Evaluating Axioms

Date: 07/09/2002 at 21:54:44
From: Leslie
Subject: Proving Axiom I-2 not valid

One of my homework problems for summer school states the following:

Consider the model: 

  Points {1,2,3,4,5}
  Lines  {1,2,3,4}, {1,5}, {2,5}, {3,5}, {4,5}
  Planes {1,4,5}, {2,4,5}, {3,4,5}

  axiom I-2: 3 non-collinear points determine a plane

  axiom I-3: if 2 points lie in a plane, then any line 
             containing the points lies in the plane.

Questions:

  Why is axiom I-2 not valid?
  Why is axiom I-3 not valid?


I am thinking that I-3 is not valid because planes {2,4,5} and 
{3,4,5} can not contain the lines {1,2,3,4} or {1,5}.

Am I correct?

But I just can't see how to show that axiom I-2 is not valid because 
it looks as if the 3 planes don't have 3 collinear points when drawn 
like this with lines extending down from 1,2,3, and 4 to point 5 to 
create lines {1,5}, {2,5}, {3,5}, {4,5}.  

  -------------------------
  1       2      3       4

                 
              5

Am I drawing it wrong and is that the problem?  Thank you for any 
help!

Date: 07/09/2002 at 22:57:52
From: Doctor Peterson
Subject: Re: Proving Axiom I-2 not valid

Hi, Leslie.

Ordinarily you don't say an axiom is not valid; axioms are assumed 
without proof. But here you mean that it does not agree with the 
model; you might better say the model does not fit the axioms. You 
need to find a single counterexample for each.

I don't like all of what you said about I-3, but you are partly 
correct. Looking at plane {2,4,5}, the axiom says that any line 
containing 2 and 4, or 4 and 5, or 2 and 5, lies entirely in that 
plane. This is true of lines {4,5} and {2,5}, but not of {1,2,3,4}, 
which contains 2 and 4. Line {1,5} is irrelevant, because it does not 
contain two points in the plane. But the proof is that if the axiom 
were true, then since 2 and 4 are in the plane {2,4,5}, {1,2,3,4} must 
be in that plane, and it is not.

As for Axiom I-2, you want to find three points that are not in the 
same line, but are not in exactly one plane. The only sets of non-
collinear points you have are 5 together with more than one of 
{1,2,3,4}. 

Choose such a set, say {1,4,5}. Are these in a plane? Only 
if this is true for ALL such sets of non-collinear points can you 
say that the axiom is true.

Your reasoning is backwards. It's true that all the planes consist 
of non-collinear points; but the axioms don't require that. What 
they require is that ANY set of non-collinear points must DETERMINE 
a plane. The logic in this kind of problem can be tricky, and you 
have to be careful to see exactly what each axiom says.

You can't really draw a problem like this, because it is not a model 
of Euclidean geometry. The points may LOOK as if they determine a 
plane (and in fact they will, because these axioms are true of any 
geometry you can draw), but are they DEFINED to be a plane? Only 
the definitions can tell you.

What you can do is to draw ALL the lines, and ALL the planes, and 
keep in mind that only those count, not what looks like a line or 
plane. You did this with the lines:

    1-----2-----3-----4
       \   \   /   /
          \ \ / /
             5

You might draw a curve around each set of points that forms a plane. 
As long as you are careful, this kind of visualization can help you 
see what is happening. But you have to go back to the definition in 
order to be sure of anything.

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/