The first number shown to be irrational in our world was supposedly the square root of 2. I know that the irrationality of π is likely out of reach for a civilization that hasn’t discovered irrationality at all, and e probably requires limits to even understand, but are there other, easily accessible irrational numbers that may civilization may have stumbled on before square root 2? How would they have encountered that number’s irrationality before root 2’s? I’d put some money on the golden ratio, but I can’t be sure, and I can’t think of the mechanism
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2$\begingroup$ It is just as easy to show that $\sqrt 5$ is irrational as it is to show that $\sqrt 2$ is irrational. (And, among others, $\varphi = (1 + \sqrt 5) / 2$ is the ratio between the diagonal of a regular pentagon and the length of a side, just like $\sqrt 2$ is the ratio between the diagonal of a square and the length of its side.) $\endgroup$– AlexPCommented Nov 4, 2021 at 2:19
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4$\begingroup$ ... And of course it is equally easy to show that $\sqrt 3$ is irrational, and it may have come from the ratio between the height of an equilateral triangle and the length of a side. $\endgroup$– AlexPCommented Nov 4, 2021 at 2:27
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1$\begingroup$ @AlexP $\sqrt{3}$ is also the ratio of the body diagonal of a cube to its side length. $\endgroup$– NyraCommented Nov 4, 2021 at 8:22
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$\begingroup$ @AlexP: I don't understand your argument. How does the fact that $\sqrt{2}$, $\sqrt{5}$, and $\sqrt{3}$ are related to regular polygon imply that they are equally easy to be proven irrational? The classic proof does not rely on geometry but on algebra. The fact that these three numbers are algebraic of degree 2, or the fact that $2$, $3$, and $5$ are prime, on the other hand, leads to some simple classic proofs. $\endgroup$– TaladrisCommented Jun 27, 2023 at 14:48
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$\begingroup$ @Taladris: The argument is twofold. First, it is equally easy to prove that the square root of 3 or the square root of 5 are irrational as it is to prove that the square root of 2 is irrational, using the same arithmetic proof. (Not algebra; the Grecians did not have algebra, only arithmetic, geometry and trigonometry.) Second, just as the square root of 2 is interesting because its link to a square, the square root of 3 or the square root of 5 are equally interesting. Which provides a direct and trivial answer to the question "are there other, easily accessible irrational numbers". $\endgroup$– AlexPCommented Jun 27, 2023 at 15:00
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3 Answers
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Via differences between converging series
I think it is quite a hard question to answer with the hard science tag, but here are my two cents.
So Pythagoreans before Hippapus of Metapontum believed that all numbers could be expressed as were all rational until Hippapus proofed otherwise according to wikipeadia. (Sorry for the bad reference but I can't read old greek) There is apparently some discussion whether he was the first but the time frame is relatively the same.
What is important here is that we nowadays call it the irrationality of the square root of 2 but that it was done in pure geometry. See Euclids Elements of Geometry for examples. So for proof of irrationality, you don't need algebra. Seeing as most proofs revolve around a logical proof of contradiction by making sides simultaneously even and uneven you have some prerequisites.
You need a society with knowledge of logical proofs, specifically contradictions, and a notation of even and uneven numbers. Furthermore, it would probably help if your society wrongly thinks that all numbers a rational and that you have some small indivisible unit as the base of numbers.
Besides a geometric proof, I would think a similar proof could probably be made with comparisons of converging series. This is conjecture but I think something like Zeno's paradox could be made to prove that numbers can be irrational. Finding proof that not all numbers are rational in this way seems quite logical since you are already dealing with increasingly smaller rational numbers.
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Loads more Square Roots
If you have discovered Pythagoras' Theorem then you encounter many irrational numbers as hypotenuses.
Draw a right triangle with whole number side lengths $a,b$ and the hypotenuse has length $\sqrt{a^2 + b^2}$. This number is usually but not always irrational.
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$\begingroup$ I don't think Pythagoras knew algebra. I thought he did the "original" argument for Pythagorian Triplets purely in geometry terms. $\endgroup$ Commented Nov 4, 2021 at 12:15
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$\begingroup$ @D.J.Klomp I don't know about Pythagoras but Euclid's elements has the $a^2 + b^2 = c^2$ version of the theorem written verbally in terms of side lengths. $\endgroup$– DaronCommented Nov 4, 2021 at 16:03
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$\begingroup$ Sorry, but as said the old Greeks didn't have algebra but geometry. Their proofs in geometry looked completely different and are usually translated to our modern algebra. If you are referring to Euclid's Elements of Geometry, Book 1, Proposition 47 (p46-47). ([This is the translation I found:][farside.ph.utexas.edu/books/Euclid/Elements.pdf], I don't think there is a single squared number in the book, certainly not in proposition 47. Calling a verbally written square the same as a value to the second power is a leap and far away from anything resembling a square root, in my opinion. $\endgroup$ Commented Nov 4, 2021 at 19:36
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$\begingroup$ @D.J.Klomp Yes a second power is very different from a square root. $\endgroup$– DaronCommented Nov 4, 2021 at 19:44
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$\begingroup$ The statement that $\sqrt{a^2+b^2}$ can be irrational needs to be proved, and I don't see why it would be simpler than proving that $\sqrt{c}$ can be irrational. $\endgroup$– TaladrisCommented Jun 27, 2023 at 14:51
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As others have mentioned, the standard proof that $\sqrt{2}$ is irrational works just as well for $\sqrt{3}$, $\sqrt{5}$, ... so your civilization would likely be pondering $\sqrt{2}$ first, since it naturally comes first in the sequence.
The only other plausible candidate I could imagine for a "first irrational" would require a much larger leap of imagination to conceive of: $\sqrt{-1}$.
But it has a proof that could arguably be considered easier, or at least slightly different:
Suppose $\sqrt{-1}$ were rational. This means that $$\sqrt{-1} = \frac{p}{q}$$ for some integers $p$, $q$. Squaring both sides and clearing denominators, we get $-q^2 = p^2$. That is, $p^2 + q^2 = 0$.
This is only solvable in integers by $p = q = 0$, but this solution is unacceptable since $q$ is a denominator and must be non-zero.
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1$\begingroup$ $\sqrt{-1}$ is not an irrational number. (At least, it is not an irrational number in our mathematics and in any kind of mathematics which makes sense.) (If is a Gaussian integer a.k.a. a complex integer.) $\endgroup$– AlexPCommented Nov 4, 2021 at 9:45
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1$\begingroup$ @AlexP It looks like you don't know what a rational number is. Fortunately, I spelled this out in the answer. Read it carefully. An irrational number is a number which is not a rational number. $\endgroup$– PriskaCommented Nov 4, 2021 at 10:16
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2$\begingroup$ @Priska You could define it that way, but this is in opposition to how it is defined and used all over the world: An irrational number is any real number which is not a rational number. $\endgroup$ Commented Nov 4, 2021 at 11:07
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2$\begingroup$ @Priska Don't play word games.The official definition taught in mathematics courses is that irrational numbers are any real number which isn't rational. There's a whole lot of difference between that and "it's not rational, so it's irrational." $\endgroup$ Commented Nov 4, 2021 at 11:54
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$\begingroup$ @Muschkopp OP is asking this question in the context of initial encounters with exotic numbers. I wouldn't assume that his use of "irrational" is intended to follow the modern convention, unless he says so. The discovery of the first irrational happened thousands of years before the reals were even defined. I will leave this for OP to clarify. $\endgroup$– PriskaCommented Nov 4, 2021 at 21:45