Please do not down vote this question. It may be stupid, but I wonder. Why is it that we cannot multiply $3.99999\cdots$ by $4$ and write $16,....$?
Asked
Active
Viewed 217 times
0
-
1Yes that was what I meant. But it is late here so I am sleepy :) – Marion Crane Nov 23 '14 at 06:19
-
Shall I edit it? – Marion Crane Nov 23 '14 at 06:20
2 Answers
4
Because it's not $12$ but $16$ ( Previously it was written as $12$ )
$$$$
Way 1
$$\begin{align}x&=3.99999....\\ 10x&=39.99999....\\ 9x&=36\\ x&=\frac{36}{9}\\ x&=4\\ 4x&=4\times4\\ 4x&=16\\ \end{align}$$
Way 2
$$\begin{align}x&=3.9999\cdots\\x=3+\frac{9}{10}+\frac{9}{100}+\cdots\\ x-3&=\frac{9}{10}+\frac{9}{100}+\cdots\\ x-3&=\frac{\frac{9}{10}}{1-\frac{1}{10}}\\ &=1\\ x&=4\\ \end{align}$$

Aditya Hase
- 8,851
-
So you can multiply $3,9999...$ by 10? That was what was interesting to me – Marion Crane Nov 23 '14 at 06:06
-
-
I do not know. I thought there was something like an epsilon delta proof that 3,999... was approximately 4, so that intrigued me – Marion Crane Nov 23 '14 at 06:10
-
1There's also that proof, but it's easy to prove that shifting the decimal point multiplies the value by $10$ for non-terminating decimals - essentially each "partial sum" (i.e. truncating the decimal) is $10$ times the truncation of the original string, so the latter will converge to $10$ times the former. – Milo Brandt Nov 23 '14 at 06:13
4
Hint:
$u_n=3.\underbrace{999\ldots9}_{n- \text{times}}$
$U=\{u_n\mid n\in \mathbb{N}\}\implies\sup U=4$