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Let d be a line, M a point on the line, and n a positive integer. Why is there exactly two points at n distance from M on d? How to prove it with Euclidian axioms (without algebra) ?

Jor
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    Wouldn't it simply follow from the fact that a circle and a non-tangent line passing through the circle intersect exactly twice? – M Turgeon Aug 23 '13 at 20:52
  • @Jor: How you prove it depends on exactly how you've axiomatized Euclidean geometry. There are many axiomatizations, which one are you working with? – Jim Aug 23 '13 at 21:01
  • @MTurgeon: Depends on whether that theorem has been proven yet... – Jim Aug 23 '13 at 21:02
  • @Jim : I am seeking a proof using Euclide or if necessary Hilbert axiomatization. – Jor Aug 23 '13 at 21:53
  • The fact that there cannot be three such points isn't really a statement you can prove with Euclids axioms, it would just be something that Euclid assumed was true. If you accept Hilberts axiomatization then your statement is a special case of Hilbert's congruence axiom 1. – Jim Aug 23 '13 at 23:07
  • Thanks, I didn't read all Hilbert axioms. – Jor Aug 24 '13 at 07:58

2 Answers2

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Consider a line (say $d$ is y=0) and a point ($M$=0), then we have two directions, say positive and negative. Then we have two points at a distance $n$ from M because there are only two directions. Suppose a third point, were at a distance $n$ on the line $d$ distinct from the two previous points. Then this point must not be in the direction of the other two point (or else it would be equal to one of them) or it is not on line $d$. In both cases, such a point contradicts its definition and hence, cannot exist.

jdwhitfield
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I think the main problem in applying Euclid's original axioms lies in combining the definition of "straight line" with that of distance (check this post).

I would use the 1:1 correspondence given in the first postulate of Birkhoff's axioms to directly find the two points corresponding to the two real numbers which are $n$ units away from the real number corresponding to $M$.

Community
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JRG
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