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.


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!


1 Answer 1


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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