Squaring the circle
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Squaring the circle is the attempt to construct, using straightedge and compass, a square with an area equal to the area of a given circle. The word "attempt" is used above because the task has been proven impossible using classical tools.^{[note 1]} This has been known for over 100 years, but it had been suspected for much longer.
Naturally, such a minor obstacle as impossibility has not stopped people from making attempts to square the circle. A person who attempts to square the circle is called a moron circle-squarer, and the term, by metaphorical extension, may be applied to any practitioner of similar recreational impossibilities.
So how can you do it?
Why would you want to square the circle?[edit]
Squaring the circle (in a finite number of steps) is a problem that has not been solved since the time of the ancient Greeks. So it follows that if you can solve it, you must be smarter than anyone since the time of the ancient Greeks. Also, you will probably get widespread recognition for knocking off such a long-standing (and therefore, extremely important) problem. Maybe you'll win a Fields medal^{}!
On a more serious note, squaring the circle would require constructing the length Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt{\pi}} . (A circle with radius Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r} has area Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi r^2} . Hence a square with the same area must have a side of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt{\pi}r} ) If this number could be constructed, that would prove that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi} is an algebraic number, meaning there is some possible set of rational numbers you can use to calculate it.
For a variety of (essentially subjective) reasons, the very thought that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi} was somehow inaccessible through "normal" numbers REALLY seems to bother some people. Legend says that Pythagoras murdered the person who discovered that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt2} was irrational, so the thought that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi} itself was completely inaccessible via the integers would have been anathema. One particular objection is based on passages in the Bible, as 1 Kings 7:23-26 is believed (by some literalists) to imply that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi} has to be rational and equal to 3.
Also for no good reason, during the 1700s the belief arose that squaring the circle would somehow solve the "Longitude" problem (the inability of sea vessels to determine where they were on the east-west axis). As there were some enormous cash prizes on offer (in 1714 the British government offered a prize of £20,000), this fired up every amateur mathematician in Europe. Circle squaring is actually irrelevant; all that was needed to solve the longitude problem was the ability to observe the sun and a really good clock.
In the mathematics world the question was put to bed in 1882 when Ferdinand von Lindemann proved that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi} is not algebraic (in technical jargon, it is "transcendental"). Because there are definitely no rational numbers that can calculate Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi} , it is impossible to construct Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt{\pi}} in Euclidean space.
However true believers won't be deterred by anything as flimsy as "proof". They persist because they believe that there is an ideological bias against circle-squarers whose brave investigations threaten the comfortable orthodoxy of Western deconstructionist mathematics.
In actuality, the only ideological bias in effect is real mathematicians' not being bothered to waste their time with cranks.
Sketch of the proof[edit]
In a compass and straightedge construction one is free to define the unit length from any pair of given points. Additionally, only points that are given and intersections of previously constructed circles and lines may be considered, and lines and circles may only be constructed from previously defined points.
Finding the intersections of a line/circle and another line/circle involves simultaneously solving a system of two equations each of which is either quadratic or linear. These lines and circles in turn depend on the points which define them, therefore, with a little algebra, it can be seen that defining a point from some given ones is equivalent to solving a quadratic equation whose coefficients are either integers, or are the result of repeated applications of this method.
Say for example we wanted to determine the points where a line with slope of four intersects a circle with a radius of four centered at the point Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (2,0)} . To find the points of intersection, we would need to set up a system of equations where the circle is given by the equation Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x-2)^2+y^2=16} and the line is given by the equation Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=4x} . Then we would substitute the equation for the line into the equation for the circle, expand and simplify.
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align}(x-2)^2+(4x)^2&=16\\x^2-4x+4+16x^2&=16\\17x^2-4x+4&=16\end{align}}
To find the roots we rearrange this to equal 0:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 17x^2-4x-12=0}
Note that this is indeed a single-variable polynomial with integers as coefficients, as would be expected from compass and straight edge construction. Since it won't factor easily we can use the quadratic formula:
For any quadratic in the form Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ax^2+bx+c=0} , the following formula is applicable:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}
This is the "quadratic formula."
Using our equation Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 17x^2-4x-12=0} the following is true:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align}x&=\frac{-(-4)\pm\sqrt{(-4)^2-4(17)(-12)}}{2(17)}\\x&=\frac{7\pm8\sqrt{13}}{34}\end{align}}
Which gives roots Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=0.966,-0.731} .
To find the Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y} values we substitute the above roots into the equation for the line:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align}&f(0.966)=4(0.966)=3.864\\&f(-0.731)=4(-0.731)=-2.923\end{align}}
Thus it follows that the line Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=4x} intersects the circle Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x-2)^2+y^2=16} at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P(0.966,3.864)} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Q(-0.731,-2.923)} .
In elementary analysis, numbers which satisfy some polynomial equation Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_n x^n + a_{n-1}x^{n-1} + \dotsb + a_2 x^2 + a_1 x + a_0=0}
where the coefficients Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_0, a_1,.....,a_n}
are integers (i.e. the quadratic equation above) are what is known as algebraic numbers. Moreover they form what is known as an algebraically closed field, that is, all roots of polynomials with algebraic coefficients are themselves algebraic numbers. Therefore, all numbers that it is possible to construct with compass and straightedge must be algebraic, which Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi}
(and therefore its square root) are not. Thus the construction is impossible. In fact, the mathematical constants e (2.71828...) and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi}
(3.14159...) belong to a class of numbers known as transcendental numbers, numbers which are not roots of non-zero polynomials with integer coefficients. The full, formal proof of this is known as the Lindemann–Weierstrass theorem.^{[1]} Unlike other fields (eg. science, law) the concept of "proof" in mathematics is absolute, i.e. once a valid proof is provided of something, there is absolutely nothing which can disprove it within the axiomatic base that it is worked on.
Cheat[edit]
A common way to square the circle is to cheat. (Mathematicians call this approximation.) Recall that the problem statement is to construct a square of the same area as a circle using straightedge and compass. Any of the terms in italics should be considered merely optional.
For example, given a circle, it is simple to construct a square having an area equal to 3.2 times the square of the radius of the given circle. This square does not have the same area of the circle, but it will look awfully close. That should be good enough for the mathematicians.^{[note 2]}
Or, instead of starting with a circle, we could start with a polygon with, say, 96 sides. That's close enough to a circle — right, everyone? It is possible to "square the polygon" (as was known to the Greeks), so it's basically possible to square the circle. Alternatively, you could show how to square a polygon with 96 sides, a polygon with 192 sides, a polygon with 384 sides, and so on. Therefore, passing to the limit, we can square the circle.
Cheating in multiple ways at the same time[edit]
The following process involves a calculator. It's not exact, but can be refined up to the accuracy of the tools you have.^{[note 3]}
- First, calculate the area of the circle.
- Then, take the square root of the area, to get the length of the edge of the square.
- If you got good drawing tools, you can even draw the square now that you have the length of the edge.
Cheating with a physical aid[edit]
- Create a wheel of the same size as the circle and which is half as wide as the circle's radius.
- Cover the side in wet paint and make it revolve over a flat surface exactly once.
- This leaves a painted rectangle with the same surface as the circle.
- Finish up by squaring this rectangle (this step can be done even with straightedge and compass).
Warning[edit]
If you develop an urge to talk with or debate circle-squarers, you should immediately seek medical attention. Circle-squarers are not, for the most part, interested in having their ideas critiqued. They are not convinced by "proof" — if they were, they wouldn't have started on the problem. See Keith Devlin's take on this for more.^{[2]}
The Classical Family of Unsolvable Problems[edit]
Squaring the circle^{}, doubling the cube^{} and trisecting an angle^{} may be called the trinity of classical unsolvable problems in Euclidean geometry. Since all three have been proven to be impossible, using nothing but a ruler and a compass, it is of course irresistible for cranks to square, double and trisect anyway. Another problem, physical this time, is inventing a perpetual motion machine, which is equally impossible. The time and effort wasted on this defies belief, but if cranks stick to these futile attempts, an argument could be made that they are at least not doing any harm while engaged in these endeavors.
See also[edit]
Notes[edit]
- ↑ By the determination^{} that π is a transcendental number.
- ↑ So good they actually tried to make π = 3.2 so they could do this.
- ↑ An ordinary compass and straightedge, both available at the dollar store.
References[edit]
- ↑ Roger Lipsett and Warren Buck. "proof of Lindemann-Weierstrass theorem and that e and π are transcendental." PlanetMath.org. 2012 November 19.
- ↑ Squaring the circle. Devlin's Angle, February 2003.