# piecewise and non-piecewise isomorphic?

• Apr 22nd 2008, 07:04 PM
wil
piecewise and non-piecewise isomorphic?
Ok, let me give the problem first:

Quote:

Consider the following interpretation of these terms: "points" are ordered pairs (x,y) in the xy-plane, "lines" are the functions

$
y = \begin{cases}
mx+b & x < 0 \\
2mx+b & x \ge 0
\end{cases}
$

And say a "point" "lies on" a "line" if the point (x,y) satisfies the equation. Show that this interpretation is a model for incidence geometry and determine which, if any, parallel property it has.
When visualizing the problem, I usually imagine the y-axis is the surface of a body of water, and the lines are light rays being refracted.

I'm already nearing a solution for the problem. However, there is one related thing I'd like to do for my own edification: prove that $
y = \begin{cases}
mx+b & x < 0 \\
2mx+b & x \ge 0
\end{cases}
$
is isomorphic with $
y = mx+b
$
Intuitively it seems true, but I don't have a clue as to how to go about proving it. Can anyone provide me with such a clue? (Thinking)

Notes:

1. If anyone wants to see my solution to the actual problem (once I finish it) let me know and I'll make an edit to this post.
2. The actual problem says "undefined terms" instead of "these terms" in the first sentence.
• Apr 22nd 2008, 07:30 PM
ThePerfectHacker
This is not incidence geometry. Consider $(0,1)$ and $(0,2)$. What is the "line" joining them?
• Apr 22nd 2008, 07:57 PM
wil
Quote:

Originally Posted by ThePerfectHacker
This is not incidence geometry. Consider $(0,1)$ and $(0,2)$. What is the "line" joining them?

Doh! (Doh) Thanks! I was starting to bog down in the middle of my proof. (Clapping)