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Old 2007-03-21, 20:31   #12
hhh
 
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Quote:
Originally Posted by R.D. Silverman View Post
Highly discontinuous? Indeed!! It is nowhere continuous.
That I know. What I still don't know if of which of the two propositions I gave or what else you were talking. H.
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Old 2007-03-21, 21:18   #13
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That the floor is continuous (no steps) is one of my assumptions.
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Old 2007-03-23, 11:58   #14
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Quote:
Originally Posted by davieddy View Post
That the floor is continuous (no steps) is one of my assumptions.
No, it wasn't. Go re-read what you wrote.
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Old 2007-03-23, 13:27   #15
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Quote:
Originally Posted by R.D. Silverman View Post
No, it wasn't. Go re-read what you wrote.
Bob,

I think that he means that "the floor is continuous" is one of the "two simple assumptions" that he used in his answer.

"However, with two simple assumptions, you can rectify the
problem by rotating the table through an angle <90 degrees.
How come? and what are the assumptions?"

Last fiddled with by Wacky on 2007-03-23 at 13:28
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Old 2007-03-23, 13:32   #16
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I suspect that "The feet of the table are coplanar" is sufficient, but not necessary, to serve as the other assumption.
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Old 2007-03-23, 14:41   #17
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This was my other assumption (feet at corners of a square).
It is most definitely necessary though.

David
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Old 2007-03-23, 15:21   #18
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Quote:
Originally Posted by Wacky View Post
I suspect that "The feet of the table are coplanar" is sufficient, but not necessary, to serve as the other assumption.
Actually, this assumption is not needed under my solution.

I can also replace "height is a u.r.v." with "height is a continuous function",
but the function will be nowhere differentiable. The height will be a
lacunary function. All that is required is that within a circle of radius epsilon
of each point, the height of the floor varies sufficiently. "sufficiently"
depends on how far the 4 feet of the table depart from a plane. If the
feet are coplanar, then I believe that the height varying by k * epsilon
for any k > 1 in the neighborhood of the point on which the 4th leg will
sit is sufficient to let one apply the Ham Sandwich Theorem.

However, while continuous, the floor will not be *smooth*
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Old 2007-03-23, 16:06   #19
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Quote:
Originally Posted by davieddy View Post
This was my other assumption (feet at corners of a square).
It is most definitely necessary though.

David
Unless you place a minimum limit on the "unlevelness" of the floor.
Wacky, did you get necessary and sufficient the wrong way round?

Last fiddled with by davieddy on 2007-03-23 at 16:07
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Old 2007-03-23, 16:27   #20
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Quote:
Originally Posted by davieddy View Post
Unless you place a minimum limit on the "unlevelness" of the floor.
Wacky, did you get necessary and sufficient the wrong way round?
No, I don't think so. I believe that the requirement is that the floor be more uneven than the feet of the table.

If my assertion is correct, then coplanar feet is sufficient. However, if would not be necessary.
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Old 2007-03-23, 16:43   #21
mfgoode
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Thumbs down Uneven floor !



All that 'higher maths' when all one has to do is place a wedge under the leg that's off the floor. Plain common sense!

Mally
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Old 2007-03-23, 17:16   #22
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To be fair, I think it is only RDS who has introduced the higher maths.
I suppose your solution (if they provided a sufficiency of paper napkins)
would be at least as practical as rotating the table a bit.
I was in a restaurant whe I first had this insight.
David
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