0.999...

0.999... (or 0.9̅) is a real number equal to 1. The equality 0.999... = 1 is the subject of confusion among mathematics students who find it counterintuitive.

Proofs

Although not rigorous, this is possibly the simplest proof of the equation:

$0.333... = 1/3$

$0.999... = 3/3 = 1$

Another one relies on algebra:

$x = 0.999\ldots$

$10x = 9.999\ldots$

$10x - x = 9.999\ldots - 0.999\ldots$

$9x = 9$

$x=1$

To make a rigorous statement, we treat $0.999\ldots$ as a sum of a geometric series:

$0.999\ldots = 0.9 + 0.09 + 0.009 + \ldots = \sum_{i = 1}^\infty 9 \cdot 10^{-i} = \frac{9(\frac{1}{10})}{1 - \frac{1}{10}} = 1.$

Or Cauchy sequences:

$0.999\ldots$ is defined by the Cauchy sequence $\left(0, \frac{9}{10}, \frac{99}{100}, \frac{999}{1000}, \frac{9999}{10000}, \ldots\right)$, which converges to $1$. Therefore $0.999\ldots = 1$.

Convincing students

The fact that 0.999... = 1 — that one real number has multiple decimal expansions — can appear paradoxical. Some students will assert that 0.999... "gets infinitely close to 1" but never reaches it despite being constant. The best way to counterargue this belief is to ask, what is 1 - 0.999...? Here are some common responses:

• 0: This is the correct answer. It implies that 1 and 0.999... are equal.
• 0.000...1: This is a malformed expression. One cannot have an infinite string of zeroes with a 1 at the end — an infinite string means it has no end!
• The smallest real number: There is no such thing. The real numbers are densely ordered — between any two real numbers is another one.
  • If the student stubbornly insists that there is a smallest real number, then let $\varepsilon = 1 - 0.\bar{9}$. What is $\varepsilon / 2$? Is that a real number?
    • If it is, then $\varepsilon / 2$ is smaller than $\varepsilon$, and the latter is not the smallest real number.
    • If it is not, then the real numbers are not closed under multiplication, which violates a fundamental property in the construction of the real numbers.
  • Also, what is $1 / \varepsilon$? If $\varepsilon > 0$, then the reciprocal of $\varepsilon$ exists. "Infinity" is not a real number and does not constitute a valid answer.