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A geometric proof of the irrationality of $\sqrt{2}$ works by constructing two right isosceles triangles with legs $n$ and hypotenuse $m$, and finding in the construction similar triangles with legs $m-n$ and hypotenuse $2n-m$. This shows that $m$ and $n$ can't both be integers as it would lead to infinite descent.

Diagram

I realise this is really a proof that the hypotenuse-leg ratio of a right isosceles triangle is irrational, and doesn't use the fact that this ratio is equal to $\sqrt{2}$, but that's an aside.

I've been trying to find a geometric construction to prove the irrationality of $\sqrt{3}$ in a similar way. I would expect this to involve 90°–60°–30° triangles. But I keep hitting dead ends.

Can it be done?

Stewart
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  • there is an isosceles triangle with large angle $120^\circ$ and edges $1,1\sqrt 3.$ Because then $c^2 = a^2 + ab + b^2 $ Also https://en.wikipedia.org/wiki/Integer_triangle#Integer_triangles_with_a_120.C2.B0_angle – Will Jagy Aug 13 '23 at 16:24
  • A geometric construction alone does not tell us whether the number we construct is rational. We need to quantify it to decide that. – Peter Aug 13 '23 at 16:46
  • @Peter That's true, but this equally applies to the construction for $\sqrt2$. So I suppose what I'm really looking for is a geometric construction that can be quantified so as to prove that $\sqrt3$ is irrational in pretty much the same way as the construction I've given can be quantified so as to prove that $\sqrt2$ is irrational. – Stewart Aug 13 '23 at 16:57

1 Answers1

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With a figure like the one on the left, Apostol demonstrated that an $(n,n,m)$ right triangle gives rise to a necessarily-smaller $(m-n,m-n,2n-m)$ triangle. Thus, there is no smallest triangle with integer sides, hence no such triangle at all, implying that $m/n=\sqrt{2}$ cannot be rational.

Likewise, the figure on the right shows that an $(n,m,2n)$ right triangle yields a necessarily-smaller $(2n-m,2m-3n,2(2n-m))$ triangle, so that $m/n=\sqrt{3}$ cannot be rational.

Blue
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  • It seems so obvious now! I suppose the main difference between the two is that for $\sqrt2$, $m-n$ is pretty much given and $2n-m$ as the hypotenuse is deduced from the construction, whereas for $\sqrt3$, the hypotenuse $2(2n-m)$ is pretty much given, and the other leg $2m-3n$ is deduced from this. – Stewart Aug 15 '23 at 16:01
  • @Stewart: Indeed, I had pretty-much the same thought: The inherent shape of each triangle gives us one other side "for free", and the construction calculates the third. ... It's worth noting that that Apostol's construction can be adapted for an $(n,pn,m)$ triangle (integer $p\geq1$), and my variant for an $(n,m,qn)$ triangle (integer $q\geq2$), effectively showing that $\sqrt{p^2+1}$ and $\sqrt{q^2-1}$ are irrational. – Blue Aug 15 '23 at 16:55