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I've started reading Introduction to Algebra by Cameron, and I'm stuck on the first exercise.

Q. Prove any line passes through at least two points using the axioms given below.

Definitions:

Collinear means lying on a common line

Lines are parrallel if no point lies on both

Three axioms:

  1. Any two points lie on a unique line

  2. If the point P does not lie on line L there is exactly one line, L', passing through P and parallel to L.

  3. There exist three non-collinear points.

My thinking: Any two points lie on a unique line, but the converse is not necessarily true. Given the axioms provided, could a line equal a point?

Line 1 passes through just one point A (is point A).

Line 2 does not pass through 'A' and joins two points (B and C) (axiom 3 and 2). Since it doesn't share any points with Line 1, Line 2 and 1 are parrallel to each other.

My thinking is that it might be possible to prove that there can be another line which is parallel to 2 which passes through point A, or that point A must also be parallel to another line which passes through either B and C, but without assuming a fourth point, or something additional about the nature of lines, I'm a bit stuck on how to do it!

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Suppose by contradiction there exists a line $a$ passing through a single point $A$. By axiom 3 there exist two other points $B$ and $C$ and by axiom 1 they belong to a line $r$ parallel to $a$.

By axiom 1 there exists a line $s$ passing through $A$ and $B$ and by axiom 2 there exists a line $t$ passing through $C$ and parallel to $s$.

Line $t$ does not pass through $A$, so $t$ is parallel to $a$. But then we have two different lines $r$ and $t$ passing through $C$ and parallel to $a$, which contradicts axiom 2.

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