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

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 Neighbor Posts: 57 From: It doesn't matter Registered: 1/26/14
Re: A question about straight lines
Posted: Jan 28, 2014 3:38 PM

> All sets of points have boundaries. The question is
> whether the boundary is part of the set of points or
> not.
>
> Think about a number line.
> Now think about the set of all numbers greater than
> 4. (all the points to the right of 4 on the number
> line)
> Now tell me what is the smallest number in that set?
> (the point closest to 4 on the number line)
> You can?t answer the question. Even though the set of
> all numbers greater than 4 has a lower boundary,
> which is 4, it isn?t in the set itself.

If you consider only natural numbers, then next to four
there will be 5. But my question is not about numbers and their proprieties, but about straight-lines in Euclid's geometry with points on them.

> You defined such a set of points by saying that G is
> part of AG and that BG is all the points left over.
> BG has a boundary (G) but the boundary itself isn?t
> part of BG (by your choice). Thus, BG is an open
> segment at G. At B it is a closed segment because B
> is part of BG.

I will show you in the following example what I can't really understand about division of a straight-line:
Let's have a finite straight-line AB, and we want to cut it in half. If I'm not mistaken we do so by cutting AB with a point on it(let the point be M) into two equal straight-lines AM and MB (AM coming before M, and MB coming after). Now suppose that I want to add AM to a straight-line, and BM to another, therefore I need to separate AB into AM and MB, and being M one, it will be either on AM, MB, or neither. If we leave M and take A(M) and (M)B without it, then the two parts are not really halves of AB, for the sum of the two halves in which one thing is divided is the thing itself, but A(M)+(M)B will be less than AB, because AB = A(M)+M+(MB). Now if we take M on either AM or MB, let it be taken on AM, then when we separate AB we will have straight-lines AM and (M)B, but they will be unequal, inasmuch as A(M)=(M)B ==> AM>(M)B.
The inference seems to be that is not a point to divide a straight-line in half but something in between, I mean to say that when we want to divide AB in half, we should divide it in AM and NB, with M and N two consecutive points, and this won't cause troubles in the separation. Therefore what cuts AB in half is something between the two consecutive points M and N. But how two points can be consecutive if between them, as you say, there is another point ever.
And this is the very thing concerning division in half of a straight-line that I cannot understand.

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