If algebraic $a$ has degree $n$, so does $-a$ I feel like the best way to move forward is to use a contradiction proof.
Since $a$ is algebraic, and is of degree $n$, it has a minimal polynomial of degree $n$, so we can write
$$f(a)=\sum_{k=0}^n{c_ka^k}=0$$
Now suppose $-a$ has degree less than $n$, say $n-1$.  Then there exists a unique polynomial $g(x)$ such that 
$$g(-a)=\sum_{k=0}^{n-1}{d_k(-a)^k}=0$$
Adding a nonzero term to both sides,
$$g(-a)+d_n(-a)^n=\sum_{k=0}^{n}{d_k(-a)^k}=\sum_{k=0}^{n}{d_k(-1)^ka^k}$$
Letting $c_k=d_k(-1)^k$
$$g(-a)+d_n(-a)^n=\sum_{k=0}^{n}{c_ka^k}=f(a)=0$$
But $g(-a)=0$.  Thus
$$g(-a)+d_n(-a)^n=f(a) \Rightarrow d_n(-a)^n=0$$
However, $d_n(-a)^n\neq 0$ by construction which leads to a contradiction.  Therefore, $-a$ has degree greater than or equal to $n$.  Using the same style argument for a new polynomial $h(x)$ we arrive at a similar contradiction and therefore, $-a$ has degree $n$ also.
I also need to show this for algebraic numbers $a^{-1}$ and $a-1$.  Is this the method best used?  And is the proof correct (outside the handwaving degree $h > n$)?
 A: It's much simpler to let $p(x)$ be the minimal polynomial for $-a$. Then $p(-x)$ is a rational polynomial satisfied by $a$, hence must have degree at least $\deg a$. Conversely $q(x)$, the minimal polynomial for $a$ has that $q(-x)$ is a polynomial satisfied by $-a$, hence
$$\deg(-a)\le\deg a\le\deg(-a)$$
so equality is everywhere.
A: A simpler way is to convert this to field extension.
Let $F$ be the field. Since $-a,a^{-1}\in F(a)$ we have $F(-a),F(a^{-1})\subset F(a)$ so $[F(-a):F],[F(a^{-1}):F]\leq [F(a):F]$. Hence degree of $-a,a^{-1}$ each are no more than that of $a$.
How about equality? Notice that the above argument also work with $a$ being $-a$ instead, or $a^{-1}$ instead. Hence you have the opposite inequality.
Note: also work for $a-1$ too, very much for the same reason. In fact, work also for $2a,a+\frac{1}{2}$ or any other such combination.
A: Hint: For $a-1$, let $p(x)$ be a minimal polynomial for $a$. Argue that $p(x+1)$ is a minimal polynomial for $a-1$.
For $a^{-1}$, let $p(x)=a_0x^n+a_1x^{n-1}+\cdots +a_n$ be a minimal polynomial for $a$. Argue that $a_nx^n+a_{n-1}x^{n-1}+\cdots +a_0$ is a minimal polynomial for $a^{-1}$. Just divide through by $x^n$.
