What does this prove? Some of the most gorgeous visual "shrink" proofs ever invented

What people said

Strong rec Peter Kagey · 2021-03 · Mathematics Stack Exchange · on “When is it possible to find a regular $k$-gon in a centered $n$-gon?”

I knew of the square grid argument (and by analogy, the triangular/hexagonal grid argument) from this wonderful Mathologer video [this video] , but those blog posts are really lovely!

Strong rec D.R. · 2025-03 · Mathematics Stack Exchange · on “What is the most unusual proof you know that $\sqrt{2}$ is irrational?”

Mathologer irrationality with triangular squares 2018 [LINK] presents a similar geometric visualization of infinite descent (using the geometric fact that one can tile an equilateral triangle △ by $n^2$ smaller equilateral triangles with orientations △,▽) shows that $\sqrt 3, \sqrt 2, \sqrt 6, \sqrt 5$ are irrational. He also shows this with other shapes besides triangles (e.g. pentagons for $\sqrt 5$). He also inverts this infinite descent to get solutions to Pell's equation. More gorgeous visual "shrink" proofs [this video]

Recommends Servaes · 2020-08 · Mathematics Stack Exchange · on “Existence of regular $n$-gon whose vertices are arbitrarily close to integer coordinates”

Hint: Let $p_1,\ldots,p_n$ be the vertices of a regular $n$ -gon in $\Bbb{R}^2$ . For each $p_i$ take a point $q_i\in\Bbb{Q}^2$ that is sufficiently close to $p_i$ (depending on $\varepsilon$ ). Now scale up all the coordinates of the $q_i$ to get vertices in $\Bbb{Z}^2$ without changing the angles $\alpha_i$ . This video [this video] illustrates the idea for $n=3$ .

Recommends Servaes · 2020-08 · Mathematics Stack Exchange · on “Existence of regular $n$-gon whose vertices are arbitrarily close to integer coordinates”

@RobertIsrael In fact there is no regular $n$-gon with vertices in $\Bbb{Z^2}$ for any $n>2$ other than $n=4$. Here is a nice animation [this video] illustrating why. A detailed explanation is given from 7:10 [this video] onwards, I suggest speeding up the video a bit.

Recommends M. Winter · 2021-09 · Mathematics Stack Exchange · on “What does the additive subgroup of $\mathbb{R}^3$ generated by the vertices of a regular polyhedron look like?”

While this is a good answer (+1) the explanation of the non-lattice nature of most $n$-gons is quite advanced. More low-level (visual) explanations are a available, e.g. in this very nice video [this video] by Mathologer.(Burkhard Polster).

Recommends D.R. · 2023-07 · Mathematics Stack Exchange · on “Proving that the ratio of the hypotenuse of an isosceles right triangle to the leg is irrational”

Just a fun note: the heart of user Oscar Lanzi's answer [LINK] can be phrased as "a square integer lattice grid can not have infinitely shrinking shapes". See this beautiful Mathologer video "What does this prove? Some of the most gorgeous visual "shrink" proofs ever invented" for other examples of this proof technique: [this video] . More explicit explanation of Oscar Lanzi's proof in terms of geometry and lattice points : First you suppose that an isosceles right triangle $T$ can be put on the integer lattice grid in 2 ways: (1) where the legs are vertical and horizontal and all the vertices are on the integer lattice points, and (2) where the diagonal is horizontal and the endpoints of the dia

Recommends D.R. · 2023-07 · Mathematics Stack Exchange · on “Proving square roots irrational using middle school level math”

For future visitors, this geometric proof can also be phrased in terms of "shrinking shapes on an integer lattice grid" (as I wrote up [LINK] ); this proof technique has some other beautiful incarnations: see this beautiful video of Mathologer "What does this prove? Some of the most gorgeous visual "shrink" proofs ever invented" [this video]

Cites Servaes · 2020-10 · Mathematics Stack Exchange · on “Finding a regular 16-sided polygon whose vertices are lattice points”

Let $L$ be a lattice and let $v_1,\ldots,v_{16}\in L$ be the vertices of such a regular $16$ -sided polygon of minimum side length $m$ , in order, so that $|v_{i+1}-v_i|=m$ for all $i$ . Because $L$ is a lattice also $v_{i+1}-v_i\in L$ , and the points $v_{i+1}-v_i$ together form a regular $16$ -sided polygon. The diameter of this polygon is $2m$ , so its side length is strictly smaller than $m$ , contradicting the minimality of $m$ . Hence no such polygon or lattice exists. This idea is illustrated quite clearly for the regular octagon in this animation [this video] .