3-sat to 2-sat reduction
You can prove it by contradiction: 

Suppose that $P \neq NP$ and there is a polynomial-time reduction from 3-SAT to 2-SAT; then 2-SAT is NP-complete, but 2-SAT is also solvable in polynomial time, so
for all decision problems $A \in NP$ you can decide $x \in A$ reducing $x$ to the corresponding 2-SAT instance in polynomial time and solve it in polynomial time (and the total time is still polynomial); so $A \in P$ and hence $NP \subseteq P$. But we also have $P \subseteq NP$, so $P = NP$ which is a contradiction. :-)