Axiom proof by contradiction

**slaypullingcat** · June 30th 2010, 03:00 AM

Hi there, I am new to proofs and I need to choose one Euclidean axiom and demonstrate that it holds true on the Cartesian plane. The proof must be by contradiction. Any help or links to examples would be greatly appreciated.

Thank you

**Bruno J.** · June 30th 2010, 03:20 AM

I'm not sure what you're asking for! The point of axioms is that you don't prove them.

**Zaph** · June 30th 2010, 03:29 AM

Hello.

For instance you could take a look on a line $l$ and a point $p$
where $p$ does not lay on the line. Then only one line can be drawn through $p$ parallel to $l$

Why is that? Cause any other lines through $p$ not parallel with $l$ will intersect with $l$ .
That shouldn't be too hard to see/proove..

**HallsofIvy** · June 30th 2010, 05:20 AM

Originally Posted by Bruno J.

I'm not sure what you're asking for! The point of axioms is that you don't prove them.

You don't prove the axioms of Euclidean geometry in Euclidean geometry. You certainly can prove that the axioms of Euclidean geometry hold in the Cartesian plane, thus showing that the geometry of the Cartesian plane is Euclidean.

**Bruno J.** · June 30th 2010, 06:25 AM

Originally Posted by HallsofIvy

You don't prove the axioms of Euclidean geometry in Euclidean geometry. You certainly can prove that the axioms of Euclidean geometry hold in the Cartesian plane, thus showing that the geometry of the Cartesian plane is Euclidean.

Of course! Silly me.

**HallsofIvy** · June 30th 2010, 11:34 AM

Originally Posted by Zaph

Hello.

For instance you could take a look on a line $l$ and a point $p$
where $p$ does not lay on the line. Then only one line can be drawn through $p$ parallel to $l$

Why is that? Cause any other lines through $p$ not parallel with $l$ will intersect with $l$ .
That shouldn't be too hard to see/proove..

Yes, it shouldn't be. But remember that you have to use "Cartesian plane" properties.

I recommend that you write the given line as, say, ax+ by= 1 (every line in the Cartesian plane can be written that way- NOT every line can be written "y= ax+ b"), and take the point to be $(x_0, y_0)$ such that $ax_0+ by_0\ne 1$ . Now, suppose $cx+ dy= 1$ is a line through that point that does NOT intersect that line. What can you say about c and d? Well, first, what must be true of c and d so that $cx_0+ by_0= 1$ . Now try solving for the intersection of ax+ by= 0 and that line. What would prevent you from finding a solution?

If you wanted the easiest example, you might try showing that "given any two points, there exist exactly one line passing through both".

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