1 and .999 repeating are the same quantity. Exactly equal.

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Description

I argue that .999r IS NOT approaching the number 1. Does NOT estimate or round to the number 1. But is in fact the same as the number 1.

By round or estimate to the number one I do not mean the syntactic changing of one number to another. And that any rounding that may occur is no different than rounding 2.000 to 2. They are the same number.interchangeable.

Round 1

09.27.2018 11:59PM

Pro

Hello, I appreciate my challenger taking me up this debate. Good luck to you Sir/Ma'am

I'll begin first by stating that in Math if the inputs are the same, the operators the same, then the outcome must be identical. This is what establishes Math as a repeatable operation for everyone. If I add 1 plus 1 and get 2. So will anyone properly performing the same equation.

Take for instance the addition of one third plus two thirds:

1/3 + 2/3 = 1.

A third plus two thirds equals a whole, ie 1.

I can also treat the fractions as division sub problems.

1/3 = .333... Repeating

2/3 = .666... Repeating

Adding them together we arrive at .999 repeating.

The equal signs are the correct signs for the conclusion of the division of the 1 divided by 3 and it's compliment. Because they are equal we can use the substitutive property to rewrite the first equation. This would be substituting 1/3 with .333r and 2/3 with .666r. Thereby giving us the equation written as:

.333r + .666r = 1

Which we established in the second equation also equaling .999 repeating. Affirming 1 equals .999r

09.28.2018 12:30AM

Con

1/3 + 2/3 = 3/3

3/3 = 1 (you will never ever conclude that 3/3 = 0.9 recurring without the context that Pro is tricking you into taking as valid reasoning).

1/3 when rounded to the nearest thousandth is 0.333

2/3 when rounded to the nearest thousandth is 0.667

So if we do actually round correctly, the answer even with decimals replacing fractions is 0.333 + 0.667 = 1.000 =/= 0.999

What Pro is arguing is that if you never ever rounded the 2/3 to end up with a 7 at the end of it (since you round up 0.6 recurring to end with a 7 no matter what) then you'd never end up with the answer of 1 as opposed to 0,9 recurring as the result of 1/3 + 2/3. This 'I'm so smart' quip made by people who think they are math geniuses fails to admit that if a number is recurring, you never ever could finish writing or delivering the answer in any way at all. What I mean by this is that the millisecond you stop typing '6' and '9' you're betraying yourself as you're rounding the answer and if you round the answer you never ever, ever, ever will get anything but 1.000000 (to whatever decimal point you rounded to).

Now let me give you actual 'I am smart and good at math' sums that make the resolution impossible.

0.9 *3 = 2.7

0.99 *3 = 2.97

So if one is to ever conclude that 3/3 = 0.9 recurring there is at some point a '3' that they are ignoring needs to be added on to the '7' in order to ever make this true.

Therefore if we are ever tricked by the formatting of sum to conclude that 1/3 + 2/3 = 3/3, we must remind the one tricking us to remember the '7' that never can end up being a '10' so as to make this answer true.

Round 2

09.28.2018 12:52AM

Pro

Thank you Mad for the quick reply.

However your argument is mute in that you changed the terms. At no point did I round 1 divided by 3. Nor asserted that a rounded number was the same as it's original number.

You performed a change of the number 1/3 or .333 repeating when you rounded it. It became a different quantity.

We can confirm this because of the additive identity. identity property of addition, which simply states that when you add zero to any number, it equals the number itself. So if the difference between 2 numbers is not zero. They are not equal.

When you rounded .333 repeating to .333 and we subtract them we get .00099999.. repeating 9's. That is not zero so your rounded number is not the same as the actual number 1/3 or .333 repeating.

However..

If we take 1 and subtract .999 repeating we are quick to say the answer is an infinite set of zeros then a 1. ie .00000..infinity..somehow ends in a 1.

But those zeroes go on for infinity. That singular "1" never appears. Making the answer to what is "1" minus ".999 repeating" equal to an infinite set of zeroes.

It's important to note that 2.0000 equals 2.00 is not rounding. It's dropping a place holder (which is a semantic use of "0") not the alteration of the value "two".

So just as 2.0000 equals 2.

And

2.0000 infinite 0's equal 2

Then so does

0.00000 infinite 0's equals 0

Therefore, there is no quantity between 1 and .999 repeating. The difference between them is "zero". Because of this the identity of addition implies that they are the same number, or the same quantity, value and exactly equal to each other. Confirmed again that 1 equals .999 repeating.

https://www.factmonster.com/math/numbers/rounding-numbers-rules-examples-fractions-sums

09.28.2018 12:58AM

Con

If you really didn't round any number involved in 1/3 + 2/3 then your answer could never ever be typed as a decimal given that you'd have to endlessly type '3333333' infinitely but even more so because you'd be lying to type '666666' and not end it with a 7 since you round the last 6 up to 7 given rules of rounding:

When rounding a number, you first need to ask: what are you rounding it to? Numbers can be rounded to the nearest ten, the nearest hundred, the nearest thousand, and so on.

Consider the number 4,827.

4,827 rounded to the nearest ten is 4,830
4,827 rounded to the nearest hundred is 4,800
4,827 rounded to the nearest thousand is 5,000
All the numbers to the right of the place you are rounding to become zeros. Here are some more examples:

34 rounded to the nearest ten is 30
6,809 rounded to the nearest hundred is 6,800
1,951 rounded to the nearest thousand is 2,000

You're completely deceiving the reader when you imply that 1/3+2/3 = 0.9999(recurring)

The reason is that the answer cannot possible be the same as the answer of (1+2)/3 given that 3/3 is 1.0000(recurring obviously) and that the answer of 0.999recurring when multiplied by 3 has to have a 3 added to it at some point because it will always end with a '97' and that 7 can never be the 10 it needs to be without ignoring it.

Your resolution is impossible because:

EITHER

We round and get 1 and/or 3.

We don't round and admit that '7' which is missing in that 0.9recurring * 3 so it can never truly be 3 and thus 3/3 can't be 0.9recurring.

Checkmate.

Round 3

09.28.2018 01:09AM

Pro

I'm not rounding.

1/3 is equal to .3 with the 3 repeating ad nauseam to infinity.

And 2/3 is equal to .6 with an infinite series of 6's to infinity.

I can add them together easily. because the addition never carries to another place. Meaning I can say it equals .9 repeating to infinity.

To say they equal anything else would be mathematically wrong.

The reason the numbers repeat is a side effect of the decimal system. That ten is not divisible by 3. There's no digit that ends the math without remainder.

But just because the numbers are written differently does not mean they are different quantities or different numbers.

16 is equal to 4^2 (4 squared). Those are 2 different ways of writing the same exact number. They are equal in all senses and completely interchangeable.

In the same token, I've shown that without changing the numbers with rounding (as you have), that there is no difference between .999 repeating and "one". The subtraction of the 2 is equal to "zero".

The two equations:

1/3 + 2/3

and

.333r + .666r

Are the same quantities added together. Because 1/3 is the equal to .333 repeating. They are 2 different ways of writing the same number.

As is 2/3 equals .666 repeating.

Because they are the same quantities with the same operation "addition" the result must be the same. Just like adding 4 squared to 4 makes 20. or adding 16 to 4 makes 20. Because 4 squared and 16 are the same number.

Therefore 1 and .999 repeating are the same number. Equal and interchangeable as any 2 identical quantities are.

09.28.2018 03:00PM

Con

Pro concedes that without rounding, 0.9 recurring never ever is equal to 1.

Pro further concedes by not proving me wrong that 3/3 can never ever equal 0.9 recurring given that there will be a 3 which the final 97 needs in order to have 0.9recurring*3 equal 3.

David

10.10.2018 05:54PM

#90

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@Juubi_Wolf

*******************************************************************
>Reported Vote: Juubi_Wolf// Mod action: Removed

>Points Awarded:7 points to con

>Reason for Decision: 1=1.000000000
1 does not equal .9999999

>Reason for Mod Action: The voter does not explain any of the points that we awarded. .
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bsh1

10.10.2018 02:50PM

#89

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@whiteflame

*******************************************************************
>Reported Vote: Whiteflame // Mod action: Not Removed

>Points Awarded: 3 points for argument

>Reason for Decision: Pro sets up a rather clear equation on which to base his comparison, explaining that by turning each of the fractions he's presented into a decimal, you can find that adding them together leads to a number that is not 1, in spite of the fact that adding those two fractions together does result in 1. The difference is infinitesimally small, but it does exist. He's essentially stating that the number 0.000r is equivalent to 0 for the same reason. While I understand Con's responses regarding the need to round in order to get a real number, I don't think that's necessary when you're comparing what is, effectively, an unmeasurable quantity. That's what Pro is doing with his argument, and while I think he could have defended it better, I don't think just railing against the lack of rounding suffices as a reason for me to vote Con. I do think there are ways to challenge this that involve more complex math, but those aren't presented, leaving me with little choice but to vote Pro.

>Reason for Mod Action: The voter surveys the main arguments and counterarguments, assess the strength of these arguments, and weighs them to produce a result. This meets the basic standard of sufficiency for argument points.
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bsh1

10.09.2018 10:14PM

#88

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@3RU7AL

*******************************************************************
>Reported Vote: 3RU7AL // Mod action: Removed

>Points Awarded: 1 point to Pro for Conduct

>Reason for Decision: I would call this a tie on arguments, because CON fails to make a clear case, but I'd like to make a case to award points to PRO for conduct.
Round one PRO - "Hello, I appreciate my challenger taking me up this debate. Good luck to you Sir/Ma'am" which is polite.
Round one CON - "...Pro is tricking you..." which is a negative characterization ad hominem strongly suggesting that PRO is intentionally deceptive.
Round one CON - "'I am smart and good at math'" which is not only a bald assertion but also an indirect ad hominem directed at PRO.
Round two PRO - "Thank you Mad for the quick reply." which is polite.
Round two CON - "You're completely deceiving the reader..." which is a negative characterization ad hominem strongly suggesting that PRO is intentionally deceptive.
Round two CON - "Checkmate." which is a rush-to-declare-victory fallacy.
Round three PRO - No positive or negative comments, just arguments.
Round three CON - "Pro concedes..." and "Pro further concedes..." which is another rush-to-declare-victory and by using the term "concedes" falsely suggests that PRO actually conceded the debate.

>Reason for Mod Action: In order to award conduct points, the voter must show that one debater was "excessively rude, profane, or unfair, or broke the debate rules, or forfeited one or more rounds in the debate without reasonable and given cause." The voter listed as evidence of misconduct statements which are not conduct violations. Rush-to-judgement fallacies are faults of logic, not of conduct. Similarly, boasting about one's own abilities is not itself misconduct, unless the voter can contextualize it. The remaining two or three acts of misconduct the voter cites do not rise to the level of "excessive."
*********************************************************************

10.05.2018 11:48PM

#87

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@3RU7AL

What is so hard in understanding that an infinite amount of zero's equals zero.

1 = 1.0 = 1.00000 = 1.000r

Do you Maths at all?

10.05.2018 11:21PM

#86

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@drafterman

@Mhykiel

New arguments on this topic here https://www.debateart.com/debates/146

10.05.2018 07:35PM

#85

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@Mhykiel

Ok, here you go -

So, if you take the remainder of 1 - 0.99999(r), lets call it 1/infinityith and you then multiply 1/infinityith by infinity then you end up with a 1 with infinite zeroes behind it. By contrast, if you multiply 0 by infinity you always end up with 0. That's a pretty big difference.

Therefore, 1 =/= 0.99999(r) (without rounding) because 1/infinityith does not equal zero.

10.05.2018 05:54PM

#84

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@3RU7AL

I'm not interested in a semantics debate. The statement is an inherently mathematical one. If you are not interested in addressing it within that context, then the conversation is over. Let me know what the case is.

10.05.2018 05:49PM

#83

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@drafterman

@Mhykiel

First of all, the debate resolution, which is, once again, "1 and .999 repeating are the same quantity. Exactly equal." says absolutely nothing about mathematics or rules or axioms or authority or popularity. So technically, and I do like to get technical, "1" is only one character and ".999 repeating" is fourteen characters including the space, therefore they are not the same quantity of characters and the resolution is technically defeated.

Second of all, for Mhykiel, if you wanted to use google info, you should have made those references explicit within the actual debate. Appealing to a third party will get you nowhere at this point, either make your own case with logic or I remain unconvinced.

And to drafterman, when you say, "All numbers are "finite" because "infinity" isn't a number. The second number would require an [][]infinite number[][] of digits to represent in written form, but we needn't worry about that because we have the [][]appropriate symbols[][] (r) to account for those infinite digits."

How is that substantively different from the idea that "There is an infinitesimal difference of 0.0000(r)1, which is a non-zero value."?

If we have, as you say, "...the appropriate symbols (r) to account for those infinite digits." it does not follow that the symbol referenced "(r)" could not have a number after it or that we couldn't use some other form of notation to identify the infinitesimal.

I'm also not sure how you can say, "infinite number" right after you say, "infinity is not a number".

"Them's the rules" is missing from the debate resolution.

I would cordially like to invite each of you to present your preferred definitions of "infinitesimal" for further examination.

10.05.2018 04:45PM

#82

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@3RU7AL

No the 3 and 6 repeat an infinite amount of times.

Have you googled the resolution? Because I you could benefit from a more solid understanding of fundemental mathimatical concepts such as the difference between rational and irrational numbers. And how number notation define quantities.

10.05.2018 04:45PM

#81

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@3RU7AL

"8/9 is finite."
"0.8888(r) is hypothetically infinite."

All numbers are "finite" because "infinity" isn't a number. The second number would require an infinite number of digits to represent in written form, but we needn't worry about that because we have the appropriate symbols (r) to account for those infinite digits.

"While I am willing to grant you they are practically identical, even functionally identical, but they are not perfectly identical."

Incorrect.

"There is an infinitesimal difference of 0.0000(r)1, which is a non-zero value."

It isn't a value. It isn't a number.

"You are making an axiomatic equivocation, which is fine, but logically, this is the same as a bald assertion or an appeal to dogma."

Whatever you call it is part of mathematics and that is the domain in which this resolution is being stated. If you dismiss it, you aren't talking about math anymore and you are off topic.

10.05.2018 04:41PM

#80

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@Mhykiel

So when you say in round 1,

"1/3 = .333... Repeating
2/3 = .666... Repeating"

You don't mean implicitly "repeating an infinite number of times"?

Do you in fact more precisely mean "repeating an unknown yet finite number of times"?

10.05.2018 04:36PM

#79

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@drafterman

8/9 is finite.
0.8888(r) is hypothetically infinite.

The difference is "infinite".

While I am willing to grant you they are practically identical, even functionally identical, but they are not perfectly identical.

There is an infinitesimal difference of 0.0000(r)1, which is a non-zero value.

You are making an axiomatic equivocation, which is fine, but logically, this is the same as a bald assertion or an appeal to dogma.

10.05.2018 04:18PM

#78

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@3RU7AL

.888r is not infinite.

It can written as the fraction 8/9

Being writeable as a fraction is one quality that means .888r is not an irrational number.

What is infinite is the representation of the number in decimal form. That's not to say the quantity is infinite.

10.05.2018 04:07PM

#77

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@3RU7AL

"While 0.8888(r) may be a very very very very very very close approximation of 8/9, it is not identical."

Then this is the problem. Mathematically 0.8888(r) isn't an approximation of 8/9, it is equal to 0.8888(r) exactly. They are different written representations of the exact same number. If you disagree, then I wonder what you say the answer to 0.8888(r) + 1/9 equals.

Now, I'm not saying you can't construct a form of mathematics that denies this, but it isn't the same mathematics that is in use today. This isn't a "precision" problem or a "flaw" of the decimal system.

Mathematically. 8/9 = 0.8888(r). Literally. Exactly. Precisely.

10.05.2018 04:03PM

#76

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@drafterman

8/9 is finite.
0.8888(r) is hypothetically infinite.

While 0.8888(r) may be a very very very very very very close approximation of 8/9, it is not identical.

It is virtually identical and in practice, practically identical but not perfectly identical.

This is a limitation (flaw) of the decimal system, not a limitation of common fractions.

If the debate resolution was, "9/9 = 1" then I'm absolutely certain there would be no dispute whatsoever.

10.05.2018 03:47PM

#75

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@3RU7AL

Do you assert that 8/9 =/= 0.8888(r) and that 1/9 =/= 0.1111(r)?

10.05.2018 03:45PM

#74

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@drafterman

8/9 + 1/9 = 1

This was never in dispute when written precisely in this format.

0.8888888888(r) + 0.1111111111111(r) =/= 8/9 + 1/9

The problem identified by PRO (highlighted by the resolution itself) is a limitation (flaw) of the decimal system, not a limitation of common fractions.

This is fundamentally a precision problem.

10.05.2018 03:14PM

#73

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@3RU7AL

And what does 8/9 + 1/9 equal?

10.05.2018 02:55PM

#72

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@drafterman

"...0.8888888888(r) + 0.1111111111111(r) = 0.99999999999999(r)..."

10.05.2018 02:53PM

#71

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@3RU7AL

Ok. I'll bite.

What does 0.8888888888(r) + 0.1111111111111(r) equal?

10.05.2018 02:51PM

#70

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@drafterman

My stated supporting argument directly addresses the debate resolution.

RationalMadman seemed to think that, "Your resolution is impossible because: EITHER We round and get 1 and/or 3. OR We don't round and admit that '7' which is missing in that 0.9recurring * 3 so it can never truly be 3 and thus 3/3 can't be 0.9recurring. Checkmate." closed their case.

However, RationalMadman's argument does not directly address the debate resolution.

The following steel man actually does directly address the debate resolution.

(IFF) 0.8888888888(r) + 0.1111111111111(r) = 0.99999999999999(r) (THEN) 0.99999999999999(r) =/= 1.0000000000(r)
0.99999999999999(r) =/= 1.0000000000(r)

Therefore, "1 and .999 repeating are the same quantity. Exactly equal." is false.

I have no idea why you found my previous comments "annoying and non-productive" since both of these are negative characterizations (ad hominem) and not strictly legitimate objections to my logic.

I really and truly would like to clear up any questions or misunderstandings you or anyone else might have.

Destroying your opponents examples (supporting arguments) alone does not make your case.

That would be like merely blocking your opponent from making a goal.

You actually have to make a goal yourself in order to get a point.

10.05.2018 02:26PM

#69

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@3RU7AL

In your own words:

"Focus on the actual debate resolution and not the supporting arguments." Also, I find it funny you accuse me of being a broken record when I hadn't even said it as many times as you had yet.

If you agree that this response is annoying and non-productive, then perhaps we can proceed with an actual discussion of the topic. If you disagree, then I'm afraid I'm going to have to ask you to stick to the debate resolution, please.

10.05.2018 02:24PM

#68

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@drafterman

It's called a supporting argument. Supporting arguments are actually integral to the very concept of a debate.

(IFF) 0.8888888888(r) + 0.1111111111111(r) = 0.99999999999999(r) (THEN) 0.99999999999999(r) =/= 1.0000000000(r)

0.99999999999999(r) =/= 1.0000000000(r)

Therefore, "1 and .999 repeating are the same quantity. Exactly equal." is false.

10.05.2018 02:04PM

#67

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@3RU7AL

Your question is off-topic. You seem to have abandoned the debate resolution. Stick to the debate resolution, please. The debate resolution doesn't say anything about 0.8888888888(r) or 0.1111111111111(r).

10.05.2018 02:01PM

#66

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@drafterman

0.8888888888(r) + 0.1111111111111(r) = 0.99999999999999(r) =/= 1.0000000000(r)
Therefore, "1 and .999 repeating are the same quantity. Exactly equal." is false.

If you disagree, please present your argument that disproves my argument or in some other way affirms the debate resolution of "1 and .999 repeating are the same quantity. Exactly equal."

Do you for instance, honestly believe that 0.8888888888888(r) + 0.1111111111111(r) = 1.000000000000(r) ?????????????????(r)

I really and truly would like to clear up any questions or misunderstandings you might have, but at this point you sound like a broken record.

10.05.2018 01:52PM

#65

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@3RU7AL

Stick to the debate resolution, please.

10.05.2018 01:45PM

#64

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@drafterman

0.8888888888(r) + 0.1111111111111(r) = 0.99999999999999(r) =/= 1.0000000000(r)

Therefore, "1 and .999 repeating are the same quantity. Exactly equal." is false.

10.05.2018 01:44PM

#63

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@3RU7AL

Stick to the debate resolution, please.

10.05.2018 01:42PM

#62

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@drafterman

0.8888888888(r) + 0.1111111111111(r) =/= 1.0000000000(r)

Therefore, "1 and .999 repeating are the same quantity. Exactly equal." is false.

10.05.2018 01:37PM

#61

@drafterman

https://www.debateart.com/debates/146

10.05.2018 01:24PM

#60

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@3RU7AL

"Therefore, "1 and .999 repeating are the same quantity. Exactly equal." is false."

Prove it.

10.05.2018 01:06PM

#59

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@Logical-Master

"...could one infinitely increase the height of the second rock without ever making the second rock 5 feet tall?"

The realistic and very practical answer is no.

There is a practical limit on how many decimal places you can actually type.

Reality is not infinitely divisible. There is a smallest possible unit of space-time.

This would be the Planck length. - https://duckduckgo.com/?q=Planck+length&atb=v79-2&ia=web

10.05.2018 01:01PM

#58

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@drafterman

Therefore, "1 and .999 repeating are the same quantity. Exactly equal." is false.

10.05.2018 01:00PM

#57

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@Death23

When you say, (IFF) "He's not saying that there's a 1 at the "end". He's saying that it's a mistake to think that there's a 1 at the end." (THEN) how do the "9's" ever change? Why don't they stay exactly 0.9999999999999999(r) if there is no way to convert them into anything else?

When I first read debate, I thought of the classic A =/= A argument. However this seems to be a case of A = A.00000000000(r)0000000000001 which is quantifiably false.

10.05.2018 12:55PM

#56

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@3RU7AL

Stick to the debate resolution, please.

10.05.2018 12:53PM

#55

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@drafterman

Therefore, by PRO's own logic in quote number one, since an infinite set of zeros with a 1 at the end is not exactly and precisely the same as zero, they magically round up to the target value of "1". Because if 1.0 and 0.999(r) were eXactly the same, then the result would obviously be actual zero.

Therefore, "1 and .999 repeating are the same quantity. Exactly equal." is false.

Not to mention 0.8888888888(r) + 0.1111111111111(r) =/= 1.0000000000(r)

10.05.2018 12:50PM

#54

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@Mhykiel

This is literally rounding 0.0000(r)0001 to 0.0000(r). PRO stakes their case on rounding. This defeats the debate resolution.

10.05.2018 03:00AM

#53

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@Logical-Master

Yes, provided that each incremental increase wasn't equal. In the beginning you may increase the shorter rock's height by 0.0000009 feet to make 4.99999909 feet tall, and next you may increase it by 0.00000009 to make it 4.999999099 feet, and so on. If you do this an infinite number of times you will end up with a rock that is 4.9999990999999999(repeating) feet tall or, expressed differently, as 4.9999991 feet tall.

Logical-Master

10.05.2018 02:47AM

#52

Question.

Suppose we have two rocks. One rock is 5 feet tall and the other rock is 4.999999 feet tall. If one were to keep increasing the height of the second rock by small enough increments, could one infinitely increase the height of the second rock without ever making the second rock 5 feet tall? :P

10.05.2018 02:40AM

#51

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@3RU7AL

The difference between 1 and .999r is not .111r
It's .000... infinity. Never a difference. 2 things with no difference are ergo the same thing.

10.05.2018 12:07AM

#50

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@3RU7AL

The round 2 quote you reference - this part - "we are quick to say the answer is an infinite set of zeros then a 1. ie .00000..infinity..somehow ends in a 1. But those zeroes go on for infinity. That singular "1" never appears." - He's not saying that there's a 1 at the "end". He's saying that it's a mistake to think that there's a 1 at the end.

10.04.2018 10:27PM

#49

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@3RU7AL

Stick to the debate resolution, please.

10.04.2018 10:14PM

#48

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@Death23

And even if I quote from your previous comment, "Pro's case was that 0.999r must be 1 because 1/3 = 0.333r and 2/3 = .666r and 1/3 + 2/3 = 3/3 = 1. It's a pretty weak case but it doesn't have anything to do with rounding." this is also provably false because 0.333(r) and 0.666(r) approach 1/3 and 2/3 respectively but they require rounding.

Since you can never type out infinite anything, anything ending in an infinite sequence must (EITHER) be typed out explicitly (OR) rounded (like pi).

10.04.2018 10:13PM

#47

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@Death23

What PRO actually argued was what PRO said.

And I quote from round 2, (1)"We can confirm this because of the additive identity. identity property of addition, which simply states that when you add zero to any number, it equals the number itself. So if the difference between 2 numbers is not zero. They are not equal."

And I quote again from round 2, (2)"If we take 1 and subtract .999 repeating we are quick to say the answer is an infinite set of zeros then a 1. ie .00000..infinity..somehow ends in a 1. But those zeroes go on for infinity. That singular "1" never appears. Making the answer to what is "1" minus ".999 repeating" equal to an infinite set of zeroes."

This is literally rounding 0.0000(r)0001 to 0.0000(r). PRO stakes their case on rounding. This defeats the debate resolution.

HOwever, I did not give the "arguments" points to CON because CON failed to clearly identify this error in logic.

You also make a good parallel case, and I applaud your excellent steel man, but I am not fabricating a new argument, I have based my critique on CON's actual statements.
When you said, and I quote from your RFD, "Pro correctly pointed out that no rounding was being supposed.", this statement is provably false.

10.04.2018 09:01PM

#46

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@3RU7AL

What Pro actually argued was what Pro said. Pro's case was that 0.999r must be 1 because 1/3 = 0.333r and 2/3 = .666r and 1/3 + 2/3 = 3/3 = 1. It's a pretty weak case but it doesn't have anything to do with rounding. Con's attack on it was an obvious strawman. A better attack would have been something like "Pro's argument is based on the premises that 1/3 = 0.333r and 2/3 = 0.666r. These premises haven't been supported by Pro and I challenge them as unsubstantiated. The burden of proof is on Pro to show that these premises are true. If Pro has not met his burden, then Pro's argument fails."

10.04.2018 02:05PM

#45

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@RationalMadman

@Tejretics

@Mhykiel

When Death23 makes the bare assertion, "Pro correctly pointed out that no rounding was being supposed." this is provably false.

What PRO actually argues is that the difference between 1.00000(r) and 0.999999(r) is 0.00000(any number)00000010000(r) which is according to PRO "impossible" and therefore virtually equivalent to zero. However, by notating 0.00000(any number)00000010000(r) it would seem to be no more "impossible" than 0.999999(r) or any other "infinite" sequence (like pi).

0.8888888888(r) + 0.1111111111(r) =/= 1.00000(r)

I would call this a tie on arguments, because CON fails to make a clear case, but I'd consider making a case to award points to PRO for conduct.

Round one PRO - "Hello, I appreciate my challenger taking me up this debate. Good luck to you Sir/Ma'am" which is polite.
Round one CON - "...Pro is tricking you..." which is a negative characterization ad hominem strongly suggesting that PRO is intentionally deceptive.
Round one CON - "'I am smart and good at math'" which is not only a bald assertion but also an indirect ad hominem directed at PRO.

Round two PRO - "Thank you Mad for the quick reply." which is polite.
Round two CON - "You're completely deceiving the reader..." which is a negative characterization ad hominem strongly suggesting that PRO is intentionally deceptive.
Round two CON - "Checkmate." which is a rush-to-declare-victory fallacy.

Round three PRO - No positive or negative comments, just arguments.
Round three CON - "Pro concedes..." and "Pro further concedes..." which is another rush-to-declare-victory and by using the term "concedes" falsely suggests that PRO actually conceded the debate.

Tejretics

10.04.2018 02:59AM

#44

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@Death23

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>Reported vote: Death23 // Moderator action: NOT removed<

3 points to Pro (arguments). Reasons for voting decision: The instigator's position is a truism. There is no serious debate in mathematics as to whether or not 0.999r = 1. The reason people have a difficult time grasping how 0.999r = 1 has to do with difficulty grasping an infinite series. We are accustomed to the finite and the infinite is something that simply isn't part of our everyday experience, and this is where Pro's position is weakest. Pro merely posits that 1/3 = 0.333r and that 2/3 = 0.666r. This is a bare assertion, but is later supported rather weakly. This assertion is attacked by Con with his rounding argument. However, Con's attack fails because Pro correctly pointed out that no rounding was being supposed. Con's other attack with the 2.9999(r)7 also fails because Pro correctly pointed out that this was a change in terms (i.e. off topic, and this is true - The topic is 0.9999(r).

[*Reason for non-removal*] While the voter does violate the principle of tabula rasa judging by intervening in the debate with their own opinion, they do reference specific arguments in the debate and explain why Pro's counterarguments to Con's case were sufficient to grant Pro the victory in terms of arguments. Thus, the vote is sufficient.
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10.03.2018 11:12PM

#43

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@RationalMadman

1 - 0.999(r) = 0.000(r) = 0 ergo 1 = 0.999(r)

There is no 0.000(r) and then a 1 after because there is no "after" with infinity. It doesn't work that way.

10.03.2018 10:16PM

#42

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@Death23

therefore an infinite string of 9's can't ever be said to approach 1

10.03.2018 10:09PM

#41

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@RationalMadman

Infinity isn't a number. It's an idea. By definition, infinity has no end. Your arguments about having to round (e.g. "0.9 *3 = 2.7
0.99 *3 = 2.97 So if one is to ever conclude that 3/3 = 0.9 recurring there is at some point a '3' that they are ignoring needs to be added on to the '7' in order to ever make this true. ") can be soundly rejected because they're not consistent with the idea of infinity. There's no "at some point". At what point? That point doesn't happen with an infinitely long sequence.

whiteflame

10.08.2018 02:31AM

Criterion

Pro

Tie

Con

Points

Better arguments

3 point(s)

Better sources

2 point(s)

Better legibility

1 point(s)

Better conduct

1 point(s)

Reason:

Pro sets up a rather clear equation on which to base his comparison, explaining that by turning each of the fractions he's presented into a decimal, you can find that adding them together leads to a number that is not 1, in spite of the fact that adding those two fractions together does result in 1. The difference is infinitesimally small, but it does exist. He's essentially stating that the number 0.000r is equivalent to 0 for the same reason. While I understand Con's responses regarding the need to round in order to get a real number, I don't think that's necessary when you're comparing what is, effectively, an unmeasurable quantity. That's what Pro is doing with his argument, and while I think he could have defended it better, I don't think just railing against the lack of rounding suffices as a reason for me to vote Con. I do think there are ways to challenge this that involve more complex math, but those aren't presented, leaving me with little choice but to vote Pro.