see what $a,x$ specifically have to be in order for $(a,x)$ to qualify as an identity element: $a = \frac{5}{2}$ and $x = -\frac{3}{4}$. But you still need to check that $\left(\frac{5}{2},-\frac{3}{4}\right)$ is truly an identity element.