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I know that rational numbers can be represented with two integers $\frac{a}{b}$. But is there any way to represent irrational numbers with an finite amount of integers?

My best guess is $\frac{a}{b} ^ \frac{c}{d}$. it can represent any root of any number, but I don't know if it can represent things like $\sqrt{2}^\sqrt{2}$.

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There are uncountably many irrational numbers but there are only countably many finite sets (or lists) of integers. So it is impossible to represent every irrational number using only finitely many integers.

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  • $\begingroup$ Great minds think alike (also, in this case, normal minds like mine :)) $\endgroup$
    – Clement C.
    Mar 4, 2021 at 23:03
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    $\begingroup$ @ClementC Sure, but you must type faster than me! $\ddot{\smile}$. $\endgroup$
    – Rob Arthan
    Mar 4, 2021 at 23:08
  • $\begingroup$ but... if you take normal ways of representing irrational numbers (with infinite integers), it is a contably amount of integers, can a uncountably set of numbers (irrationals) be represented as a countably set of countably sets? $\endgroup$
    – gdor11
    Mar 5, 2021 at 10:46
  • $\begingroup$ A countable union of countable sets is countable, e.g., see math.stackexchange.com/questions/603456/…. $\endgroup$
    – Rob Arthan
    Mar 5, 2021 at 13:04
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Suppose you have an encoding scheme allowing you to express a subset $C\subseteq \mathbb{R}$ of real numbers using, for each, a finite number of integers. Then you have a bijection between $C$ and $$ \cup_{k\in \mathbb{N}} \mathbb{N}^k $$ and since this is countable, then so is $C$. But the irrational numbers are uncountable.

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If ${\frac{a}{b}}^{\frac{c}{d}}=\sqrt{2}$ then $(\log{\frac{a}{b}})*{\frac{c}{d}}=\frac{log{2}}{2}$, which implies $\frac{a}{b}=2, \frac{c}{d}=\frac{1}{2}$.

But if ${\frac{a}{b}}^{\frac{c}{d}}=\sqrt{2}^\sqrt{2}$ then $\frac{c}{d}=\sqrt{2}$, and we would have found a rational number equal to $\sqrt{2}$. This is impossible, consequently there is no solution for ${\frac{a}{b}}^{\frac{c}{d}}=\sqrt{2}^\sqrt{2}$ among the rational numbers and, more generally, there can be no way of writing every real number using only a finite number of rational numbers and operations.

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  • $\begingroup$ The issue with this is that it doesn't address other possibles schemes of "writing a real number with a finite number of integers." You showed that for this scheme (rational to the rational) it doesn't work, but... one could then suggest another scheme. $\endgroup$
    – Clement C.
    Mar 5, 2021 at 0:13
  • $\begingroup$ following the same logic, $(\sqrt{2}^{\sqrt{2}})^{\sqrt{2}}$ is not $(2^{1})^{1}$ $\endgroup$
    – gdor11
    Mar 16, 2021 at 22:10

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