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Squaring the circle

  1. Mar 10, 2012 #1
    The ancient problem of squaring the circle is to find a square with the exact area of a given circle. I don't see why this problem is so hard.

    Say a circle has radius 5. All you have to do is


    Then that's the length of the side of the square. Help me out here. Why is that not a solution to the problem?

    I put this in the math discussion section because this is not a homework section.
  2. jcsd
  3. Mar 10, 2012 #2
  4. Mar 10, 2012 #3


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    To square the circle, you must construct a square having the same area as a given circle using compasses and a straightedge. And you encapsulated the difficulty when you said "find a square with the exact area of a given circle". Yes, you could do that by constructing a line segment with length exactly [itex]5\sqrt{\pi}[/itex]. How do you do that? If you are given a circle with radius exactly 5 you could use that to construct a segment of length exactly 1 (dividing a segment into n equal parts is easy). But then you need to get to [itex]\sqrt{\pi}[/itex].

    The point is that not all numbers are constructible. In fact, there exist relatively few numbers, x, such that given a line segment of length 1, we can construct a line segment of length x. Given a segment of length 1, we can, of course, duplicate it any number of times and so "construct" a segment of length any integer n. And because we can divide a segment into any number of equal parts, we can construct a segment of line 1/n for any postive integer n. It is classic to construct a segment of length [itex]\sqrt{2}[/itex] and that can be extended to constructing any square root of an integer.

    To cut this short (perhaps the site dijkarte links to gives the details but they are very deep) the "constructible numbers" are precisely the real numbers that are "algebraic of order a power of 2". A number is said to be "algebraic of order n" if it satisfies a polynomial equation, with integer coefficients, of degree n but no such equation of lower degree. For example, every rational number x= n/m, satisfies mx-n= 0, a polynomial equation of degree 1, with integral coefficients. That is, the numbers that are "algebraic of order 1" are exactly the rational numbers. [itex]\sqrt{2}[/itex] satisfies [itex]x^2- 2= 0[/itex] and so is "algebraic of order 2" (and so constructible) but [itex]\sqrt[3]{2}[/itex] is algebraic of order 3, not a power of 2, and so is not constructible.

    [itex]\pi[/itex], and so [itex]\sqrt{\pi}[/itex], is not algebraic of any order (such numbers are called "transcendental"). Since [itex]\sqrt{\pi}[/itex] is not algebraic of order 2, it, and any multiple such as [itex]5\sqrt{\pi}[/itex] is not "constructible"- you cannot construct a line segment, using compasses and straightedge, of length [itex]5\sqrt{\pi}[/itex].
  5. Mar 10, 2012 #4
    It seems like what you're saying is rather obvious, ie, you cannot come up with an exact number for pi, you can only come up with an exact number for integers. Well, why can't people understand that? why are so many people wasting their time refusing to listen to what you're saying? It's almost as if between your lines, you have developed a negative proof for squaring the circle.

    By the way, I really like your post. It's real dense with mathematical jargon. I like that. I'm a big fan of jargon, irregardless of what field. Very well written and thanks for the time you put in to writing it.
  6. Mar 10, 2012 #5
    It has been proved impossible, no mathematician is wasting their time with it, and an undergraduate math major can understand it (many more people can also get an intuitive feel for it). Did you hear somewhere that this is an open question? It was proved in the late 19th century.
  7. Mar 10, 2012 #6


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    I will try to enlarge on what halls said with fewer math terms if possible. You are right there is a segment of length sqrt(pi). As he said the problem is whether one can obtain this length in a rather restricted way, as a straightedge and compass construction.

    Those tools can only construct lines and circles, and must begin from 2 given points, whose distances are already known. That means one begins from a single segment of length one and then repeats this to obtain segments of any integer length, and circles of any integer radius.

    But one can also subdivide lengths evenly so one gets also segments of any rational length (too much jargon?, a rational number is a fraction of integers like 3/7), and circkles of any rational radius.

    Then one can intersect all these to get more points, and see what lengths one can get. E.g. one CAN get a segment of length sqrt(1/2), by finding the intersection point of the line at 45 degrees through the origin and the circle of radius one about that same origin. Then dropping a perpendicular on the x axis (I hope not too much jargon here, perpendicular make right angles, i.e. 90 degree angles).

    The whole idea is that the points one can get involve only square roots at worst, since a circle has a quadratic equation, so finding the point where it meets a line involves only taking square roots. So "constructible" numbers, i.e. the ones you can construct, are precisely those numbers that can be formed starting from rational numbers (fractions) and taking repeated square roots, plus of course adding and multiplying and subtracting and dividing.

    These numbers might be called repeated quadratic numbers. Now a sqrt of a sqrt satisfies a 4th degree algebraic equation, and taking another sqrt gives a number satisfying an 8th degree algebraic equation. So the degree of the algebraic equation for the constructed number is always a power of 2, as halls said.

    In particular one cannot construct the length cuberoot(2) in this way, much less sqrt(pi). I.e. although one can construct square roots, one can only construct square roots of numbers one already has constructed, so one could construct sqrt(pi) if one could first construct pi, but that is the hard part here.

    I.e. one can construct sqrt(pi) starting from pi, but one cannot construct it starting only from 1, as is the challenge here.

    By the way sarcasm is not a plus here, especially when people are donating their time to help you. Next time I suggest you just ask the person whose post puzzles you, for further definitions. It is not their fault that they cannot read your mind and see which words you know and which you do not. After reading your rude response it was only with difficulty that I brought myself to continue the answer so kindly provided before me.
  8. Mar 11, 2012 #7
    thanks for your reply, mathwonk, i really appreciate the time you spent working on this reply.
  9. Mar 11, 2012 #8


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    No, I said nothing of the sort. We certainly can "come up with an exact number for pi"- it is [itex]\pi[/itex]. If you mean it cannot be written as a terminating decimal, well neither can [itex]\sqrt{2}[/itex] but we certainly can construct a line segment of length [itex]\sqrt{2}[/itex].
    "Mathematical Crankery" is a common sport!

    jargon= precise.
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