Deleting digits from an irrational number

Question

Is it true that by deleting infinitely many appropriate digits out of the decimal representation of any positive irrational number, we can always get back the original number?

Answer 1

Yes, because you can skip past all the digits that occur finitely many times (say inifinitely many digits are 1,2,5,7,9 and finitely many digits are 0,3,4,6,8$.

After that you are left with a digit where all digits that occur occur infinitely often.

Say it is:

$$.a_1a_2a_3\dots$$

Then you keep $a_1$, and then delete from the $a_2$ to right before the next $a_2$. Keep the next $a_2$ and delete from that $a_2$ to the next occurrence of $a_3$. Repeat forever. You can do so because you know that each $a_i$ occurs infinitely often (since you've skipped past the digits that occur finitely often.)

For rational numbers, you can delete a finite number of digits and get the same value back. That is not possible for an irrational...

So you need to know that in any irrational number, there is at least one digit that occurs infinitely often. This is actually true for all real numbers, if you count repeating $0$s as "infinitely often."

For example, if the digits were:

$$0.123412345345534555345555\dots$$

with the number of times the $5$ occurs increasing each time, we keep the digits $123412$ because digits $1,2$ occur finitely often.

Then we delete from $3$ to next $3$ yielding:

$$0.123412[345]345534555345555\dots$$

Then you keep that next $3$ and delete from $4$ to the next $4$:

$$0.123412[345]3[4553]4555345555\dots$$

The you need the next digit to be $5$, so you delete one $5$ and the next digit is $5$:

$$0.123412[345]3[4553]4[5]55345555\dots$$

You need the next digit to by $3$, so you delete one fie:

$$0.123412[345]3[4553]4[5]5[5]345555\dots$$

The next digit has to be four, so we delete $455553$ etc.

At every step past that initial fixed set of digits, we delete at least one letter, and we ensure the next digit is "correct."