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In the book on Linear Algebra that I am using, the author defines a line in an arbitrary vector space $V$, given direction $ 0 \neq d \in V $ and passing through $ p \in V$ as $ l(p;d)= \lbrace v\in V| \;\exists t \in \mathbb R,v=p+td\rbrace $.

He then proceeds to show that $ l(p;d)=l(q;d) $ iff $ (q-p) $ is a multiple of $d $. and further that any two distinct points determine a unique line. He later defines 2 lines $ l(p;d_1) , l(q;d_2)$ as parallel if $d_1= \alpha d_2$ for some $ \alpha \in \mathbb R$.

He finally proves Euclid's parallel postulate using these tools. My question is how does this setup still allow for a Non-Euclidean Geometry? Does it have to do anything with the fact that the definition of line here uses a "$ t \in \mathbb R$" which inhibits the validity of this definition? If so, how does one define a straight line using an arbitrary field? I hope I have been able to convey my thoughts precisely enough to warrant an answer.

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    @AndréNicolas Please consider expanding your comment into an answer, so that this question gets removed from the [unanswered tab](http://meta.math.stackexchange.com/q/3138). If you do so, it is helpful to post it to [this chat room](http://chat.stackexchange.com/rooms/9141) to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see [here](http://meta.stackexchange.com/q/143113), [here](http://meta.math.stackexchange.com/q/1148) or [here](http://meta.math.stackexchange.com/a/9868). – 2013-09-14

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