EDIT: Don't think about this. The problem statement is flawed. (See comments)
Let $K$ be a field, and let $K(T)$ be the quotient field of polynomials over $K$.
Then I define $v(f/g) = \deg(f) - \deg(g)$ if $f/g\neq 0$ and $v(0) = \infty$. I need to show that this is a discrete valuation.
The properties needed:
(i) $v(f/g) = \infty$ if and only if $f/g = 0$.
(ii) $v\Bigl((f/g)\cdot (p/q)\Bigr) = v(f/g) + f(p/q)$
(iii) $v\Bigl((f/g) + (p/q)\Bigr) \geq \min(v(f/g),v(p/q))$
The first two properties were easy, but I run into the following difficulty with the 3rd. Assume wlog that $v(f/g)\leq v(p/q)$.
I have $v\Bigl((f/g) + (p/q)\Bigr) = v\Bigl((fq + pg)/gq\Bigr) = \deg(fq + pg) - \deg(gq)$.
I somehow need to get $\deg(fq + pg) \geq \deg(fq)$, and then I can finish the chain to get what I want.
I feel like the following argument "almost" works.
Since $\deg(f/g)\leq \deg(p/q)$, then I think that $\deg(fq)\leq \deg(pg)$.
If the latter inequality were strict, then I would KNOW that (1) $\deg(fq)\leq \deg(fq + pg)$.
But if equality holds, I don't know how to deal with something like the following possibility.
$f(x)q(x) = x^2 - 1$, $p(x)g(x) = -x^2$, then this violates $(1)$.