0.999...

Part of a convergent series on Mathematics

1+1=11

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We are number one

John Polkinghorne JBS Haldane W.V.O. Quine James A. Garfield v - t - e

“”Number nine, number nine, number nine, number nine, number nine, number nine, number nine, number nine, number nine, number nine, number nine, number nine, number nine, number nine, number

—"Revolution 9", The Beatles

0.999... is a number that has created a lot of confusion on the Internet due to the fact that it equals 1. Not "very close" or "infinitely close", but equal to 1. 0.999… is a never-ending sequence of 9s after the decimal point.

Number nine… number nine… number nine… like that Beatles song!

Understanding the real numbers

A real number is any number that is not an imaginary number (so none of those square roots of negative numbers). So this includes any number you learnt in primary school.

• -1/2
• π
• Square root of 2

All real numbers can be placed on an infinite number line.

The number line: can you guess where the numbers above will be placed?

However, there are different types of real numbers. The main difference is between "rational numbers" and "irrational numbers" (For now, let's forget about the negative numbers).

• A rational number is a number that can be described by a ratio of two whole numbers. So for example, 0.666… is a rational number because it can be described as a ratio of two positive whole numbers (which is 2/3).

• An irrational number is a number that cannot be described by a ratio of two whole numbers. So for example, the square root of 2 is irrational. There is no way you can describe this number as a ratio of two positive whole numbers. Seriously. Don't believe me? Try it yourself. Struggling? Yeah, it's impossible.

Decimal notation

This of course raises an interesting question: What would happen if I write those numbers above in decimal notation?

Three things could happen:

• The number has a string of digits and a decimal point but eventually ends.

  1/2 in decimal notation is 0.5

  1/25 in decimal notation is 0.04

• The number could originally start with arbitrary string of digits but eventually repeats over and over again (bold is for the repeating digit(s)).

  2/3 in decimal notation is 0.66666666666666…

  1/7 in decimal notation is 0.142857142857142857…

  7/30 in decimal notation is 0.233333333…

• The third way is that the number has a non-repeating string of digits that goes on forever. This is the case with irrational numbers.

  Square root of 2 in decimal notation is 1.41421356237…

  π in decimal notation is 3.1415926535…

Rational numbers

So what group is 0.999… in? 0.999… is an infinite string of numbers that repeats over and over again. So we can assume that 0.999… is the second group above. However we haven't proven that 0.999… is a rational number. So that leaves the interesting conjecture that this article is trying to prove:

PROVE that 0.999… is a rational number, meaning that it can be represented by p/q where p and q are both integers.

Proof that 0.999… is a rational number

Let's define 0.999… as x.

x = 0.999…

Let's multiply both sides by 10. Note that all this does is move the decimal point one place.

10x = 9.999...

Now let's split off the right hand side to two parts.

10x = 9 + 0.999¬

Now we can substitute 0.999… for x.

10x = 9 + x

Now subtract x from both sides.

9x = 9

Now divide both sides by 9.

x = 1

Oh look! If x equals 0.999… and x equals 1, then that means…

1 = 0.999…

Rigorous proof using real analysis techniques

Recall that the value of an infinite series is determined by what the sequence of its partial sums converge to.

Now, we wish to consider the convergence of the series $\sum _{i=1}^{\infty }{\frac {9}{10^{i}}}$. This is clearly a monotone sequence, bounded above by 1 and below by 0, so it must converge by the Monotone Convergence Theorem. In particular, it converges to 1, because 1 is the supremum of the set of partial sums.

Now, notice that the nth partial sum, which we will define as $a_{n}$ is given by $a_{n}=\sum _{i=1}^{n}{\frac {9}{10^{i}}}$

To see that 1 is indeed the supremum, suppose for contradiction that $a<1$ is the supremum of $\{a_{n}:n\in \mathbb {N} \}$. Let $d=1-a$. Let $N\in \mathbb {N}$ be such that ${\frac {1}{10^{N}}}<d$. Now, notice that the term $a_{N}$ is actually larger than the supremum, which is a contradiction. The proof of the case that $a>1$ is immediate.

Refusal to accept proof

Despite being mathematically proven that 0.999… equals 1, there are still people out there who refuse to accept that 0.999… equals 1. There have been tons of other proofs on the internet that show that 0.999… equals 1. However, some internet proofs claim to show that 0.999… does not equal one as well. This is due to them making the claims that 0.9 + 0.09 + 0.009 + 0.0009 + … approaches 1, but never hits it. Some also claim 1/3 isn't exactly 0.333… but just above it. This is, however, due to a failure to understand the higher-math concept of limits, which for example are essential to understanding calculus.

See also

• Science and Math Defeated — a denialist site or Poe's Law

External links

• Every PROOF you've seen that .999... = 1 is WRONG