Is non-paralellism transitive?

Question

I've been trying to check if non-paralellism is transitive. At the moment, I know it's symmetric. But I have no idea on how to prove that it's transitive. I did the following:

$$(a \not\parallel b) \wedge (b\not\parallel c)$$

And then, if $a\not\parallel b$, then $a\sim b$ and then:

$$((a \not\parallel b) \not\parallel c)$$

As $a\sim b$ and $b\sim c$, then:

$$(a \not\parallel b\not\parallel c)$$

I'm not sure if this proves what I want. I'm also not sure if it's needed to be proved (perhaps we just assume that it needs to be?). I'll try to prove it with means of analytic geometry because it sounds easier.

$a$ is not parallel to $b$, so $b$ is not parallel to $a$ but... — lulu, Aug 31 '15 at 00:48

score 4 · Accepted Answer · answered Aug 31 '15 at 01:10

4

no, see the diagram: $$ \not\parallel $$

the first vertical line is not parallel to the diagonal line. The diagonal line is not parallel to the second vertical line. However..........

answered Aug 31 '15 at 01:10

Will Jagy

139,541

nice use of the symbol as a diagram, +1 :) – Alan Aug 31 '15 at 01:14
Oh yes. Sorry, I'm stupid. – Red Banana Aug 31 '15 at 02:18

Is non-paralellism transitive?

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