## Is 0.9999… (ending with indefinitely many 9s) a real number?

Is 0.9999… (ending with indefinitely many 9s) a real number?

I think it is. However from Knuth’s
Fundamental Algorithms, I got the definition to ** Real Number** as
below:

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***is a digit between 0 and 9, and the sequence of digits doesn’t end with infinitely many 9s.**

*i*I was very wondering why there is a rule that those digits cannot be ending with infinitely many 9s, an interesting one but odd to me. Because according to this definition, 0.9999…. and any other number in this form are ruled out from real number set however, from any perspective of the math knowledge stuffed me, I think those numbers really are real numbers. Ummm… wired…

Okay then, let’s now first look into the nature of those numbers in this kind. Take 0.9999…(ends with indefinitely many 9s) as an example:

Let n = 0.9999….

then we have: 10 * n = 9.9999….

(10 * n) – n = 9.9999… - 0.9999…

9 * n = 9

Finally we have: n = 1

So from the proving
above, we could see 0.9999… is actually the number of 1, an integer!! Things are
getting clear, 0.9999… and 1 are same number, two representations of same entity
value. For Set theory, each element in a set must be unique, so if we are
talking about the **set** of real numbers, we have to remove the extra
duplicate representation of one real entity, and in this case, we abandon the
form of 0.9999… and the form of 1 successes.

Turn back to the first line of this post, is 0.9999… a real number? Yes off course it is. Because 0.9999… just is 1 and Knuth’s definition was just ruling out A representation of number 1 to be align to Set theory…(at least I think so..)

Sum it up: 0.9999… and 1 represent same number, the value of this number is 1 and this number is a real number.

posted on 2007-10-25 22:40 allen geng 阅读(...) 评论(...) 编辑 收藏