I have a series of exercises to do, but one of these seems to be really easy to do, that it even seems strange to be asked to do.
The exercise is:
Use only the field axioms to show that:
$$(x + y)(x - y) = (x \cdot x) - (y \cdot y)$$
for any $x$, $y$ $\in \mathbb{Q}$.
Well, without using the field axioms, we can immediately say that the statement it's true.
Can we simply use the distributivity field axiom:
$$x\cdot (y + z) = (x\cdot y) + (x \cdot z)$$
to demonstrate the equality, or maybe I am jumping important passages and I am not showing all the steps to achieve the result?
I did it like this:
Replace $(x + y)$ with a $k$, since addition is closed for $\mathbb{Q}$: $$k(x - y)$$
Use distributivity field with $k$ $$k \cdot x - k \cdot y$$
Replace $k$ with the original value $(x + y)$
$$(x + y) \cdot x - ((x + y) \cdot y)$$
Use distributivity field for the first and second subtrahend (can we say subtrahend?)
$$(x \cdot x + y \cdot x) - (x \cdot y + y \cdot y) = x \cdot x + y \cdot x - x \cdot y - y \cdot y = x \cdot x - y \cdot y$$
If this is not correct, could you tell me why?
Your work is good, but note that in the last step you use the property: $ y\cdot x =x \cdot y$ to eliminate two terms. So to find the result you need commutativity of the product, that is a typical property requested for a field. In a non commutative ring the identity $(x+y)(x-y)=x^2-y^2$ is not valid.