Verifying properties of a discrete valuation. 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 $(1)$.
 A: 0) It is  to your credit  that you couldn't prove  iii) because it is false! Indeed
$ v((x^2)+(x-x^2))\geq \operatorname {min}(v(x^2),v(x-x^2)) \;$ is equivalent to  $1 \geq \operatorname {min}(2,2)$  
1) So $v$ is not a valuation. The correct definition of the valuation is
$$w(f/g)=\operatorname {deg} g-\operatorname {deg} f \; \text {for} \; f\neq 0 \quad (\text {and} \;w(0)=\infty)$$
Of course it can't hurt to realize that $w$ just computes the order of vanishing of  fractions at infinity . For example  $w( \frac {T^3-1}{T^7+T^2+1})=4$  
2) The proofs of i) and ii) are trivial and in iii)  you may reduce (by taking a common denominator) to the case of non zero fractions $\frac {f}{g}, \frac {F}{g}$ .
You then have to show that:    
$w( \frac {f+F}{g})= \operatorname {deg}(g)-\operatorname {deg} (f+F)\stackrel {?}{\geq} \operatorname {min}(w(\frac {f}{g}),(w(\frac {F}{g}))=\operatorname {min}(\operatorname {deg}g -\operatorname {deg}f,\operatorname {deg}g -\operatorname {deg}F)$ 
which boils down to the obvious $\operatorname {deg}(f+F)\leq \operatorname {max}(\operatorname {deg}(f),\operatorname {deg}(F)) $
