kramberol

2022-06-29

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

Expert

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
$\sum _{n=k}^{\mathrm{\infty }}\left({d}_{1}...{d}_{r}\right){10}^{-rn}$
where ${d}_{j}\in \left\{0,...,9\right\}$ are the digits in the period and r its length. Clearly this expression is rational if and only if $\sum _{n=0}^{\mathrm{\infty }}\left({d}_{1}...{d}_{r}\right){10}^{-rn}$ is rational and
$\sum _{n=0}^{\mathrm{\infty }}\left({d}_{1}...{d}_{r}\right){10}^{-rn}=\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{{d}_{1}...{d}_{r}}{{10}^{r}}\right)}^{n}=\frac{1}{1-\frac{{d}_{1}...{d}_{r}}{{10}^{r}}}$
is rational as a geometric series. Note $\frac{{d}_{1}...{d}_{r}}{{10}^{r}}<1$. Since there are (quite sophisticated) proofs that π 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.

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