Instigator / Pro
1
1500
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8
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50.0%
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Topic
#4409

# 0.99999... = 1

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Finished

The debate is finished. The distribution of the voting points and the winner are presented below.

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1
0

After 1 vote and with 1 point ahead, the winner is...

Math_Enthusiast
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Standard
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5
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Two days
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Two weeks
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Contender / Con
0
1500
rating
1
debates
0.0%
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Description

Unfortunately, the last time I did this debate, I was a bit sloppy, and no one voted on it anyway. Let's see how things go this time... Put any questions on the topic or on definitions in the comments!

BoP is on pro.

Round 1
Pro
#1
Before I get in to my argument, I will note that it may be helpful for anyone who may be reading this to review your inequality rules.

Consider the following two properties:

1. x is greater than any number of the form 0.999...9. (With finitely many nines.)
2. x is the smallest number that satisfies (1).

0.99999... satisfies these two properties. I expect con to inform me if she disagrees with this. I ideally want to keep things simple, so I will only defend this if con challenges it, but for now, I will assume that it is sufficiently obvious. Now, certainly:

1 satisfies (1)

I intend to demonstrate:

1 satisfies (2)

Suppose that x is less than 1, and consider 1 - x, which will be a positive real number. By the Archimedean property, 1/(1 - x) < n for some natural number n. Then, taking reciprocals, 1 - x > 1/n. We certainly have n < 10^n and so 1/n > 1/10^n. Combining the inequalities 1 - x > 1/n and 1/n > 1/10^n, we have 1 - x > 1/10^n. It follows from that that x < 1 - 1/10^n = 0.999...9 with n nines. It follows that x does not satisfy (1), and therefore, 1 is the smallest such number, that is, 1 satisfies (2).

Now, every set has no more than one minimum, so since both 1 and 0.99999... are the smallest numbers for which (1) holds, they must be equal.
Con
#2
Cool. We'll start with this issue: assuming that there exists a value y (I suppose you're working from the real plane) that is the smallest value greater than 0.99..9
The most obvious rebuttal I can think of is a cute property of the real number system - it has no "gaps" [Completeness property or something https://en.wikipedia.org/wiki/Completeness_of_the_real_numbers]. That meaning, there is no smallest value y greater than 0.99..9 as there will always be a smaller value than that. It goes without saying that this is the case for 0.999...
It is an infinite decimal expansion, meaning, we can't speak of it as a minimum but a mere approximation to a well-defined minimum.

We can even work with the proof you've laid out. If we let n --> ထ, then it is obvious that 0.999... surely would not satisfy property 1. Thus, making your deduction partially flawed as we have not only stated from the beginning that x < 1 (thus excluding any possibility that x = 1) but also that x cannot satisfy property (1) - leading to  contradiction.
Round 2
Pro
#3
I can tell that con was a bit confused as to what my argument actually was, and perhaps it was a bit difficult to understand, so let's clear some things up.

Cool. We'll start with this issue: assuming that there exists a value y (I suppose you're working from the real plane) that is the smallest value greater than 0.99..9
The most obvious rebuttal I can think of is a cute property of the real number system - it has no "gaps" [Completeness property or something https://en.wikipedia.org/wiki/Completeness_of_the_real_numbers].
Now, to quote my own argument:

Consider the following two properties:

1. x is greater than any number of the form 0.999...9. (With finitely many nines.)
2. x is the smallest number that satisfies (1).
Observe the use of the phrase "of the form." I am not referring to one specific number but all numbers of a particular form. In particular, I am referring to the numbers 0.9, 0.99, 0.999, etc. Thus, property 1 can be phrased as follows:

1. x is greater than each of 0.9, 0.99, 0.999, etc.

Hopefully that is a bit clearly. As to the existence a number satisfying both (1) and (2), as I demonstrated in my previous argument, 1 is such a number.

We can even work with the proof you've laid out. If we let n --> ထ, then it is obvious that 0.999... surely would not satisfy property 1.
What is n here? I'm not sure what con is arguing.

Thus, making your deduction partially flawed as we have not only stated from the beginning that x < 1 (thus excluding any possibility that x = 1) but also that x cannot satisfy property (1) - leading to  contradiction.
I guess we need a lesson on use of variables. That is not the same "x." The "x" referenced in (1) and (2) is a local variable. It is simply standing in for any arbitrary number. The x that is defined to be less than 1 is a global variable. The point is, con is confusing the two uses of "x." From this point forward, I will avoid reusing variables, as clearly it causes confusion.

Conclusion:

I will end by making sure that there is no remaining confusion by presenting a revised version of my original argument. It is effectively the same argument, but it avoids the things which appear to have caused confusion. (e.g. reusing variables)

Consider the following two properties:

1. x is greater than each of 0.9, 0.99, 0.999, etc.
2. x is the smallest number that satisfies (1).

0.99999... satisfies these two properties. I expect con to inform me if she disagrees with this. I ideally want to keep things simple, so I will only defend this if con challenges it, but for now, I will assume that it is sufficiently obvious. Now, certainly:

1 satisfies (1)

I intend to demonstrate:

1 satisfies (2)

Suppose that y is less than 1, and consider 1 - y, which will be a positive real number. By the Archimedean property, 1/(1 - y) < n for some natural number n. Then, taking reciprocals, 1 - y > 1/n. We certainly have n < 10^n and so 1/n > 1/10^n. Combining the inequalities 1 - y > 1/n and 1/n > 1/10^n, we have 1 - y > 1/10^n. It follows from that that y < 1 - 1/10^n = 0.999...9 with n nines. It follows that y does not satisfy (1), and therefore, 1 is the smallest such number, that is, 1 satisfies (2).

Now, every set has no more than one minimum, so since both 1 and 0.99999... are the smallest numbers for which (1) holds, they must be equal.

Con
#4
"What is n here? I'm not sure what con is arguing."

Recall that I have said:
"We can even work with the proof you've laid out. "

"Suppose that x is less than 1, and consider 1 - x, which will be a positive real number. By the Archimedean property, 1/(1 - x) < n for some natural number n. Then, taking reciprocals, 1 - x > 1/n. We certainly have n < 10^n and so 1/n > 1/10^n. Combining the inequalities 1 - x > 1/n and 1/n > 1/10^n, we have 1 - x > 1/10^n. It follows from that that x < 1 - 1/10^n = 0.999...9 with n nines. It follows that x does not satisfy (1), and therefore, 1 is the smallest such number, that is, 1 satisfies (2).

"I guess we need a lesson on use of variables. That is not the same "x."

Undermining my intelligence is not going to make your argument any more correct, if that's the strategy you're using.
It's interesting that's you're using the terms local variables and global variables in the context of mathematics. These are not defined mathematical terms as far as I know. These terms are more familiar in the field of Computer Science, so if we're going to speak of variables in a mathematical analysis setting, we can't invoke strange terms like these.
I'm starting to doubt you've done any sort of analysis course at university, if not, then I won't be continuing with this debate and you can take the "winning" streak. As arguing with someone who is ignorant of math is like arguing with a kid about the moon not being cheese.

Now, from this inequality:

x < 1 - 1/10^n = 0.999...9

If we let n ---> ထ, then we can make x (or "y" in your updated proof) as arbitrarily close to 1 as possible. If you're familiar with the concept of small and large infinities, then it means that we can even let x (or "y") be 0.999... as there will be a larger infinity allowing for the inequality to still hold. To specify, we can get these cases and eventually land at the final inequality:

0.99 < 0.999, 0.999999 < 0.9999999, 0.99999999999999999999 < 0.9999999999999999999999, ... , 0.999... < 0.999...

So, since we concluded in the end that x (or "y") does not satisfy (1), it follows that 0.999... does not satisfy (1) and so it surely cannot satisfy (2). Thus, there is no equality with 1 and 0.999...

This could've been easily seen with a mere consideration to the completeness property of real numbers.

"I intend to demonstrate:

1 satisfies (2)"

No, 1 cannot be the smallest number that satisfies (1) if you're speaking of values 0.9, 0.99, ... individually as there is no such "smallest" number greater than any of these. However, if you're referring to the set {0.9, 0.99, ...} then the smallest natural number satisfying both (1) and (2) would have to be 1 - and this is obvious and requires no proof.
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Now to conclude, it is worthwhile remembering that your very hypothesis, being 0.999... = 1, is false. It has already been well-established by professional mathematicians that this can simply not be true. You're trying to ignore the very fundamental properties of numbers and, perhaps unintentionally, misleading other potential mathematicians. Worse, you're misleading yourself. My suggestion is to first build on your foundational knowledge of mathematics and then proceeding to try and prove what is more logical. There are many open problems in mathematics. Messing with the very basics and foundations of mathematics will get you nowhere - trust me, I've been there too.

Round 3
Pro
#5
"What is n here? I'm not sure what con is arguing."

Recall that I have said:
"We can even work with the proof you've laid out. "

"Suppose that x is less than 1, and consider 1 - x, which will be a positive real number. By the Archimedean property, 1/(1 - x) < n for some natural number n. Then, taking reciprocals, 1 - x > 1/n. We certainly have n < 10^n and so 1/n > 1/10^n. Combining the inequalities 1 - x > 1/n and 1/n > 1/10^n, we have 1 - x > 1/10^n. It follows from that that x < 1 - 1/10^n = 0.999...9 with n nines. It follows that x does not satisfy (1), and therefore, 1 is the smallest such number, that is, 1 satisfies (2).
Okay, at least now I know what "n" con was referring to, but there is no reason that "as n goes to infinity it is obvious that 0.999... surely would not satisfy property 1." Con will need to explain how this makes it clear that 0.99999... does not satisfy property 1, as I don't see any reason this should be true, and I am doubtful voters will either.

"I guess we need a lesson on use of variables. That is not the same "x."

Undermining my intelligence is not going to make your argument any more correct, if that's the strategy you're using.
It's interesting that's you're using the terms local variables and global variables in the context of mathematics. These are not defined mathematical terms as far as I know. These terms are more familiar in the field of Computer Science, so if we're going to speak of variables in a mathematical analysis setting, we can't invoke strange terms like these.
I'm starting to doubt you've done any sort of analysis course at university, if not, then I won't be continuing with this debate and you can take the "winning" streak. As arguing with someone who is ignorant of math is like arguing with a kid about the moon not being cheese.
\This isn't any sort of formal mathematical term, and I didn't intend it as such, I am just trying to make it clear that the two uses of "x" were separate. Nonetheless, this becomes entirely irrelevant in my revised argument where I avoided this issue altogether.

Now, from this inequality:

x < 1 - 1/10^n = 0.999...9

If we let n ---> ထ, then we can make x (or "y" in your updated proof) as arbitrarily close to 1 as possible. If you're familiar with the concept of small and large infinities, then it means that we can even let x (or "y") be 0.999... as there will be a larger infinity allowing for the inequality to still hold. To specify, we can get these cases and eventually land at the final inequality:

0.99 < 0.999, 0.999999 < 0.9999999, 0.99999999999999999999 < 0.9999999999999999999999, ... , 0.999... < 0.999...
Larger and smaller infinite numbers are in reference to cardinalities. Infinite decimals always contain a countably infinite number of digits, so this "small and large infinities" thing is irrelevant.

This "final inequality" also by no stretch of the imagination shows that 0.99999... does not satisfy (1), and in fact it is exactly why it does. As these inequalities show, less nines --> smaller number.

"I intend to demonstrate:

1 satisfies (2)"

No, 1 cannot be the smallest number that satisfies (1) if you're speaking of values 0.9, 0.99, ... individually as there is no such "smallest" number greater than any of these. However, if you're referring to the set {0.9, 0.99, ...} then the smallest natural number satisfying both (1) and (2) would have to be 1 - and this is obvious and requires no proof.
That is exactly what I meant, except for that, as I demonstrated in my argument, 1 is in fact the smallest real number (and not just the smallest natural number) that has this property.

Now to conclude, it is worthwhile remembering that your very hypothesis, being 0.999... = 1, is false.
Simply saying "what you're trying to prove is false" isn't an argument.

It has already been well-established by professional mathematicians that this can simply not be true.
I beg to differ:

Even Terrence Tao, generally considered the worlds best mathematician, thinks so:

https://math.unm.edu/~crisp/courses/math401/tao.pdf (See page 390 of the textbook. Page 157 of the pdf file. Proposition 13.2.3 states that 0.99999... = 1.)

You're trying to ignore the very fundamental properties of numbers and, perhaps unintentionally, misleading other potential mathematicians. Worse, you're misleading yourself. My suggestion is to first build on your foundational knowledge of mathematics and then proceeding to try and prove what is more logical. There are many open problems in mathematics. Messing with the very basics and foundations of mathematics will get you nowhere - trust me, I've been there too.
Firstly, I'm not "messing with" anything. 0.99999... = 1 is well established among mathematicians. Secondly, math isn't reliant on what we think is "more logical," it's reliant on what can be objectively proven through logical steps. I have demonstrated that 0.99999... = 1 through such logical steps.
Con
#6
Please vote Pro. I need to go and learn more mathematics, it seems. 😞🧐 [I probably won't, my brain is hurting too much rn]
Round 4
Pro
#7
Extend.

Thanks to con for participation in this debate and for the concession.
Con
#8
Forfeited
Round 5
Pro
#9
Extend.
Con
#10
Forfeited