# Visualizations of the (potential) irrationality of $\sqrt{2}$

The following statement is equivalent to Euclid's statement that $\sqrt{2}$ is irrational but has a rather different flavour.

Consider the straight line through two points $0$ and $1$ with the natural numbers $[\mathbb{N}]$ constructed as points

$[n] = 1 + 1 + \dots + 1$ ($n$-times):

Consider the point $1'$ constructed like this:

Define $[\mathbb{N}]'$ as the set of points constructed as

$[n]' = 1' + 1' + \dots + 1'$ ($n$-times):

Then the statement goes:

For all $n,m$: If $[n] = [m]'$ then $m = n = 0$.

This means, the sets $[\mathbb{N}]$ and $[\mathbb{N}]'$ have only the point $0$ in common. For no $n,m \neq 0$ do the points $[n]$ and $[m]'$ happen to coincide.

I wonder if this visual and conceptual perspective on the irrationality of $\sqrt{2}$ has been taken and discussed before, or if it's even a standard perspective - and if not so: if it's an enlightening perspective?

Added: Another - and more straight forward - way to state the irrationality of $\sqrt{2}$ geometrically is by stating that there are no $n,m \in \mathbb{Z}$ such that the point $n/m$ constructed this way:

coincides with the point $1' = \sqrt{2}$:

Note that this statement is much harder to draw accurately and to catch and believe visually.

Note further, that this picture does not imply a visual proof of the irrationality of $\sqrt{2}$, neither.

Added: Another - even simpler but less straight forward - way to state the same geometrically is by this construction over two Gaussian integers $n = p + q\ i$, $m = r + s\ i$, which gives the rational point $x(n,m) = p - q\frac{p-r}{q-s}$:

The statement is: For no two Gaussian integers $n = p + q\ i$, $m= r + s\ i$ does $x(n,m)$ coincide with $\sqrt{2}$.

• +1) In essence it is Dedekind's notion of cut: if you consider the horizontal line with only rational numbers and you "project the diagonal of the unit square on it, the compass will not "cut" the line, because there is no intersection (rational) point. – Mauro ALLEGRANZA Sep 4 '18 at 9:11

The first picture proves the constructibility of $\sqrt2$.

The second illustrates the concept of irrationality.

There is nothing here about the irrationality of $\sqrt2$.

• I admit that! A proof that the tick marks never coincide doesn't immediately follow from the construction. But it's a way to define and visualize irrationality: A number $\alpha$ is irrational when $m\alpha, m \in \mathbb{Z}$ only coincides with $n \in \mathbb{Z}$ when $m = n = 0$ (see Ennar's answer). – Hans-Peter Stricker Sep 4 '18 at 10:34
• I admit there is no visible link between the two constructions, but on the other side: there is one, it's only not visible immediately. – Hans-Peter Stricker Sep 4 '18 at 10:36
• You are perfectly right: It's just a visualization of irrationality (as a concept), and certainly no "visual proof" of the irrationality of $\sqrt{2}$. Thank you for making this clear. (But I won't change the title of my question, OK?) – Hans-Peter Stricker Sep 4 '18 at 10:39
• I don't think the visualization was intended as any sort of proof, just a geometric restatement of the definition of irrationality. IMO, such visualization could be very beneficial for educational purposes for the students first encountering irrational numbers compared to the dull "decimal representation of irrational number is infinite and non-periodic". – Ennar Sep 4 '18 at 10:39
• @Ennar: Thank you for your comment. That's exactly how I intended my question. (I should have mentioned it). What you say about the dull "decimal representations of irrational number are infinite and non-periodic" perfectly hits the point. – Hans-Peter Stricker Sep 4 '18 at 10:43

It is not just $\sqrt 2$, you can actually characterize all irrational numbers in such a way.

So, let $\alpha\in\mathbb R\setminus\{0\}$ and consider set $\{k\alpha\,\mid\, k\in\mathbb Z\}$. That set will contain non-zero integers if and only if $\alpha\in\mathbb Q$. This is essentially what you wrote. However, this is nothing new from the usual definition of (ir)rational number.

We can make things more interesting, though, by considering not $\{k\alpha\,\mid\, k\in\mathbb Z\}$ on a line, but on a circle instead (by wrapping the real line on a circle). Formally, for a non-zero real $\alpha$ consider the set of integer multiples of $\alpha$ on a circle $\{k\alpha + \mathbb Z\,\mid\, k\in\mathbb Z\}\subseteq \mathbb R/\mathbb Z \cong \mathbb S^1.$

One interesting thing to note is that for irrational $\alpha$, $k\alpha + \mathbb Z,\ k\in\mathbb Z$, are all distinct. Moreover, the following is true:

The set of integer multiples of non-zero real $\{k\alpha + \mathbb Z\,\mid\, k\in\mathbb Z\}$ is finite if and only if $\alpha\in\mathbb Q.$ Otherwise, it is dense in $\mathbb R\setminus \mathbb Z\cong \mathbb S^1.$

There is an elementary argument as to why this is true that you can find here.

Not a direct answer to your question - but perhaps what you are looking for.

Assuming $a^2=2b^2$, with positive integers $a$ and $b$, one can easily establish that also $(2b−a)^2=2(a−b)^2$. The impossibility of the former lies in the fact that $a>2b−a$ (which shows that we are just at the beginning of an infinite descent.) This is one of the proofs of the irrationality of $\sqrt{2}$.

https://www.cut-the-knot.org/proofs/GraphicalSqRoots.shtml

• To be a "proof without words" this would require that you can literally "see" that $(2b-a)^2 = 2(a-b)^2$ which is the content of the carpets theorem. But I found this quite hard - and the proof is involved. – Hans-Peter Stricker Sep 4 '18 at 15:13
• Thanks for the hint! But to be honest: It's Tom Apostol's proof that is really a perfect "proof without words" - without much explanation you immeadiately see, why infinite descent would be possible – Hans-Peter Stricker Sep 4 '18 at 15:37
• @HansStricker I agree. That is in fact the proof I vaguely remembered when I searched for the one I eventually posted. You should provide another answer to this question. – Ethan Bolker Sep 4 '18 at 16:01

Following Ethan's advice I want to draw your attention to Apostol's proof of the irrationality of $\sqrt{2}$ which is as visual as a proof can be (in my opinion): One can literally see at a glance that it proves what it's supposed to prove: the impossibility of a isosceles triangle with integer side length (by infinite descent):

Note that it's not a proof completely without words. It helps a lot to read the comments of the author:

Each line segment in the diagram has integer length, and the three segments with double tick marks have equal lengths. (Two of them are tangents to the circle from the same point.) Therefore the smaller isosceles right triangle with hypotenuse on the horizontal base also has integer sides.

But through own thinking one could come up with this by oneself (having in mind what's to be proved).