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How can I prove that $\pi$ has $2$ digits repeating an infinite number of times in its decimal expansion? Proving that $1$ digit repeats an infinite number of times is easy – $\pi$ is irrational, and by the definition of irrational number, a digit must be repeated an infinite number of times?

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HINT: Assume only one digit appears infinitely many times, then after finitely many decimal places we will have only the repeating digit appearing. This means that $\pi$ is periodic (has a repeating decimal pattern), hence it's rational, which is a contradiction.

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Well, if only one digit appears an infinite number of times, then $\pi$ is rational.


Suppose that only $d$ appears an infinite number of times.

Split $\pi$ into the following parts:

  • $A=$ the integer part, i.e., $\lfloor{\pi}\rfloor$
  • $B=$ the fraction part's until only $d$ appears

Let $N$ denote the number of decimal digits in $B$, then:

$$\pi=\frac{9A\cdot10^{N}+9B+d}{9\cdot10^{N}}$$

Hence $\pi$ is rational.


For example, suppose $\pi=3.14159\overline{2}$, with $d=2$:

  • $A=3$
  • $B=14159$
  • $N=5$

Then:

$$\pi=\frac{9\cdot3\cdot10^{5}+9\cdot14159+2}{9\cdot10^{5}}=\frac{2827433}{900000}$$

Hence $\pi$ is rational.

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  • $\begingroup$ S. Peter: Please see a more detailed explanation + an example in the updated answer. $\endgroup$ – barak manos Nov 2 '16 at 10:20

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