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Re: Alternative area postulate for geometry
Posted:
Jun 29, 2012 6:02 PM
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I think this all turns out to be a geometry version of Haim's accounting joke. The only difference is that the misdirection begins right at the beginning with "Instead of the unit square, assume that the unit of area is a unit circle..."
The joke has already been set in motion. The unit square is not a unit and thus not a unit of area.
Bob Hansen
On Jun 29, 2012, at 5:19 PM, Robert Hansen <bob@rsccore.com> wrote:
> First...
>
> 1. Math is dimensionless (there are no units).
> 2. Thus, there are no units to choose from.
> 3. Even saying that the "unit square" is a "unit" is wrong. It is not a unit. It is a square with sides of length = 1 that has an area of 1. We call it a unit square because its sides are 1 just like we call a circle a unit circle because its radius is 1. We don't mean its a unit. And the area of a unit square is itself NOT A UNIT SQUARE, it is 1.
>
>
> Now the contradiction...
>
> 1. If the area of the unit circle (for example) is 1 then the formula for the area of (any) circle would have to be r^2 (not Pi*r^2).
> 2. But, if I compute the definite integral of y = sqrt(r^2 - x^2) from -r to +r then I get the area (by integration) to be Pi*r^2.
> 3. Contradiction.
>
> You cannot fix that integral without putting a 1/Pi somewhere. But where? What if my integral is to compute how far an object falls in a gravity field rather than area (as if there is even a difference)? Am I to say now that it actually falls less (slower)?
>
> Gauss's law relates to things like that fact that in the middle of the earth there is no net force (due to gravity) resulting from any of the mass that is further from the center than your are. Thus, inside a hollow (uniform) shell (of any shape), you would be weightless, even if you were right next to the (inside) shell. I find that to be very convenient, otherwise something like the earth would be pulling its mantle towards its surface. Mathematically, this results from how the integration works with any force that varies inversely by the square of the distance. The definite integral just cancels itself out because even though you might be near one side of the shell, there is just that much more shell on the other side, further away, and what is lost in the force inversely by the square of distance is gained directly by the area (the square of distance). It all cancels out.
>
> If the integral of 1/x^2 is no longer -2/x then Gauss's law folds and the earth pulls itself apart..:)
>
> Unless what you mean to be doing is converting ALL of the values to another unit, not just the area. But then it has to be stated like this...
>
> 1. Given a circle with a radius of 1 inch.
> 2. Given 1 Gary Unit equals Pi inches.
> 3. What is the area of the circle in gary units?
>
> Bob Hansen
>
> On Jun 29, 2012, at 2:23 PM, Joe Niederberger wrote:
>
>> R. Hansen says:
>>> Mathematically, the area of the unit squad [sic] is 1, period. It is not a choice because there are no other possible choices.
>>
>> OK - how do you know that? I mean, don't just give some wild examples about the earth falling apart - really reason through step by step to where the contradictions appear should anything but the unit square be chosen as a unit of area.
>>
>> Joe N
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