# 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:

Because we assumed 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:

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.

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:

Proof 1:

This is probably the most easy to understand.

Let's assume:

x = 0.999...

Next we'll multiply both sides by 10, so we have:10 x = 9.999...

Let's subtract x

on both sides.10 x - x = 9.999... - x

Since x = 0.999...

we get:9 x = 9.999... - 0.999... = 9

We now can divide each side by 9 because it's not equal to zero and we get:x = 1

Because we assumed that

x = 0.999...

and we have that x = 1

that means that 1 = 0.999...

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:

(a + b) / 2

Let's assume that 0.999... ≠ 1

. That means that with the arithmetic mean we can find a number in between, with that we would get 0.999...5

? 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.

0.999... = 0.9 + 0.09 + 0.009 + ...

We can rewrite that further to= 9 * 0.1 + 9 * 0.01 + 9 * 0.001 + ...

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:= 9 * 1 + 9 * 0.1 + 9 * 0.01 + 9 * 0.001 + ... - 9

Let's rewrite it again= 9 * (1/1) + 9 * (1/10) + 9 * (1/100) + 9 * (1/1000) + ... + 9

= 9 * (1/10)^0 9 * (1/10)^1 + 9 * (1/10)^2 + 9 * (1/10)^3 ... + 9

We can pull 9 outside of the sum to get:= 9 * (1/10)^0 9 * (1/10)^1 + 9 * (1/10)^2 + 9 * (1/10)^3 ... + 9

= 9 * [(1/10)^0 + (1/10)^2 + (1/10)^3 + ...] - 9

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:

1/(1-q)

In our case q = 1/10 which is smaller than one, so we can use that formula here and we get:= 9 * [1 / (1 - q)] - 9

with q = 1/10 we have:= 9 * [1 / (1 - 1/10)] - 9

When we simplify it we get= 9 * [1 / (9/10)] - 9 = 9 * [10 / 9] - 9 = 10 - 9 = 1

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.