Let u and v be two verticies in the tournament T. Let u and v be two verticies in the tournament T. Prove that if U and V do not lie  on a  common cycle then the out degree  of u does not equal the outdegree of v.
So I will try to do the contra positive. So assume that the out degree of u equal the out degree of v.
The we assume (u,v) is an arc in the tournament T. The since $deg^+(v)=deg^+(u)$ there must also be be a (v,u) arc in the tournament. Hence u,v lie on a common cycle.
The only thing which confuses me is that in a tournament you can only have (u,v) or (v,u) as an but not both.
 A: 
since $\deg^+(v)=\deg^+(u)$ there must also be be a $(v,u)$ arc in the tournament.

Why is that so? What theorem/argument are you using? Just having $\deg^+(v)=\deg^+(u)$ doesn't seem to imply that $(v,u)$ is also an arc in the tournament (in fact this isn't the case precisely because of your "confusion", i.e., in a tournament, you cannot have both $u\rightarrow v$ and $v\rightarrow u$)

Here's one approach to prove it:
Let the tournament $T$ be on $n$ vertices.
WLOG say $u\rightarrow v$ and assume that $u,v$ have equal outdegrees.
Now, define $I(u):=\{x\in T\mid x\rightarrow u\}$ and $O(v):=\{x\in T\mid v\rightarrow x\}$. We also define $O(u)$ and $I(v)$ analogously. Note that $\deg^+(u)=|O(u)|$ (analogously for $v$)
Can you show that $I(u)\cap O(v)$ is non-empty? (if it is non-empty, say $w\in I(u)\cap O(v)$, then $v\rightarrow w\rightarrow u\rightarrow v$ forms a directed cycle)
Hint: Suppose it is empty and obtain a contradiction.
Hint #2: Every vertex other than $u$ must either be in $I(u)$ or $O(u)$ (analogously for $v$) since $T$ is a tournament.
A: As the previous answer by Prasun Biswas suggests , consider w.l.o.g, that you have $u\rightarrow v$(which means $u$ beat $v$ in the tournament). Consider all the people beaten by $v$, which number $deg^{+}(v)$. If even one of them, say $w$, beats $u$, then we get the cycle $u\rightarrow v\rightarrow w \rightarrow u$, and we are done. Otherwise, if all of these people beaten by $v$ are also beaten by $u$, then $deg^{+}(u)\geq deg^{+}(v)+1$, since $u$ had also beaten $v$, which gives a contradiction.
