I want my proof writing skills to get better. I am trying to do this through proving theorems from Hilbert's axioms for Euclidean Geometry. I found Hilbert's The Foundations of Geometry here, a wonderful resource. After I got through the axioms of connection, a couple of theorems (pg. 3) were presented that could be proved from those seven axioms.

I tried proving the statement: "Two straight lines of a plane have either one point or no point in common."

This is the axiom I use in my proof: "1. Two distinct points A and B always completely determine a straight line a. We write AB = a or BA = a."

What follows is my attempt at a proof (by contradiction).

Let $a, b$ be distinct straight lines of a plane $\alpha$. We assume (towards a contradiction) that $a,b$ have the points $A_1, \dots, A_n$ in common with $n\ge 2$. That is, the points $A_1, \dots, A_n$ lie on both lines. Without loss of generality, let us pick two of these points, say, $A_1$ and $A_2$. Then, by axiom 1, we know that $a=A_1A_2$ and $b=A_1A_2$, i.e. $a=b$. However, this contradicts the distinctness of $a$ and $b$. Therefore, the lines must have either one point or no point in common. Q.E.D.

Tips on the writing and correctness of the proof would be appreciated. Also, I was reluctant to use a proof by contradiction because at first it seemed like one could prove it directly, but I'm not sure. This is not that interesting of a theorem or anything, but if you have any other ways to prove it please share. Thank you for your time!


You should not say, "assume ... that $a$, $b$ have the points $A_1, A_2, \ldots, A_n$ in common", because the notation "$A_1, A_2, \ldots, A_n$" suggests a finite list. Based on the assumption for your proof by contradiction, you only know that $a$ and $b$ have at least two (which includes the possibility of infinitely many!) points in common. So pick 2 of those points and call them $A_1$ and $A_2$. Now the rest of your proof goes through.

  • $\begingroup$ Thank you for that! Not only does your answer make the proof correct, but it lets me get rid of some silly notation. Very helpful. $\endgroup$ – Saudman97 Jan 21 '16 at 8:22

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