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

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- Jun 30th 2010, 03:00 AMslaypullingcatAxiom proof by contradiction
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 - Jun 30th 2010, 03:20 AMBruno J.
I'm not sure what you're asking for! The point of axioms is that you don't prove them.

- Jun 30th 2010, 03:29 AMZaph
Hello.

For instance you could take a look on a line $\displaystyle l$ and a point $\displaystyle p$

where $\displaystyle p$ does not lay on the line. Then only one line can be drawn through $\displaystyle p$ parallel to $\displaystyle l$

Why is that? Cause any other lines through $\displaystyle p$ not parallel with $\displaystyle l$ will intersect with $\displaystyle l$.

That shouldn't be too hard to see/proove.. - Jun 30th 2010, 05:20 AMHallsofIvy
- Jun 30th 2010, 06:25 AMBruno J.
- Jun 30th 2010, 11:34 AMHallsofIvy
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 $\displaystyle (x_0, y_0)$ such that $\displaystyle ax_0+ by_0\ne 1$. Now, suppose $\displaystyle cx+ dy= 1$ 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 $\displaystyle cx_0+ by_0= 1$. 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".