Steven G
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I understand what you are saying but in my opinion, nothing in math is assumed to be understood, one needs to state any restrictions.
split - Convergence Test Help (unnecessary arguments)
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Steven G
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Yes, you and your professor are absolutely correct----but only if you have more 0's before than 1 than you have 9's after the decimal.
I think that I already proven you wrong by stating that 1.00001*.99999999999 > 1
AvgStudent
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Since [imath]0.99\bar{9}=1 [/imath], then [imath]1.00001\times 0.99\bar{9}=1.00001\times 1 =1.00001>1?[/imath]
Steven G
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Really? There are functions/series that have a bounded value/sum but at infinity the function/series approaches infinity.
Unfortunately, I can't think of any at the moment. Hopefully someone else will post some functions for us to see.
My mind keeps going to the union and intersection of sets for my example above.
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Saying that you have already proven me wrong, means that you have not read my long post.
1.00001*.99999999999 > 1 is one of my professor cases!
Don't forget that this method was used only when,
\(\displaystyle \lim_{x \rightarrow \infty} f(x)g(x) = 1\)
What you are talking about is another thing.
Believe it or not,
\(\displaystyle 0.99\bar{9} \neq 1\)
When \(\displaystyle 0.99\bar{9}\) is written as \(\displaystyle 0.99\bar{9} = 1\), it means that the limit of the sequence of partial sums as we add more and more terms is approaching \(\displaystyle 1\).
\(\displaystyle 0.99\bar{9} = \frac{9}{10} + \frac{9}{100} + \frac{9}{1000}..... = 9\left(\frac{1}{10}\right) + 9\left(\frac{1}{10^2}\right) + 9\left(\frac{1}{10^3}\right).....=\sum_{k=1}^{\infty}\frac{9}{10^k} = 1\)
This is not even a sum in the usual sense of the word, but rather the limit of the sequence of partial sums.
-------
\(\displaystyle 1.00001\times 0.99\bar{9} = 1.00001\bar{0}\times 0.99999\bar{9}\)
This is case d.
The position of the digit \(\displaystyle 1\) is not the last digit as you are comparing,
\(\displaystyle 0.99999\bar{9}\)
\(\displaystyle 1.00001\bar{0}\)
\(\displaystyle 1.00001\times 0.99\bar{9} > 1\)
topsquark
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Ummmm..... Yes it does. [imath]\displaystyle \sum_{k= 1}^{\infty} \dfrac{9}{10}[/imath] is perfectly well defined as a limit. Please let's not get into a big argument about this here. There are plenty of other threads on the site that do that.
-Dan
Steven G
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Sorry topsquark, but I just have to ask this question to nasi112.
If you think that 0.999ˉ <1, then please tell me any number of your choice that is in between 0.999ˉ and 1.
If you think that 0.999ˉ >1, then please tell me any number of your choice that is in between 1 and 0.999ˉ.
Unless you can do that, the debate is over.
If you think that 0.999ˉ <1, then please tell me any number of your choice that is in between 0.999ˉ and 1.
If you think that 0.999ˉ >1, then please tell me any number of your choice that is in between 1 and 0.999ˉ.
Unless you can do that, the debate is over.
I will answer your question if you can tell me how many 9s after the decimal point.
Steven G
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There are infinitely many 9s after the decimal point.
Steven G
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You, not me, wrote that \(\displaystyle 0.99\bar{9} \neq 1\)
So either \(\displaystyle 0.99\bar{9} < 1 or 0.99\bar{9} > 1\)
Please pick one and give me a number in between.
topsquark
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Yes. \(\displaystyle 0.99\bar{9} \neq 1\).
And \(\displaystyle 0.99\bar{9} = 1\) (means the limit of the sequence of the partial sums is approaching \(\displaystyle 1\) as you add more and more terms)
If you cannot answer my question, how am I supposed to answer your question?
No need.
I will stop here.
And don't forget that professor Steven G that the main discussion was about \(\displaystyle f(x) \times g(x) > 1\) or \(\displaystyle f(x) \times g(x) < 1\) when \(\displaystyle \lim_{x \rightarrow \infty} f(x)g(x) = 1\), but you jumped outside the box talking about a side post.
Steven G
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A number can not equal two different values.
Since you have not answered my question about finding a number in between I will not respond to any of your posts until you do so.
Cubist
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I think that a number with a repeating decimal (click) is just a number with a single value as @Steven G says. In particular read converting repeating decimals to fractions on that page (click)
Can you provide any reference to books/ sites that treat repeating/ recurring decimals as limits? (BTW: I'm not being argumentative or condescending, I'm just trying to help you out
)A simple
Not because \(\displaystyle \neq\) is wrong, but because they decided that any limit of partial sums will be treated as \(\displaystyle =\).
So by definition, it is accepted to write directly,
\(\displaystyle 0.99\bar{9} = 1\)
In fact,
\(\displaystyle 0.99\bar{9} = ***... = 1\)
What are these stars and what is going on behind the scene? (They are just the explanation of telling you that: Hey professor Steven, the sums are approaching the limit of \(\displaystyle 1\)).
Since \(\displaystyle 0.99\bar{9} = \sum_{k=1}^{\infty}\frac{9}{10^k}\).
And \(\displaystyle \sum_{k=1}^{\infty}\frac{9}{10^k}\) is the limit of the sequence of partial sums.
Then, \(\displaystyle 0.99\bar{9} = 1\), means the sums are approaching the limit of \(\displaystyle 1\).
Reference is any Calculus Book.
A simple definition that can be accepted worldwide by great mathematicians can force \(\displaystyle \neq\) to become \(\displaystyle =\).
Not because \(\displaystyle \neq\) is wrong, but because they decided that any limit of partial sums will be treated as \(\displaystyle =\).
So by definition, it is accepted to write directly,
\(\displaystyle 0.99\bar{9} = 1\)
In fact,
\(\displaystyle 0.99\bar{9} = ***... = 1\)
What are these stars and what is going on behind the scene? (They are just the explanation of telling you that: Hey professor Steven, the sums are approaching the limit of \(\displaystyle 1\)).
We are not concerned about repeating decimals, but rather about this specific number \(\displaystyle 0.99\bar{9}\).I think that a number with a repeating decimal (click) is just a number with a single value as @Steven G says. In particular read converting repeating decimals to fractions on that page (click)
Can you provide any reference to books/ sites that treat repeating/ recurring decimals as limits? (BTW: I'm not being argumentative or condescending, I'm just trying to help you out
Since \(\displaystyle 0.99\bar{9} = \sum_{k=1}^{\infty}\frac{9}{10^k}\).
And \(\displaystyle \sum_{k=1}^{\infty}\frac{9}{10^k}\) is the limit of the sequence of partial sums.
Then, \(\displaystyle 0.99\bar{9} = 1\), means the sums are approaching the limit of \(\displaystyle 1\).
Reference is any Calculus Book.
Cubist
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You wrote your number as a repeating decimal therefore we must surely be concerned with the notation of repeating decimals
?Your "Then" doesn't follow. Surely you intend something along the lines of...
Once the limit is evaluated, \(\displaystyle \sum_{k=1}^{\infty}\frac{9}{10^k} = 1 = 1.0 = 2^0 =0.\bar{9}\)
As far as I'm aware, the notation \(\displaystyle 0.99\bar{9}\) doesn't itself imply a limit. However, there are many limits that when evaluated that do yield a value of 1
This seems a little glib ?
D
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10 * \(\displaystyle 0.\bar{X}\) - \(\displaystyle 0.\bar{X}\) = 9 * \(\displaystyle 0.\bar{X}\) .................. and
10 * \(\displaystyle 0.\bar{X}\) = \(\displaystyle X + 0.\bar{X}\)
9 * \(\displaystyle 0.\bar{X}\) = \(\displaystyle X\)
\(\displaystyle 0.\bar{X}\) = \(\displaystyle X\)/9
when \(\displaystyle X\) = 9
\(\displaystyle 0.\bar{9}\) = 9/9 =1
No need to get tangled up in/with "limits". I was taught this in high school (wayyy before I was introduced to Zeno/Liebniz/Newton).
10 * \(\displaystyle 0.\bar{X}\) = \(\displaystyle X + 0.\bar{X}\)
9 * \(\displaystyle 0.\bar{X}\) = \(\displaystyle X\)
\(\displaystyle 0.\bar{X}\) = \(\displaystyle X\)/9
when \(\displaystyle X\) = 9
\(\displaystyle 0.\bar{9}\) = 9/9 =1
No need to get tangled up in/with "limits". I was taught this in high school (wayyy before I was introduced to Zeno/Liebniz/Newton).
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Hey professors. I respect all of your opinions, and I understand why it is hard for you to follow me.
Even if some of you are not professors, I do respect your knowledge. I do respect even the opinion and knowledge of all students of all levels.
I will explain to you again, and for the last time what is this all about and why \(\displaystyle 0.\bar{9} = 1\) is a limit of \(\displaystyle 1\), rather than pure \(\displaystyle 1\) regardless of the equality sign here.
When I say
\(\displaystyle \sum_{k=1}^{3}\frac{9}{10^k} = \frac{9}{10} + \frac{9}{10^2} + \frac{9}{10^3} = 0.999\) (pure sum)
This is called a sum, and I hope that we all agree.
When I say
\(\displaystyle \sum_{k=1}^{\infty} a_k = a + a_1 + a2 +.... = \) (finite value)
This is also called a sum, but please focus on this: It is not a sum in the usual sense of the word, but rather, the limit of the sequence of partial sums.
\(\displaystyle \sum_{k=1}^{\infty} a_k \) =
This equation says that as we add together more and more terms, the sums are approaching the limit of the smiley face.
Even if some of you are not professors, I do respect your knowledge. I do respect even the opinion and knowledge of all students of all levels.
I will explain to you again, and for the last time what is this all about and why \(\displaystyle 0.\bar{9} = 1\) is a limit of \(\displaystyle 1\), rather than pure \(\displaystyle 1\) regardless of the equality sign here.
When I say
\(\displaystyle \sum_{k=1}^{3}\frac{9}{10^k} = \frac{9}{10} + \frac{9}{10^2} + \frac{9}{10^3} = 0.999\) (pure sum)
This is called a sum, and I hope that we all agree.
When I say
\(\displaystyle \sum_{k=1}^{\infty} a_k = a + a_1 + a2 +.... = \)
(finite value)This is also called a sum, but please focus on this: It is not a sum in the usual sense of the word, but rather, the limit of the sequence of partial sums.
\(\displaystyle \sum_{k=1}^{\infty} a_k \) =
This equation says that as we add together more and more terms, the sums are approaching the limit of the smiley face.