# 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?

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You should explain why it's a contradiction. –  Jim Feb 22 '13 at 16:42
I would explain more, maybe by taking $P$ to be the intersection of $l$ and $q$ and chasing down what the parallel postulate gives you. –  Louis Feb 22 '13 at 16:44

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 $\ell \neq q$.

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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.

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