3 proofs that 0.999... is the same as 1

First of all writing the number 0.999... isn't not valid but for the sake of this post let's allow it.

Proof 1:

This is probably the most easy to understand.

Let's assume:
Next we'll multiply both sides by 10, so we have:
Let's subtract on both sides.
Since we get:
We now can divide each side by 9 because it's not equal to zero and we get:

Because we assumed that and we have that that means that
Proof 2:

Because the real numbers are dense, that means that you can find always a real number between two real numbers. For example we can use the arithmetic mean:
Let's assume that . That means that with the arithmetic mean we can find a number in between, with that we would get ? Which isn't a real number, there is nothing like a "last digit". And even it there was, that number wouldn't be in between 0.999... and 1. That means that our initial assumption is wrong thus 0.999... is equal to 1.

Proof 3:

This is basically the actual proof because it's per definition the same. Let's look at what we mean when we write 0.999... it means that ever digit after the 0 is 9. Let's rewrite that.
We can rewrite that further to
We can add "0" since it doesn't change the value, or more specifically let's add 9 and subtract 9. With that we get:
Let's rewrite it again
We can pull 9 outside of the sum to get:
To write it a bit more abstract we get:
That this is 9 times the sum from i = 0 to infinity over (1/10)^i and from that we subtract 9 again.
This sort of sum is called geometric series, it's of the form:
The sum from i = 0 to infinity over q^i. There's a formula for the result of the sum if |q| < 1:
In our case q = 1/10 which is smaller than one, so we can use that formula here and we get:
with q = 1/10 we have:
When we simplify it we get
Thus by writing 0.999... it's the same as 1. And because that's ambiguous, it's not allowed to write it so that every number written as digits is unique.