transitivity property of parallel lines proof Statement: Prove, under the assumption of the parallel postulate (P-1),  parallelism of lines is transitive. That is if l||m and m||q, then l||q.
Parallel Postulate(p-1)-If l is any line and point P not on l there exists an unique line passing through P parallel to l( in the plane of  P,l).
Proof- Assume to the contrary that l is not parallel to q. Further assume the parallel postulate p-1. Sine l is not parallel to q that means both lines meet at least 1 point.But that's a contradiction since it contradicts parallel postulate p-1.
Is that correct? or do i need to explain it a bit more why it contradicts?
 A: Hypothesis: $ \ell \parallel m $ and $ m \parallel q $.


*

*Suppose that $ \ell = q $. By convention, $ \ell \parallel q $.

*Suppose that $ \ell \neq q $. By way of contradiction, assume that $ \ell \not\parallel q $. Then $ \ell $ and $ q $ intersect in at least one point $ x $, which implies that $ \ell $ and $ q $ are distinct lines parallel to $ m $ passing through $ x $. We thus contradict the Parallel Postulate that there exists only one line parallel to $ m $ passing through $ x $. Our assumption that $ \ell \not\parallel q $ is therefore false, so we conclude that $ \ell \parallel q $.

Note: In order to derive a contradiction, you need to explicitly assume that $ \ell \neq q $.
A: Having graded for a geometry class before I'll tell you that I would take a few points off for that answer.
Were I your grader I would like you to say explicitly why $P$ doesn't lie on $m$, that the parallel postulate tells you there is exactly one line through $P$ parallel to $m$, and emphasize that $l$ and $q$ are distinct lines both parallel to $m$ (and what to do if they are not distinct!).
The idea behind your proof is completely correct, all you need now is to be extra careful as you explain it.
