TITLE: Prove no field exists if $b+d\neq 0$
QUESTION [5 upvotes]: I need to prove no field $K$ exists (with minimum 3 elements) which holds:
$\forall a,c\in K, b,d\in K\setminus\{0\}: b+d\neq 0 \implies\frac a b + \frac c d = \frac{a+c}{b+d}$  
I know this can't be true, but I don't know where to start to prove it. Probably my problem is with proving things false in general.
Thanks in advance.

REPLY [4 votes]: Suppose for contradiction that the statement held.
If there are at least three elements in the field, then we can find $b\ne 0$, $d\ne 0$, and $b+d\ne 0$, since we could take for example $b=1$, $d\ne 0,-1$. Hence there are some $b,d\ne 0$ with $b+d\ne 0$ such that for all $a$ and $c$ in the field $\frac{a}{b}+\frac{c}{d}=\frac{a+c}{b+d}$.
Now multiply by $(b+d)bd$, to get $ad(b+d)+cb(b+d)=(a+c)bd = adb+ad^2+cb^2+cbd=abd+cbd$. Cancel $abd$ and $cbd$ to get $ad^2+cb^2=0$, and this is supposed to be true for all $a$ and all $c$. In particular, if $a=0$ and $c\ne 0$, then we have $cb^2=0$, but $c\ne 0$ and $b\ne 0$, so $cb^2\ne 0$. Contradiction.