Rational Numbers and Repeating Decimals
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Dr.Peterson
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What is 1/9 in base 9? How about base 3?
Surely you know that 1/9 > 1/10; what are you asking about that?
0.999... is defined as the sum of the geometric series 9/10 + 9/100 + 9/1000 + ..., whose sum is 1.
Surely you know that 1/9 > 1/10; what are you asking about that?
0.999... is defined as the sum of the geometric series 9/10 + 9/100 + 9/1000 + ..., whose sum is 1.
pka
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To understand this one needs to know and understand the theorem: Given that \(|r|<1\) then the sum \(\sum\limits_{k = J}^\infty {{r^k}} = \dfrac{{{r^J}}}{{1 - r}}\)
Now \(.9999\cdots=\)\(\sum\limits_{k = 1}^\infty {9 \cdot {{10}^{ - k}}} = \dfrac{{9 \cdot {{10}^{ - 1}}}}{{1 - {{10}^{ - 1}}}} = \dfrac{9}{{10 - 1}} = 1\)
The difference between 1/9 and 1/10 is 0.11111...... How does this relate to what you said about the sum of the geometric series??
And what is the decimal form of 1/9 in base 9??
Steven G
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Not so! Try again, how much is 1/9-1/10? Isn't 1/9 = .111...?
HallsofIvy
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No! It isn't! 1/9= 0.11111... The difference between 1/9 and 1/10 is 0.11111...- 0.1= 0.01111.... Writing x= 0.01111..., 10x= 0.1111... and 100x= 1.1111.... Subtracting, 100x- 10x= 90 x= 1 so x= 1/90.
\(\displaystyle \frac{1}{9}- \frac{1}{10}= \frac{10}{90}- \frac{9}{90}= \frac{1}{90}\).
Equivalently 0.0111111= 0/10+ 1/100+ 1/1000+ .... If we factor out 1/100, that is 1/100(1+ 1/10+ 1/100+ ...). 1+ 1/10+ 1/100+... is a geometric series, \(\displaystyle \sum_{n=0}^\infty ar^n\) with a= 1, r= 1/10. The sum is \(\displaystyle \frac{a}{1- r}= \frac{1}{1- 0.1}= \frac{1}{9/10}= \frac{10}{9}\) so multiplying by \(\displaystyle \frac{1}{100}\) we get 10/900= 1/90.
Rather than give you a formally correct answer, I’ll give you TWO intuitive answers.
[MATH]\dfrac{1}{3} = 0.333... \implies \\ 3 * \dfrac{1}{3} = 3 * 0.333... \implies \\ 1 = 0.999...[/MATH]Or
[MATH]\dfrac{1}{9} = 0.111... \implies \\ 8 * \dfrac{1}{9} + \dfrac{1}{9} = 8 * 0.111... + 0.111... \implies \\ 9 * \dfrac{1}{9} = 0.999.... \implies \\ 1 = 0.999... [/MATH]I stress these are not formal proofs. Those require standard or non-standard analysis.
The problem is infinity. If we think of any finite number n of 9’s to the right of 0., that obviously differs from 1 by 0. Followed by n-1 zeroes and then a 1. But when we get to an infinite number of 9’s, there is no place for the 1 to go.
[MATH]\dfrac{1}{3} = 0.333... \implies \\ 3 * \dfrac{1}{3} = 3 * 0.333... \implies \\ 1 = 0.999...[/MATH]Or
[MATH]\dfrac{1}{9} = 0.111... \implies \\ 8 * \dfrac{1}{9} + \dfrac{1}{9} = 8 * 0.111... + 0.111... \implies \\ 9 * \dfrac{1}{9} = 0.999.... \implies \\ 1 = 0.999... [/MATH]I stress these are not formal proofs. Those require standard or non-standard analysis.
The problem is infinity. If we think of any finite number n of 9’s to the right of 0., that obviously differs from 1 by 0. Followed by n-1 zeroes and then a 1. But when we get to an infinite number of 9’s, there is no place for the 1 to go.
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Steven G
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Try to do the division. Do it like you normally do but in base 9. 1*1=1, 2*1 = 2, 3*3 =10, 7*8=62, etc
D
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In base 9, the number symbols are 0, 1, 2, 3, 4, 5, 6, 7, 8, 10, 11.... there would not be 9 in there.....
Steven G
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What will we do without 9. That is my favorite number. John Lennon even wrote a song called number 9. Poor 9
D
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7 will go hungry - no 9 to eat.
Steven G
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Here is a base 9 joke. 7 8 10
I meant that 1/9 - 1/10 = 0.01111...
Dr.Peterson
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Technically, "decimal" means "base ten", but I know what you mean.
I asked you this question myself, because you would learn more by trying to think for yourself than just being told the answer. But it looks like someone has to tell you, since you don't seem willing to try, or even to tell us how much you know about "decimals" in other bases, on which I could base an answer.
In base 9, the place values to the left of the point are 1, 9, 81, 729, ...; the place values to the right of the point are 1/9, 1/81, ... . So in base 9, the number we call 1/9 would be written as 0.1! That's all there is to it.
This is really helpful. I see the connection to the geometric sequence somewhat. I definitely need to take more time to think about the idea of the sum of geometric sequences because I am not familiar with it.
Using what you said above, I tried seeing 9 times what number would give me 1 - multiplying in base 9. My answer was 0.1.
1/9=x
9x=1
9 X 0.1=1 (in base 9)
Therefore, I guess, 1/9 as a decimal is 0.1 in base 9. Was that what you were asking me to do?
I did not have a chance to answer your question right away. That does not mean that I'm not willing to try. I've been trying to make sense of this question, but because I have never studied the sum of geometric sequences before, I need a little more time to think about what you all are saying.
Jomo, I see that these are the place values in base 9. I also see that you wrote them as powers of nine. Is it true, that technically the decimal system is used for base ten only, therefore we wouldn't be able to use 0.1 to refer to 1/9 in base 9?