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Topic: A question about straight lines
Replies: 103   Last Post: Feb 8, 2014 8:28 AM

 Messages: [ Previous | Next ]
 Joe Niederberger Posts: 4,657 Registered: 10/12/08
Re: A question about straight lines
Posted: Feb 1, 2014 9:28 AM

R Hansen says:
>There is no way to teach an elementary student naive set theory and it is just silliness to try. You will understand what I mean when you try to answer my specific question. What insight into points, lines and infinity, at all, would a student get from glue and plugs? I suspect everything you offer will be from other exercises and not that analogy. You will realize that such an approach is nothing but an illusion.
>You can only get all of those details through experience, not a perfect explanation.

Naive set theory is exactly what is needed to cut through the fog here. (Plugs are just a bit of a silly diversion I'll agree with you there. And I meant to say earlier I was *not* creating a lesson plan.) So let's get on with the naive set theory and see if it helps. Explanations, coupled with our understandings of basic concepts like sets, unions, intersections, and combined with following the below, step by step, plus pondering, and asking questions to clarify things that aren?t clear, and working more problems, etc. etc. is what you mean by ?experience?. I agree.

I'm talking now to Neighbor, he'll have to read up on sets, subsets, unions, intersections, etc. to follow along. I may be asking far too much of him below. He might try reading the Wikipedia page on naïve set theory, and never make it to the exercise below. Oh well, he?s not in my class!

What?s been at issue here is what happens to the cut point, when the original line is ?to be taken separately?, as two.

1. We are going to view a line segment now as a set of points; ?made up of points?, if you prefer.

2. Lets pick one point though (our cut point.) We can generate a property that applies to all other points by reference to that one.
Grant me that all other points lie on one side, or the other side, of our reference cut point.

3. We will express this mathematically as follows:
Let C be our cut point, our reference point. If A is any point on the line whatsoever, either:
(1) A = C (they are the same point in fact.)
(2) A is on one side of C, which I will write as A < C
(3) A is on the other side of C, and I?ll notate this as A > C

4. Now I'm going to re-interpret your question about bisecting a line in set terms: you want:
a. to create two subsets L2 and L3, of L,
b. such that L1 and L2 still qualify as lines, and
c. L1 union L2 = L, and
d. L1 intersect L2 = {} (empty set).

5. Here are some subsets of L, I?m going to expect you to ?translate? from what I write into a more or less English form. I?ll do the first one and third one:
[1] { x in L | x > C } Translation: this is the subset of L that consists of all points, x, in L, such that x > C. Its the line segment you get by ?taking? all points that lie on one side of C.
[2] { x in L | x < C }
[3] { x in L | x > C *or* x = C } This is like [1], except I also include the point C itself in this subset.
[4] { x in L | x < C or x = C }

I?ll claim without proof that all 4 of these are still line segments of some variety.

6. Think about those 4 subsets listed above, and what they mean *as sets of points*!

7. Pick 2 of those that meet the conditions a, c, d, listed in bullet 4. We?ll forget b for later, and just assume they all are line segments in fact.

Can you do it?

Cheers,
Joe N

Date Subject Author
1/26/14 Neighbor
1/26/14 kirby urner
1/27/14 Neighbor
1/27/14 Robert Hansen
1/27/14 Bishop, Wayne
1/27/14 kirby urner
1/27/14 Neighbor
1/27/14 Robert Hansen
1/27/14 Neighbor
1/27/14 Neighbor
1/28/14 Neighbor
1/28/14 Robert Hansen
1/28/14 Joe Niederberger
1/28/14 Robert Hansen
1/28/14 Neighbor
1/28/14 Robert Hansen
1/28/14 Joe Niederberger
1/28/14 Joe Niederberger
1/28/14 Neighbor
1/28/14 Robert Hansen
1/28/14 Joe Niederberger
1/28/14 Joe Niederberger
1/28/14 Robert Hansen
1/28/14 Neighbor
1/28/14 Gary Tupper
1/28/14 Robert Hansen
1/28/14 Joe Niederberger
1/28/14 Joe Niederberger
1/28/14 Louis Talman
1/28/14 Joe Niederberger
1/28/14 Joe Niederberger
1/28/14 Joe Niederberger
1/28/14 Joe Niederberger
1/29/14 Robert Hansen
1/29/14 Bishop, Wayne
1/29/14 GS Chandy
1/29/14 Domenico Rosa
1/29/14 Neighbor
1/29/14 Joe Niederberger
1/29/14 Neighbor
1/29/14 Domenico Rosa
1/30/14 Bishop, Wayne
1/29/14 Joe Niederberger
1/29/14 Neighbor
1/29/14 Robert Hansen
1/29/14 Joe Niederberger
1/29/14 Joe Niederberger
1/29/14 Robert Hansen
1/30/14 Neighbor
1/29/14 Robert Hansen
1/29/14 Joe Niederberger
1/30/14 Louis Talman
1/29/14 Joe Niederberger
1/30/14 Neighbor
1/30/14 Jonathan J. Crabtree
1/30/14 Joe Niederberger
1/30/14 Joe Niederberger
1/31/14 kirby urner
1/30/14 Joe Niederberger
1/31/14 Neighbor
1/30/14 Joe Niederberger
1/31/14 Neighbor
1/31/14 Robert Hansen
1/31/14 Joe Niederberger
1/31/14 Joe Niederberger
1/31/14 Joe Niederberger
1/31/14 Joe Niederberger
1/31/14 Robert Hansen
1/31/14 Joe Niederberger
1/31/14 Neighbor
1/31/14 Robert Hansen
1/31/14 Joe Niederberger
1/31/14 Joe Niederberger
1/31/14 Robert Hansen
1/31/14 Neighbor
1/31/14 Robert Hansen
1/31/14 GS Chandy
1/31/14 GS Chandy
1/31/14 GS Chandy
1/31/14 Joe Niederberger
2/1/14 GS Chandy
2/1/14 Robert Hansen
2/1/14 kirby urner
2/1/14 Robert Hansen
2/1/14 Joe Niederberger
2/1/14 Robert Hansen
2/1/14 Joe Niederberger
2/2/14 Robert Hansen
2/2/14 kirby urner
2/1/14 Joe Niederberger
2/2/14 Robert Hansen
2/2/14 James Elander
2/2/14 GS Chandy
2/2/14 Robert Hansen
2/2/14 GS Chandy
2/2/14 Robert Hansen
2/2/14 Joe Niederberger
2/2/14 GS Chandy
2/2/14 Robert Hansen
2/2/14 GS Chandy
2/4/14 GS Chandy
2/4/14 GS Chandy
2/4/14 GS Chandy
2/8/14 GS Chandy