# Geometric proof of existence of irrational numbers.

It is easy, using only straightedge and compass, to construct irrational lengths, is there a way to prove, using only straightedge and compass, that there are constructible lengths which are irrational? Ie a geometric proof.

And is it possible to construct an (unending) sequence of rational lengths or areas, such that they can get arbitrarily close to the area or circumference of a circle?

If not, then does this provide evidence that the real numbers are not sufficiently refined to capture exactly the circumference or area of an idealized circle?

(The idea being that the reals can be constructed from equivalence classes of infinite sequences of rationals, so if the circumference cant be approached arbitrarily by rationals then its not necessarily a real number)

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This wording, "And is it possible to construct an unbounded sequence of rational lengths or areas, such that they can get arbitrarily close to the area or circumference of a circle?", may not be what you intend to ask. The usual meaning of "unbounded sequence" is that, for each positive integer $M,$ some of the numbers in the sequence do not belong to the interval $[-M,M].$ – Dave L. Renfro Feb 22 '12 at 20:45
In the last paragraph, is it supposed to say "rational numbers"? It's not clear to me why anything before that should tell us anything about the real numbers. – joriki Feb 22 '12 at 20:45
– WimC Feb 22 '12 at 20:51

A ratio between quantities is rational if and only if the Euclidean algorithm applied to the quantities terminates, in the process expressing the ratio as a finite continued fraction. The golden ratio $G=\frac12(\sqrt{5}+1)$ is irrational, in fact given by the continued fraction $$1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cdots}}}.$$ To prove this geometrically, draw a regular pentangle and its five diagonals (which form a pentagram). The ratio of a diagonal to a side is $G$. You can perform the Euclidean algorithm by visual inspection in this figure: If the side of the pentangle is $1$ and the diagonal $G$, each point has two sides of length $G-1$, etc., in fact any two quantities in this figure which look the same are the same, and after employing a few steps of Euclids algorithm you are left with the diagonal and side of the smaller pentangle outlined by the diagonals of the big one. So the process repeats forever.

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This is of course very closely related to the answer by André Nicolas. – Harald Hanche-Olsen Feb 22 '12 at 21:18
The diagram that you refer to, however, has a more striking visual character. Rectangles are boring. Pentagons, and pentagrams, by contrast, are full of magic, mystery, and menace. – André Nicolas Feb 22 '12 at 21:30
@AndréNicolas "...magic, mystery, and menace." We all love alliteration! And pentagrams, of course. – Pedro Tamaroff Feb 23 '12 at 3:43
The Pythagoreans also loved pentagrams, if we are to trust the testimony of someone living some $800$ years later. How would they feel about pentagram being a source of the destruction of their belief that "all is number"? Of course the report of the Pythagoreans being upset by irrationality is not necessarily to be believed. – André Nicolas Feb 23 '12 at 3:54

This proof was published in 2000 by Tom Apostol, apparently claiming it was new, but I learned it before that from a book by Otto Toeplitz that was probably published about 50 or 60 years ago.

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There are many attractive geometric arguments for irrationality. Here is one for the Golden Ratio.

Construct a golden rectangle. A recipe for doing this can be found in Euclid, and was known to early Pythagoreans.

There should be a diagram to illustrate the idea. Perhaps you can draw it yourself. Say the golden rectangle is $ABCD$, with the vertices as usual enumerated counterclockwise, and let $AB$ be a long side of the rectangle.

We prove that the long side and the short side of a golden rectangle are incommensurable. Suppose to the contrary that sides $AB$ and $BC$ have a common measure $m$. Or else, in more modern language, suppose that $AB$ and $BC$ have lengths that are each an integer multiple of some common number $m$. Or, else, even more arithmetically, suppose that each side is an integer.

Cut off a square $AEFD$ from the rectangle, by finding the point $E$ on $AB$ such that $AE=AD$, and slicing straight up. That leaves a rectangle $EBCF$, which by the definition of golden rectangle, is itself golden. It is clear that $m$ is a common measure of the sides of $EBCF$.

Continue, by cutting off a square from $EBCF$, leaving an even smaller golden rectangle whose sides have common measure $m$. Clearly, this process can be continued forever. But after a while, each side of the little golden rectangle just produced will be less than the hypothesized common measure $m$, and we get our contradiction.

There has been speculation that this was the first irrationality proof. The only problem with this theory is the total lack of evidence.

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+1 for the last sentence. – JeffE Feb 22 '12 at 23:45

In my mind, irrationality is an inherently algebraic concept. Or more strongly, a number-theoretic concept. It's not clear what a geometric representation of the notion of 'irrational' would even mean, aside from the obvious superficial ones.

In fact, in a common formalization of Euclidean geometry in first-order logic (due to Tarski), you can prove it's impossible to even ask if a number is irrational.

It is possible to construct a sequence of rational lengths and areas to get arbitrarily close to the area or circumference of a circle (or any real multiple of a unit length). In fact, this very idea is the heart of the classic proof that $$\text{area} = \frac{1}{2} \text{radius} \cdot \text{circumference}$$ by comparing the circle to an inscribed regular polygon, which in turn is subdivided into triangles with a vertex at the center. (aside: the notion of length is surprisingly subtle and there are some important technical details needed to fill the holes in the proof)

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Wont these polygons have irrational area/circumference? – user1708 Feb 22 '12 at 21:49
Ah good point; I was just focused on the demonstration of limiting length/area (and I was probably thinking about the algebraic/transcendental dichotomy). That said, I find it very likely you could manage polygons with triangulations with rational sides and area. – Hurkyl Feb 23 '12 at 4:30

Indeed there is a way to geometrically show that $\sqrt{2}$ is irrational. I know the proof from the blog Gaussianos, which in turn got it from "Irrationality of the Square Root of Two – A Geometric Proof" from the American Mathematical Monthly (Nov. 2000, ps. 841–842) by Tom Apostol. You can check the images in the link or see the article, the proof is very simple.

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This proof is also presented in Rademacher and Toeplitz, The Enjoyment of Math. – Neal Feb 22 '12 at 21:12

I'm not sure how you propose to use straightedge and compass to prove something, rather than to construct something. However, maybe Proof 7 at http://www.cut-the-knot.org/proofs/sq_root.shtml comes close.

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