Using the field axioms, prove that $x^{-1} = y^{-1}$ if $x = y$ Here's the original text of the question:

Let F be an ordered field, and let $x, y \in F$ s.t. $x \neq 0$ and $y \neq 0$. Using field and order axioms (and the properties proved in class or on your homework) prove that if $x = y$, then $x^{-1} = y^{-1}$.

I think my answer should look like following, but I definitely do not trust myself.
$x = y$
$x\otimes y^{-1} = y\otimes y^{-1}$ by commutativity of multiplication, which simplifies to
$x\otimes y^{-1} = 1$ by multiplicative inverse
$x\otimes y^{-1}\otimes x^{-1} = 1\otimes x^{-1}$ by commutativity of multiplication, which simplifies to
$1\otimes y^{-1} = 1\otimes x^{-1}$ by multiplicative inverse
$y^{-1} = x^{-1}$ by multiplicative identity
Is this answer correct? If not, how do I answer this question?
 A: The two times you quote the "commutativity of multiplication" are strange, at least to me.
For instance, the first time:

$x\otimes y^{-1} = y\otimes y^{-1}$ by commutativity of multiplication

is actually just a property of equality, not a property of multiplication (and certainly not commutativity).
The second time:

$x\otimes y^{-1}\otimes x^{-1} = 1\otimes x^{-1}$ by commutativity of multiplication

is a bit more problematic.  Once again, you haven't used commutativity of multiplication, but in this case, you're implicitly using associativity of multiplication to even make the left-side make sense.  At this early stage of detail, that's likely a big no-no.
For the second one, I'd advise instead to write
$$x^{-1} \otimes (x\otimes y^{-1}) = x^{-1} \otimes 1$$
again, just as a property of equality.  Then explicitly use associativity and inverses to simplify the left-hand side.
A: Sounds good to me, but you can improve the last part:
\begin{align*}
xy^{-1} = 1 & \Rightarrow (xy^{-1})x^{-1} = x^{-1}\\\\
& \Rightarrow x(y^{-1}x^{-1}) = x^{-1}\\\\
& \Rightarrow x(x^{-1}y^{-1}) = x^{-1}\\\\
& \Rightarrow (xx^{-1})y^{-1} = x^{-1}\\\\
& \Rightarrow 1y^{-1} = x^{-1}\\\\
& \Rightarrow y^{-1} = x^{-1}
\end{align*}
Hopefully this helps!
A: Alternative approach:
In Field Theory, if I am not mistaken:

*

*Inverses are unique.

So, $0 \neq x=y \implies xy^{-1} = yy^{-1} = e.$
This implies that $y^{-1} = x^{-1}$, because, by definition, $x^{-1}$ is the unique inverse to $x$.
