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I have been looking at the popular proofs that rational numbers have limitations when trying to define real-world lengths.

For instance, there does not exist a rational $c$ such that $c^2=2$. Basically, proofs focus upon the fact that all square roots of natural numbers, other than of perfect squares, are irrational.

I would like to see other illustrations of proofs that show existence of irrational numbers without relying upon square roots of natural numbers. But on other areas of mathematics.

Veak
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  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Aug 11 '23 at 03:30
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    The issue is with how vague this is. The most natural "next step" of generalization would be all numbers of the form $\sqrt[a]{b}$, but even this doesn't cover everything: realistically, it only broaches into what we call "algebraic numbers" (those which can be the roots of a polynomial). Even beyond that (or rather, the opposite of that) you have the transcendentals, like $e$ or $\pi$ which, to my knowledge, aren't so easy to "generate" outside of the natural contexts in which they arise. (In part, this is because there is a rigorous sense in which "almost all" numbers are transcendental.) – PrincessEev Aug 11 '23 at 03:32
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    @PrincessEev *roots of a polynomial with integer coefficients (not all vanishing). Just being nitpicky :) – peek-a-boo Aug 11 '23 at 03:47
  • A valid and important nitpick nonetheless, that is my bad. – PrincessEev Aug 11 '23 at 03:48
  • The proof that $e$ is irrational is elementary and pretty easy to follow for a first-year calculus student. – Brian Tung Aug 11 '23 at 04:05
  • Would you be so kind to show me the one using roots of a polynomial with integer coefficients ? – Veak Aug 11 '23 at 04:24

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A rational number's decimal representation is terminating or eventually periodic. So, for example, consider $0.1101001000100001\ldots$, where the numbers of $0$'s between successive $1$'s are $0,1,2,3,4,\ldots$. This is neither terminating nor eventually periodic, so this number is irrational.

[EDIT] BTW, this particular example happens to be $$\frac{10^{\frac{1}{8}} \vartheta_{2}\! \left(0, \frac{1}{\sqrt{10}}\right)}{20}$$ where $\vartheta_2$ is a Jacobi theta function.

Robert Israel
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  • Can one show that rational number decimal representation is terminating or eventually periodic ? Where can I find it ? – Veak Aug 11 '23 at 04:15
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    One way to do it is via the pigeonhole principle - when you divide $p$ by $q$, there are $q$ possible remainders you can get. Therefore, when you calculate the decimal representation of $\frac{p}{q}$, once you reach the point where you're always appending a zero when you do the division, go another $q+1$ steps. Somewhere in those $q+1$ steps there must be two divisions using the same remainder, and so everything after that will form a loop and hence a repeating decimal expansion. – ConMan Aug 11 '23 at 04:31
  • Where can I find this? Would like to scrutinise it in more detail. – Veak Aug 11 '23 at 05:35
  • See e.g. Wikipedia – Robert Israel Aug 11 '23 at 14:00