There are many **proofs that the repeating decimal 0.999... is equivalent to the number one**, some more rigorous than others. Also with counter-proofs.

## Prerequisites

- The notion that 0.999... represents an infinitely repeating decimal—that is, the digit 9 repeats itself without end to the right of the decimal place.

## Sources

## Proof

### Sum/product of fractions

- $ \frac13=0.333\ldots $

Therefore:

- $ \frac13+\frac13+\frac13=0.333\ldots+0.333\ldots+0.333\ldots $
- $ \frac13=1=0.999\ldots $

Likewise:

- $ \frac19=0.111\ldots $
- $ 9\cdot\frac19=9\cdot0.111\ldots $

Multiplying each side by 9 we obtain:

- $ 1=0.999\ldots $

QED

The only refutation this example proof has is that which questions whether or not $ \frac13=0.333\ldots $ . Both one-third as an infinitely repeating set of three's and one as an infinitely repeating set of nine's are equally exact, and both must be taken on a little bit of faith. 1/3 being equivalent to a repeating threes rarely ever questioned, but is the same reasoning phenomenon - the same paradox. If one-third and one-ninth in decimal form are taken without question to be equal to their fractional counterparts, then why cant one as a decimal of nines?

### Conversion to fraction

- $ n=0.999\ldots $

Multiply by a factor of 10:

- $ 10n=9.999\ldots $

Subtract:

- $ 10n-n=9.999\ldots-0.999\ldots $
- $ 9n=9.000\ldots=9 $
- $ n=1 $

QED

The only refutation this example has is as follows. One might argue that when multiplying by ten, the right hand side shifts to the left a decimal place, leaving a terminating zero at the end of the infinite string of 9's ($ 10n=9.999\ldots9990 $). And, therefore, when subtracting $ 10n-n=9n=8.999\ldots991 $ (with a one after the last of infinite nines). But it is important to realize the meaning of infinite. There is no terminating 9 and therefore no placeholder after it. There will always be another 9.

### Infinite geometric series

$ 0.999\ldots=0.9+0.09+0.009\cdots $

$ 0.999\ldots=9\cdot0.1+9\cdot0.01+9\cdot0.001+\cdots $

$ 0.999\ldots=9\cdot10^{-1}+9\cdot10^{-2}+9\cdot10^{-3}+\cdots $

$ 0.999\ldots=9\cdot(10^{-1}+10^{-2}+10^{-3}+\cdots) $

$ 0.999\ldots=9\cdot\sum_{n=1}^\infty 10^{-n} $

$ 0.999\ldots=9\cdot\sum_{n=1}^\infty\left(\frac{1}{10}\right)^n $

Evaluating infinite geometric series is easy when utilizing the theorem:

- $ \sum_{n=1}^\infty r^n=\frac{r}{1-r} $

Therefore:

$ 0.999\ldots=9\cdot\sum_{n=1}^\infty\left(\frac{1}{10}\right)^n=9\cdot\frac{\frac{1}{10}}{\left(1-\frac{1}{10}\right)} $

$ 0.999\ldots=9\cdot\frac{\frac{1}{10}}{\frac{9}{10}} $

$ 0.999\ldots=9\cdot\frac19=1 $

QED

### Argument from averages

The average $ A $ of two numbers $ m,n $ is found by adding them and dividing by two.

- $ A=\frac{m+n}{2} $

The average is larger than the smaller number $ m $ , but smaller than the larger number $ n $ .

- $ m<A<n $

If $ m=n $ , then $ A=m=n $

Assuming that 0.999... is less than 1, the average between the two is:

$ A=\frac{0.999\ldots+1}{2} $

$ A=\frac{1.999\ldots}{2} $

$ A=0.999\ldots $

Since $ A=m $ , then so does $ A=n $ , thus $ m=n $ : $ 0.999\ldots=1 $

QED

### Argument from philosophy

The definition of the real numbers as a continuum:

If two numbers $ x,z $ exist, such $ x\ne z $ and $ x<z $ ; there must exist a third number $ y $ in between such that $ x<y<z $

This is saying that if two numbers are not equal, there is a third number that is also unequal and that can fit in between them on the number line. Regardless of the type of real number or the difficulty in computing its value or representing its value, from a purely abstract perspective, there does exist a number that is larger than one but smaller than the other. It is impossible to find a "next higher" number that is both larger to a given value but couldnt have been smaller (see argument from averages).

If $ 0.999\ne1 $ , then what number $ y $ could exist in between them such that $ 0.999\ldots<y< 1 $ ?

Since there is no conceivable number that can exist in between the two, they must be equal according to the definition of the real numbers as a continuum.

$ 0.999\ldots=1 $

QED

There could also be numbers between 0.99999999..9...... and 1: like 0.9.....................91 (the dots represent infinite "9" repeating after each other.)

### Arithmetic proof

Evaluate the difference between 1 and 0.9999...

$ 1-0.999\ldots=0.000\ldots $

One might argue that after the infinitely many zeros, there is going to be a 1 ($ 0.000\ldots0001 $). But it is important to grasp what "infinite" means. The 9's are infinite, there is no terminating number at the end. The zeros are also infinite, there is no 1 at the end. There is no "end", there will also be another 9 or another 0.

An infinite string of zeros past the decimal is still just 0:

$ 1-0.9999\ldots=0 $

Since any number subtracted from an equal value is zero: $ m-m=0 $

Algebraic rearranging:

$ 1=0.999\ldots $

QED

It is not a law in Mathematics that there could not be an end to the result of $ =0.999\Idots $