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Does the following provide a proper proof for the irrationality of $\sqrt2$? It might be a good prelude to the Pythagorean theorem and real numbers.

We know that the hypotenuse of a right isosceles triangle is $b=\sqrt{2}a$, where $a$ is the side length. To see this, we divide the area of a square into two right triangles: $$a^2 = \frac{1}{2}b\cdot \frac{b}{2}+ \frac{1}{2}b\cdot \frac{b}{2}.$$

Let us assume that the side length $a$ is the smallest possible integer such that the hypotenuse $b$ is also an integer. But if $b=\sqrt{2}a$, then we may multiply both sides by $(\sqrt2-1)$ and have $(2a-b)=\sqrt{2}(b-a)$ that represents a right isosceles triangle with the integer valued side length and hypotenuse that are smaller than $a$ and $b$. Thus, there is no integer $a$ such that $b=\sqrt{2}a$ is also an integer.

Hulkster
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    Looks like this geometric proof: https://en.wikipedia.org/wiki/Square_root_of_2#Geometric_proof – lhf Aug 07 '22 at 13:24
  • @lhf Apparently it's the same thing in another, possibly simplified way. – Hulkster Aug 07 '22 at 13:32
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    The geometry is totally irrelevant here. If you simply omit all mention of triangles and start with $\sqrt 2=b/a$ as usual the rest of the proof works. – David C. Ullrich Aug 07 '22 at 14:30
  • @DavidC.Ullrich Oh yes, very true. On the other hand, the tringle stuff shows how there is a geometric (true) length $\sqrt{2}$ that is to be investigated. Maybe this has some educational value before student are introduced to the Pythagorean theorem - and the Pythagorean tragedy of irrational ratios. – Hulkster Aug 07 '22 at 15:37
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    It is important before talking about $\sqrt 2$ to have a good reason why we think $\sqrt 2$ should exist at all. After all there is no reason to assume $orange juice + 1$, $\frac 10$, $\sqrt {-1}$ or $turtles \times blueberries$ exist so why should $\sqrt 2$? A vague concept of continuum is pretty important but vague and abstract and hard for a student to grasp what a big deal it is. Isosceles right triangles make it pragmatic (although it's really just the concept the geometry is measurable which is just the concept of continuum made plain). – fleablood Aug 07 '22 at 16:04
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    It's not a new proof, but I don't see the reason for the downvotes. We have plenty of questions about proving that $\sqrt 2$ is irrational, such as Prove that $2^{1/2}$ is irrational, but none of the ones I found take a geometric approach. Yes, Wikipedia has such a proof; this question is a good way to link to that. – David K Aug 07 '22 at 18:27
  • @DavidK Now I found this one https://math.stackexchange.com/questions/2316991/sqrt2-proved-irrational-by-denominator-descent-on-sqrt-2-1?rq=1 – Hulkster Aug 08 '22 at 03:12
  • Yes, that's another non-geometric proof. I think it's good to link to these for comparison. – David K Aug 08 '22 at 03:16
  • @DavidC.Ullrich . In the book A History Of Mathematics by Merzbakh & Boyer they suggest the Golden Ratio $\phi$ might have been the first known irrational by the method in this Q: We have $\phi=\frac {1}{\phi -1}$. If $\phi=\frac {a}{b}$ with $a,b\in \Bbb N$ and the least possible $a$, then $ a>b$ (as $\phi>1$), but then $\phi=\frac {1}{a/b-1}=\frac {b}{a-b}$ with a smaller numerator. – DanielWainfleet Aug 08 '22 at 03:51

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