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Can we find using rule and

  1. Mar 29, 2007 #1


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    coud we find using rule and compass :Confused: an exact segment of value:

    [tex] \sqrt (n) [/tex] for every natural number n

    - the same but for [tex] n^{1/m} [/tex] where n and m are positive integers

    - given a segment of length a known can we find an exact segment of length [tex] a 2^{-1/2} [/tex]

    for the case m=2 using Pythagorean theorem is easy to find but what about the other cases.
  2. jcsd
  3. Mar 29, 2007 #2
    It is quite possible to take the square root of a number, and use the four arithmetic functions. However, everything by ruler and compass is equivalent to solving linear and quadratic equations.

    In general we can not find for integers, N &M, N^(1/M).
    Last edited: Mar 30, 2007
  4. Mar 30, 2007 #3


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    More generally, "constructible" numbers are those that are algebraic of order a power of two- that is, that they satisfy an irreducible polynomial equation with integer coefficients of degree a power of 2 (1, 2, 4, 8, etc.).
    [itex]\sqrt{n}[/itex] satisfies the equation x2- n= 0 and so is either algebraic of order 2 (if x2- n is irreducible- cannot be factored with integer coefficients) or algebraic of order 1: both of which are powers of two.
    We can construct a segment of length [itex]n^{1/m}[/itex] if and only if m is a power of two.

    That, by the way, is the reason we can't trisect all angles. cos(20 degrees) satisfies the irreducible polynomial equation 8x3- 6x- 1= 0: it is algebraic of order 3 and so is not "constructible". It's easy to construct a 60 degree angle: give an unit length, strike arcs of that length from each end point and draw the line from one end point to a point of intersection of those arcs. IF it were possible to trisect that 60 degree angle, it would be possible to construct a 20 degree angle. Now strike of a unit length on one leg of that angle and drop a perpendicular to the other. You would have constructed a segment of length cos(20) which is impossible.
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