2
$\begingroup$

From Sipser's Introduction to the Theory of Computation:

A slight problem arises because certain numbers, such as 0.1999... and 0.2000..., are equal even though their decimal representations are different.

Can someone explain? Because that's news to me.

$\endgroup$
  • $\begingroup$ Yes. First prove that if for all $\epsilon > 0$, $|x-y| < \epsilon$, $x=y$. Now apply that to your situation. $\endgroup$ – MathematicsStudent1122 Sep 21 '16 at 20:06
  • $\begingroup$ Any hint please? $\endgroup$ – anonbdy Sep 22 '16 at 11:37
2
$\begingroup$

$$0.1999999...=\frac{1}{10}+\frac{9}{100}+\frac{9}{1000}+...$$ $$=\frac{1}{10}+\frac{9}{100}(1+\frac{1}{10}+\frac{1}{100}...)=\frac{1}{10}+\frac{9}{100}\frac{1}{(1-\frac{1}{10})}=\frac{1}{10}+\frac{1}{10}=\frac{1}{5}=0.2$$

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

We can write $$\begin{align} 0.1999 \cdots &= 0.1+0.09+0.099+ 0.999\cdots \\ &= \dfrac{1}{10}+\dfrac{9}{100}+\dfrac{9}{1000}+\dfrac{9}{10000} + \cdots\\ &= \dfrac{1}{10}+\dfrac{9}{10^2}+\dfrac{9}{10^3}+\cdots \\ &= \dfrac{1}{10}+9\left(\dfrac{1}{10^2}+\dfrac{1}{10^3}+\cdots\right) \\ &= \dfrac{1}{10}+\dfrac{9}{10}\left(\dfrac{1}{10}+\dfrac{1}{10^2}+\cdots\right) \\ &= \dfrac{1}{10}+\dfrac{9}{10}\left[\dfrac{1/10}{1-1/10} \right] \end{align}$$ since $$\dfrac{1}{10}+\dfrac{1}{10^2}+\cdots = \sum_{k=1}^{\infty}\left(\dfrac{1}{10}\right)^k = \dfrac{1/10}{1-1/10}$$ as given here (see equation (9)). The denominator "cancels out" the $9/10$ since $1 - 1/10 = 9/10$, and so we end up with $$\dfrac{1}{10}+\dfrac{1}{10}=\dfrac{2}{10}=0.2\text{.}$$

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

The point is that same numbers may have different expansions in decimal (or other) notation. We can use two different tricks to show why $0.1999 \dots = 0.2000 \dots $:

1) Let $x=0.1999 \dots$, thus $10x=1.999 \dots$ and then $10x-x=1.8$. Solving this simple equation you get $x=0.2=0.1999 \dots$ by previous definition.

2) Write down the expansion of the number (decimal): $0.1999 \dots= (1)10^{-1}+(9) 10^{-2}+(9) 10^{-3}+(9)10^{-4}+ \dots +(9 )10^{-n} + \dots = \frac{1}{10} + 9 \sum_{n=2}^{+\infty} 10^{-n} = \frac{1}{10} + 9 \left( \frac{1}{1-\frac{1}{10}} - 1 - \frac{1}{10} \right) = \frac{2}{10} = 0.2 $

| cite | improve this answer | |
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.