# Modern Geometry Proof

• May 14th 2011, 01:31 AM
jzellt
Modern Geometry Proof
Show that every line has an in nite number of points.

Hint: Start with A1 and A2 distinct points on the line. How do you get these two points?
Next describe how, once you have produced points up to An, you can find a new point An+1.
In order to guarantee that An+1 is not equal to any previous points, you will need to produce
An+1 carefully. For example, can you find an An+1 so that it is not in A1 * An * An+1? Can
you prove (by induction on n) that every Ai with i < n + 1 is in A1An+1, but not equal to
An+1. Finally, can you use this result to show that Ai != Aj if i != j?

Can someone show this? I havn't a clue on this one... Thanks.
• May 15th 2011, 03:40 AM
HappyJoe
Hm, when working with the line, am I allowed to add points to obtain another point? Or am I to consider the line as a purely geometrical object without any operations?
• May 15th 2011, 04:43 AM
Plato
Quote:

Originally Posted by jzellt
Show that every line has an infinite number of points.

How one does this proof depends on the set of definitions and axioms in play.
We don’t know the exact statements of the one you use.
That said, most follow some variant of Hilbert’s axioms.
INCIDENCE AXIOM II: For every line l there exist at least two distinct points on l.

BETWEENNESS AXIOM II: Given two distinct points A & B then there is a point C such that A*C*B. i.e. A, C & B are collinear and C is between A & B.

With only those two axioms we can prove this theorem.
If j is a line the there are two points on j, A & B.
Define $P_1=A~\&~P_2=B$.
The second axiom gives us $P_3$ such that $P_1*P_3*P_2$.

If $n\ge 4$ we are assured of a point such that $P_{n-2}*P_n* P_{n-1}$.