0
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.