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

Question

Salma Bradley · Accepted Answer

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^{- r n}

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^{- r n}

is rational and

\sum_{n = 0}^{\infty} (d_{1} . . . d_{r}) 10^{- r n} = \sum_{n = 0}^{\infty} {(\frac{d_{1} . . . d_{r}}{10^{r}})}^{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.

How can we know that the decimal expansion of an

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