union/intersection

• March 7th 2011, 11:04 PM
jzellt
union/intersection
Show that if A and B are distinct points, then the intersection of (ray(AB) and ray(BA)) = lineSegment(AB).

Show that if A and B are distinct points, then the union of (ray(AB) and ray(BA)) = the whole line (AB).

Can someone show these proofs. Thanks
• March 8th 2011, 01:55 AM
HallsofIvy
what are the definitions of "ray" "segment", "line" as well as "intersection" and "union"?
• March 8th 2011, 02:20 AM
emakarov
I see only one reason for proving statements that are generally clear to preschool children, and that is to justify the statements in a very specific way, according to carefully chosen definitions and laws of reasoning. This is what a quote by Bertrand Russell from this thread talks about.
Quote:

The most obvious and easy things in mathematics are not those that come logically at the beginning; they are things that, from the point of view of logical deduction, come somewhere the in middle. Just as the easiest bodies to see are those that are neither very near nor very far, neither very small nor very great, so the easiest conceptions to grasp are those that are neither very complex nor very simple (using "simple" in a logical sense). And as we need two sorts of instruments, the telescope and the microscope, for the enlargement of our visual powers, so we need two sorts of instruments for the enlargement of our logical powers, one to take us forward to the higher mathematics, the other to take us backward to the logical foundations of the things that we are inclined to take for granted in mathematics.
So, to bring people on the same page, you need to describe what type of microscope and other instruments you are using.
• March 8th 2011, 05:20 AM
TheChaz
Mostly out of curiosity, I've tried to formalize the notion of "ray":

Quote:

Given distinct points A, B, we denote ray(AB) by
$\underset{AB}{\rightarrow}$
$:=\begin{Bmatrix}
A + T(B - A): 0 \leq T
\end{Bmatrix}$

Similarly, we could arrive at segment AB by restricting $T \leq 1$ and a line would result from T ranging across the reals.

Not that these will necessarily have anything to do with the OP's definitions... just wondering what you guys think about these proposed definitions!
• March 8th 2011, 06:18 AM
Plato
Quote:

Originally Posted by TheChaz
Mostly out of curiosity, I've tried to formalize the notion of "ray":

Actually in axiomatic geometry (synthetic) we begin with a line segment $\overline{AB}=\{X:X=A,~X=B\text{ or }A-X-B\}$.
Then proceed to define a ray $\overrightarrow {AB} = \overline{AB}\cup\{X:A-B-X\}$.
That why Prof. Halls asked for definitions. If they are somewhat standard, as the one I gave then this a trivial question.
• March 8th 2011, 06:51 AM
TheChaz
I am certainly in agreement that some semantic standards be met before proceeding.
In your example, what is "A - X - B"? If it reads "X is between A and B", then how to we define betweenness?
• March 8th 2011, 07:06 AM
Plato
Quote:

Originally Posted by TheChaz
I am certainly in agreement that some semantic standards be met before proceeding. In your example, what is "A - X - B"? If it reads "X is between A and B", then how to we define betweenness?

Betweenness is not defined.
In the Hilbert/Moore axiom system the undefined terms are: point, line, lie on, between,& congruent..
There are no numbers as such in synthetic geometry.

Now Ed Moise was one of Moore's PhD students. In his book he does introduce a Ruler Postulate by which he defines between in terms of coordinates. I found that this works best with undergraduates , participially mathed majors.
• March 8th 2011, 01:00 PM
jzellt
Plato is correct about the definiton of ray that we are using. And betweenness has been defined. So, can anyone show the proof?
• March 8th 2011, 01:23 PM
Plato
These two are very straightforward.
Look at the definitions.
$\overrightarrow {AB} = \overline {AB} \cup \left\{ {X:A - B - X} \right\}$ and and $\overrightarrow {BA} = \overline {BA} \cup \left\{ {X:B -A- X} \right\}$

You know that $\overline{AB}=\overline{BA}$.
If $X-A-B$ then $X\notin\overrightarrow {AB}$.

Now you do something with $\overrightarrow {AB}\cup\overrightarrow {BA}$ and $\overrightarrow {AB}\cap\overrightarrow {BA}$.
Show us what you do.