Let $S$ be the set of all the ordered pairs in the cartesian plane. That is:
$$S=\{(x,y)|\ \ x, y \in \Bbb{R}\}$$
Then, If $a=(a_1, a_2)$ and $b=(b_1, b_2)$ are two arbitrary elements of $S$, the following operations are defined:
$$a+b=(a_1+b_1, a_2+b_2)$$
$$a\times b=(a_1\cdot b_1, a_2 \cdot b_2)$$
Also, let's state that $a=b$ iif $a_1=b_1$ and $a_2=b_2$.
Associativity and commutativity of the operations are straight forward to prove. The multiplicative neutral element is $u=(1,1)$ and the additive neutral element is $o=(0,0)$. It's trivial to show why thse two elements are the respective neutral elements.
Now, on additive inverses. For the arbitrary element $a\in S$, we can take $-a=(-a_1, -a_2)$, therefore:
$$a+(-a)=(a_1+(-a_1), a_2+(-a_2))=(0,0)=o$$
This proves there are additive inverses.
On multiplicative inverses, let's consider $b\neq o$, then if we take $b^{-1}=(\frac{1}{b_1}, \frac{1}{b_2})$, then:
$$b\times b^{-1}=\bigg(b_1\cdot\frac{1}{b_1}, b_2\cdot \frac{1}{b_2}\bigg)=(1,1)=u$$
which shows the existence of multiplicative inverses.
Finally, the distributive property:
Let $c=(c_1, c_2)$ an element of $S$. Then:
$a\times (b+c)=(a_1,a_2)\times (b_1+c_1, b_2+c_2)=(a_1b_1+a_1c_1, a_2b_2+a_2c_2)$
On the other hand,
$a\times b+ a\times c =(a_1b_1, a_2b_2)+(a_1c_1, a_2c_2)=(a_1b_1+a_1c_1, a_2b_2+a_2c_2)$. Which means that $a\times(b+c)=a\times b+a\times c$.
Did I make a mistake or fail to notice something? On the other hand, I'd love some help with the writing or style.
Best Answer
Hint: What's the multiplicative inverse of $(1, 0)$?