How can we know that the decimal expansion of an irrational number will never repeat?

Question

How can we know that the decimal expansion of an irrational number will never repeat? For example what if the value of π after some quintillions of digits start repeating.

Answer 1

Here are proofs that $\pi$ is irrational.

Here are proofs that $\sqrt2$ is irrational.

Basically, to show that $\sqrt2$ is irrational, you can assume that it has the form $\frac pq$ with $p$ and $q$ coprime; then you can write this as $2p^2=q^2$, infer that $q$ is even, write $q=2k$ and get $p^2=2k^2$, from which you can infer that $p$ is even, contradicting the assumption that $p$ and $q$ are coprime.

To see that a real number with a repeating decimal expansion is rational, note e.g. $0.\overline{abc}=\frac{abc}{999}$.

Answer 2

If there appears a period in the decimal expansion of a number then You have after subtracting a rational term an expression of the form $$\sum_{n=k}^{\infty}(d_1...d_r)10^{-rn}$$ where $d_j\in\{0,...,9\}$ are the digits in the period and $r$ its length. Clearly this expression is rational if and only if $\sum_{n=0}^{\infty}(d_1...d_r)10^{-rn}$ is rational and $$\sum_{n=0}^{\infty}(d_1...d_r)10^{-rn}=\sum_{n=0}^{\infty}\left(\frac{d_1...d_r}{10^r}\right)^n=\frac{1}{1-\frac{d_1...d_r}{10^r}}$$ is rational as a geometric series. Note $\frac{d_1...d_r}{10^r}<1$. Since there are (quite sophisticated) proofs that $\pi$ is irrational (even transcendental, i.e. no root of any polynomial with rational coefficients,) one concludes that its decimal expansion cannot lead to any period.