# Why real numbers doesn't end with infinitely many 9s?

In the text TAOCP

A real number is a quantity x that has a decimal expansion $$x = n + 0.d_1d_2d_3...$$ where n is an integer, each $d_i$ is a digit between 0 and 9, and the sequence of digits doesn't end with infinitely many 9s

Why is such a case ? Isn't all the numbers between 0 and 1 are all real ? Then the numbers which ends with infinitely many 9s should be real too, Isn't it ?

• Note that $0.9999\ldots = 1$, so those numbers are already counted using digit sequences ending in infinitely many $0$s – Hagen von Eitzen Nov 11 '17 at 11:23
• The problem is that terminating decimals, like $1$, can also be written as $.999\dots$. So if you allow your representation to end with infinitely many $9's$ then the representation isn't always unique. – lulu Nov 11 '17 at 11:23

## 1 Answer

Yes, they are real. But note that $1=0.999999\ldots$, that $1.23=1.22999999\ldots$ and so on. What whappens is that every real has one and only one decimal expansion except those numbers which can be written as $\frac a{10^n}$ with $a\in\mathbb{Z}\setminus\{0\}$ and $n\in\mathbb Z$. These ones have two decimal expansions (as in my two examples above). A natural option then consists in not to consider the decimal expansions which end with infinitely many $9$'s.

• Why not $1 \approx 0.999999...$? It looks like they are different. – Abhisek Nov 11 '17 at 11:36
• @Abhisek They or equal, because$$0.999999\ldots=\sum_{n=0}^\infty\frac9{10^n}=\frac9{10}\sum_{n=0}^\infty\left(\frac1{10}\right)^n=\frac9{10}\times\frac1{1-\frac1{10}}=1.$$ – José Carlos Santos Nov 11 '17 at 11:39