In base $\pi$, the number $\pi$ is $1\cdot \pi^1 + 0\cdot \pi ^ 0 $, which is equal to $10$.
So, is $\pi$ an irrational number in all bases or not?
In base $\pi$, the number $\pi$ is $1\cdot \pi^1 + 0\cdot \pi ^ 0 $, which is equal to $10$.
So, is $\pi$ an irrational number in all bases or not?
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The concept of rationality/irrationality (or in the case of $\pi$ also: transcendence vs algebraicity). is defined without reference to bases: A number is rational iff there exists a nonzero integer such that multiplying the given number with this integer results in an integer.
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The usual definitions of rational and irrational have nothing to do with the decimal representation of a number. Rather, a number is rational if it is the ratio of two positive integers, and irrational otherwise.
Using this definition, it turns out that a number is rational if and only if its decimal (base $10$) expansion is eventually repeating. This fact is true for any positive integer base greater than $1$, not just base $10$.
What about base $\boldsymbol{\pi}$?
Normally when we talk about bases, we mean a positive integer base greater than $1$. But of course if you allow an irrational base, then an irrational number no longer necessarily has a nonrepeating representation. Of course, you then have to define what exactly you mean by the digit representation of a number. What digits are allowed? And how do you choose which expansion to use if multiple expansions are possible?
Generally when people talk about bases, they mean integer bases. For any integer base, $\pi$ has an infinite expansion. This is because if you could write $\pi$ as a finite expansion in base $b$ (an integer), then it has the form $$\sum_{n\in\mathbb{Z}}a_nb^n$$ where only finitely many $a_n$'s are nonzero (every $a_n$ is an integer less than $b$). But this is then just a finite sum of rational numbers, hence a rational number.
Edit: As Hagen said, the notion of rationality is independent of a "base". However, you can prove that a number is irrational if and only if its expansion in every integer base is infinite. This is because for any rational number $a/b$, you can always write the integer $a$ as a finite expansion $a_nb^n + a_{n-1}b^{n-1} + \cdots + a_1b + a_0$, and hence $a/b$ has the expansion $a_nb^{n-1} + a_{n-1}b^{n-2} + \cdots + a_1 + a_0b^{-1}$.