Vtyjkyuk

The following statement is equivalent to Euclid's statement that $sqrt2$ is irrational but has a rather different flavour.

Consider the straight line through two points $0$ and $1$ with the natural numbers $[mathbbN]$ constructed as points

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

Consider the point $1'$ constructed like this:

Define $[mathbbN]'$ 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 $[mathbbN]$ and $[mathbbN]'$ 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 $sqrt2$ 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 $sqrt2$ geometrically is by stating that there are no $n,m in mathbbZ$ such that the point $n/m$ constructed this way:

coincides with the point $1' = sqrt2$:

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 $sqrt2$, 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 $x(n,m) = p - qfracp-rq-s$:

The statement is: For no two Gaussian integers $n = p + q i$, $m= r + s i$ does $x(n,m)$ coincide with $sqrt2$.

The first picture proves the constructibility of $sqrt2$.

The second illustrates the concept of irrationality.

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

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

So, let $alphainmathbb Rsetminus0$ and consider set $kalpha,mid, kinmathbb Z$. That set will contain non-zero integers if and only if $alphainmathbb Q$. 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 $kalpha,mid, kinmathbb Z$ 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 $kalpha + mathbb Z,mid, kinmathbb Zsubseteq mathbb R/mathbb Z cong mathbb S^1.$

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

The set of integer multiples of non-zero real $kalpha + mathbb Z,mid, kinmathbb Z$ is finite if and only if $alphainmathbb Q.$ Otherwise, it is dense in $mathbb Rsetminus mathbb Zcong mathbb S^1.$

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 $sqrt2$.

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

Following Ethan's advice I want to draw your attention to Apostol's proof of the irrationality of $sqrt2$ 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).