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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: Jan 29, 2014 12:47 PM

Neighbor says:
>Let's make the issue clearer: When we want to divide something into two exactly equal parts, in the process of division we, of course, don't lose parts of the thing we are dividing, but we divide the whole in two exactly equal parts. For instance, if we have 6 candies, and we want to divide them in half, we take three on one side and three on the other, not losing candies in the process of division. Now when we divide a straight-line into two equal ones, in the process of division we lose a part of that straight-line (i.e. the point through which we divide it), that is to say, initially we have AB and finally (after the division being made) we have A(C) C (C)B, and A(C) and (C)B are said to be the two equal straight-lines in which AB has been divided, for the point C cannot be divided between them, having no parts. Let's make an example like the previous: having 5 candies that we want to divide into two exactly equal parts, we take 2 on one side and 2 on the other, and the !
o!
ne candy left, if it's able to be divided in half, then we will divide the 5 candies into two exactly equal parts with success. But if the candy left cannot be divided, as the remaining point C which we cannot divide, then our purpose of dividing the 5 candies into two exactly equal parts is unreachable. And the same follows for a straight-line that is divided by means of a point on itself. Therefore, the inference seems to me to be that a straight-line cannot be divided into two exactly equal parts by means of a point on it.

OK - so I think I got your meaning exactly right when I posed (translated into something more precise than your ad-hoc notation) the problem:

Divide a (continuous) line segment S into 2 lines segments A & B (of equal lengths or not, doesn't matter really....) such that:
1. A & B are disjoint
2. A U B = S
3. A & B are both closed (or both open)

It can't be done. Too bad, but there are various things that can't be done in math. Too bad.

#3 captures, I think, what you mean by "exactly equal" -- however, if your condition is rephrased as "of equal length", then it *can* be done. I don't think you even really know, or can make very precise, what you mean by "exactly equal".

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
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1/30/14 Joe Niederberger
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1/31/14 kirby urner
1/30/14 Joe Niederberger
1/31/14 Neighbor
1/30/14 Joe Niederberger
1/31/14 Neighbor
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1/31/14 Joe Niederberger
1/31/14 Joe Niederberger
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1/31/14 Neighbor
1/31/14 Robert Hansen
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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
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2/2/14 Robert Hansen
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2/2/14 GS Chandy
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2/2/14 GS Chandy
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