Removing one 1 from real number $0.111111......$

Question

Let us say that there is a real number $0.1111......$ where $1$ gets repeated after 0. Now we remove one $1$ out of all $1$'s. Would the new real number be equal to the old real number, or would it be different? Also, what would this mean for bijection of digits of the old number and new number?

By removing 1, I mean: so we have $0.1111....$. We remove the first digit after decimal point and shift the second digit after decimal point to the first digit. Then the number will start from 1 after decimal point again.

Answer 1

Assuming that you talk about decimal expansion, this is really just $\frac19$.

What does "removing $1$" means? It means that we multiply the number by $10$, then look at the fractional part. Namely, $10\cdot\frac19=1+\frac19$. The fractional part is again $\frac19$.

Recall that the expansion of digits is indexed by integers, and we can always assume there are infinitely many digits, just that sometimes most of them are $0$ and can be omitted. Since all numbers have the decimal expansions indexed by the natural numbers, of course these are in bijection.

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Here's a similar question, write down all the integers, $0,1,2,3,4,\ldots$, in a list. Now write below that list another list with all the positive integers, $1,2,3,4,\ldots$

Do the two lists contain the same number of elements? Namely, is there a bijection between the two lists? Of course there is. $n\mapsto n+1$. And what you do here is shift the indices of the decimal digits by one, the same principle applies. However since the digits are all $1$, you don't really change the value of the number.

Answer 2

The question is essentially: when we had infinite sequence
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ …
of identical elements, and then delete the first element
 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ …
and then shift (formerly) second element to the first place, third to second, and so on
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ …
then will we obtain the same sequence?

Once I listened to a crank, and, sadly, on rather respectable venue (but he had a protection of an influential academician) who argued that we will obtain another infinite sequence, one element shorter ☺

But mainstream mathematics, with centuries-established notions of countable infinity, asserts that we will obtain exactly the same sequence.

The simplest argument is a bijective mapping of a subset onto entire (countable) set. Let $f(n) = n-1$, then it maps the subset {2, 3, 4, 5, … } onto {1, 2, 3, 4, 5, … } bijectively (note that these numbers refer to positions in the sequence, not to elements of it). Bijectivity follows both from set theory and formal arithmetics.

Answer 3

$0.1111\ldots=\dfrac{1}{9}$

If one $1$ is removed the result is $\dfrac{1}{9}-10^{(-n)}$ where $n$ is the position of the removed $1$.