5

prove that one of the digits $1,2,\ldots,9$ occurs infinitely often in the decimal expansion of $\pi$. you may use without proof the fact that $\pi$ is irrational. It is recommended using proof by contradiction.

My attempt:

Supppose that $1$ does not occur infinitely in the decimal expansion of $\pi$ and that it only occurs once.

Since the starting decimals of $\pi$ is $3.14$ then $1$ has already occurred but since $\pi$ is irrational, the decimal expansion is infinitely long it means that $1$ has to occur in $\pi$ infinitely.

By contradiction $1$ must occur infinitely long in the decimal expansion of $\pi.$

Note:

I realised my attempt is foiled since one argument to my attempt could be; what if $1$ never occurs again, but $2$ occurs infinitely often?

Question:

How do you proof what I am trying to proof by contradiction?

Michael Hardy
  • 1
user2510809
  • 535
  • 1
    In fact it is still an open question whether there are infinitely many $1$s in the decimal expansion of $\pi$ (not that anyone seriously believes there are only finitely many, but proving it is not so easy). – Erick Wong Jul 08 '16 at 05:48

2 Answers2

7

$\pi$ is irrational, so its decimal expansion does not terminate. But if every digit except $0$ occurs only finitely many times...

Arthur
  • 5,524
5

If each of the digits $1$ through $9$ occurs only finitely many times, then all of them together occur only finitely many times: the sum of nine finite numbers is finite. After all of them are done occurring you have only $0$s. But that cannot happen with an irrational number.

Michael Hardy
  • 1