is there any way to represent irrational numbers with a finite amount of integers?

Question

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 $\sqrt{2}^\sqrt{2}$.

Answer 1

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.

Answer 2

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.

Answer 3

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.