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- Thread starter JetBlckNewYr03
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selfAdjoint

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Now given the circle with its radius (easy to construct), the area in question is [pi]r

Now we assumed that x was constructible in the sense I gave above. The only length we had to start with was r. Using analytical geometry (x and y coordinates) you can prove that any length constructible from r will be the root of a polynomial in r. This is not deep, you just go through each step in the construction, express the lines and circle as functions of r - the lines will be linear functions and the circles will be square root function (roots of quadratic equations). and all the equations will multply together to get the equation of the whole construction and it will be a polynomial in r. So x, the side of tht square is the root of a polynomial in r. Plugging in that polynomial for x in our equation above and collecting terms we have a polunomial in r = 0, or r is a root of a polynomial.

But we have known since Lindemann proved it in the 1890s that [pi] is not the root of _any_ polynomial. The proof is hard. It is just possible to follow it if you had a very good two semester class in Calculus and were at the top of the class. But there is no doubt about it, [pi] is not the root of any polynomial, and every constructible number is such a root, so [pi] cannot be constructed by Euclidean moves. But squaring the circle, as we showed above, amounts to such a construction of [pi]. Therefore the circle cannot be squared.

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HallsofIvy

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I just want to point out that one can, in fact, prove that any

"constructible" length must be the root of some polynomial equation of order 2

A circle of radius 1 has area pi. A square of area pi must have sides of length sqrt(pi) so if it were possible to construct a square with the same area then it would be possible to construct a segment whose length is sqrt(pi) which, like pi, is not algebraic of ANY order.

Similarly, if it were possible to trisect ANY angle, then it would be possible to a length that is algebraic, but of order 3, not a power of 2. (This is a result of the fact that we can write sin(3*theta) as a cubic polynomial in sin(theta).)

If it were possible to "duplicate a cube"- that is, using 3 dimensional analogs of straight edge and compass (that would allow to construct the plane through 3 given points and part of a sphere with given center and radius) with twice the volume of a given cube, then it would be possible to construct a segment of length cube root of 2: again algebraic of order 3, not a power of 2.

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I have heard this question posed and answered many times over the years by learned people who answered it in terms of "Pi is irrational and no matter how accurately you measure the sides of your square the square of the sides will equal pi."

Apparently they never had a clue what they were talking about even though in their own fields they were experts. Thanks.

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