[COLOR=#323232][FONT=verdana]See the attached pdf and you will clearly
http://onedrive.live.com/view.aspx?c...=WordPdf&wdo=1

You first statement is wrong. Mathematicians cannot "trisect the angle", "duplicate the cube", or "square the circle", using only compasses and straight edge. The ancient Greeks knew how to do these using other methods, as you do.

(It was proved, in the late 19th century, that the only numbers that are "constructible with compasses and straightedge" are those that are "algebraic of order a power of two. It can then be shown that "trisecting the angle" and "duplicating the cube" are equivalent to constructing a number that is algebraic of order 2. "Squaring the circle" is equivalent to constructing " $\pi$" which is not algebraic of any order.)

Definitely a fail. His technique is to use a straight edge that has been marked with tick marks at 1, 2, 3, .... and use that to mark off distances like sqrt(pi). So no - this technique does not truly square the circle but rather gives an approximation. For exwemple, given a cicle of unknown radius he measure it with his ruler - say he gets 10.2 units radius, then 10.2 x sqrt(pi) = 18.08 (appoximately) - so he makes a square 18.08 units per side. Big whoop - the Greeks are safe.

It is probably hopeless to respond, but ....

First, you need someone expert in English to translate for you. Your English is awful. Admittedly it is infinitely better than my Serbian, but I do not try to write in Serbian.

Second, I have not tried to puzzle through all of your examples. I do not need to. The classic problem of squaring the circle involves using an UNMARKED straightedge. Now I get that you are creating a marked straightedge using a compass and an unmarked straightedge so marks derivable by that process may be within the scope of the original problem. You are creating the marks through a process of successive halving of the distance. You are then ASSUMING that this process will let you mark off one or more irrational numbers, more specifically the number $r \sqrt{\pi }.$

Third, if your assumption were valid, then it would indeed be easy to construct a square with an area equal to that of a circle with radius r, starting with only a compass and an unmarked straightedge. Your assumption, however, is invalid. You do not even attempt to prove it. You merely assert it: "continue the process ... with real numbers." The pivot on which your construction turns is that you can create markings of irrational numbers using a process of successive division by 2. Where is your proof that such a pivot exists? (Don't waste too much time on trying to prove it because it is false.)

Originally Posted by JeffM

First, you need someone expert in English to translate for you. Your English is awful. Admittedly it is infinitely better than my Serbian, but I do not try to write in Serbian.
that translates gogle, I only own Intelligence - do not possess material goods to show their math differently
that translates gogle, I only own Intelligence - do not possess material goods to show their math differently
Originally Posted by JeffM
The pivot on which your construction turns is that you can create markings of irrational numbers using a process of successive division by 2.
Only the construction of the geometry of straightedge

You keep making assertions without proof. I am not sure exactly what you mean by a "construction of the geometry of the straightedge" because it makes no sense in English. But what you seem to be doing is to cut lines in half over and over again. I agree that you can do that construction with a compass and an unmarked straightedge. You then assume (I believe) that such successive halving will eventually let you construct any length, rational or irrational. That assumption is false. The process of successive halving will not even let you construct certain very simple rational lengths such as 2/3. Your net of halves, quarters, eighths, etc. is not fine enough to snare all the rational numbers, let alone all the rational and irrational numbers.

$k\ and\ n\ are\ integers\ such\ that\ k \ge 0 \le n\ and\ \dfrac{k}{2^n} < \dfrac{2}{3} < \dfrac{k + 1}{2^n}.$

I note that there are such integers, for example k = 5 and n = 3.

Midpoint = $\dfrac{1}{2} * \left(\dfrac{k}{2^n} + \dfrac{k + 1}{2^n}\right) = \dfrac{2k + 1}{2^{(n+1)}}.$

If the midpoint equals 2/3, then their difference = 0, which is an integer.

$\dfrac{2k + 1}{2^{(n+1)}} - \dfrac{2}{3} = \dfrac{3(2k + 1)}{3 * 2^{(n+1)}} - \dfrac{2 * 2^{(n + 1)}}{3 * 2^{(n+1)}} = \dfrac{6k + 3 - 2^{(n + 2)}}{3 * 2^{(n+1)}} = \dfrac{6k + 2 - 2^{(n + 2)} + 1}{3 * 2^{(n+1)}} = \dfrac{2\left(3k + 1 - 2^{(n + 1)}\right) + 1}{2\left(3 * 2^n\right)}.$

That difference is an odd number divided by an even number, which is not an integer and so is not zero.

This is a proof that halving the distance any number of successive times will never get you to 2/3 of the distance.

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