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I'm noob in math. If 0.333(3) is 1/3, 0.666(6) is 2/3, then 0.999(9) is what?

If 3/3 and 0.999(9) is the same, then how can I express one of them without expressing the other?

N. F. Taussig
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Gintas_
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  • Then, you have 3/3 = 0.99(9) = 1. – Gregory May 21 '17 at 19:36
  • then how can I express one of them without expressing the other? – Gintas_ May 21 '17 at 19:36
  • What does that mean? They are equivalent expressions, it is similar to adding 0 or adding 5 and subtracting 5 – Gregory May 21 '17 at 19:37
  • it feels like there is some bug in the system. 0.999(9) is different from 1, at least logically. Why doesn't this logic transfer into math? – Gintas_ May 21 '17 at 19:38
  • They are provably the same. 0.99 repeating is 1 – mathfan27543 May 21 '17 at 19:42
  • Your answer there is that it is not different from one. You can happily write your infinitely long number and argue that it's different because optically, on a page, it looks different, but that doesn't actually make it so. – Matt May 21 '17 at 19:42
  • @Gintas_, $.999\cdots$ is different from 1 when using transcendental numbers, but they retain the same value. The difference between them is $0.000\cdots001$, which is $0$ – Jacob Claassen May 21 '17 at 19:42
  • @Gintas_ They are not logically different. They are apparently different, but if you think about it, an hypothetical $0.(9)$ would be, by definition, strictly larger than any number in the form $0.\underbrace{99\cdots9}_{\text{finitely many}}=1-10^{-n}$. But any number larger than all of those is, by necessity / logic / intuition (choose your favourite), larger or equal to $1$. –  May 21 '17 at 20:02

2 Answers2

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Well, notice that:

$$\text{n}_1:=0.333\dots=0.\overline{3}=\frac{1}{3}\tag1$$

So, when we muliply $\text{n}_1$ by two, we get:

$$\text{n}_2:=2\cdot0.333\dots=0.666\dots=0.\overline{6}=\frac{2}{3}\tag2$$

So, when we muliply $\text{n}_1$ by three, we get:

$$\text{n}_3:=3\cdot0.333\dots=0.999\dots=0.\overline{9}=\frac{3}{3}=1\tag3$$

Where $\overline{x}$ means a repeating decimal.

Jan Eerland
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$ 0.999 \ldots = 1 $

Proof: Let $x = 0.999\ldots $ . Now consider that $10x = 9.999\ldots$ . So then we have that $9x = 10x -x = 9.99\ldots -0.999\ldots = 9.0$ .

Thus $9x/9 = 9/9 = 1 $.

mathfan27543
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