Instigator / Pro
0
1500
rating
8
debates
50.0%
won
Topic
#4274

0.99999... = 1

Status
Finished

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

Winner & statistics
Winner
0
0

After not so many votes...

It's a tie!
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Standard
Number of rounds
5
Time for argument
Two days
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10,000
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Two weeks
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Open
Contender / Con
0
1700
rating
544
debates
68.01%
won
Description

In this debate I will demonstrate that 0.99999... = 1. To be clear, I am claiming that these are exactly the same number. I know this topic very well so I suggest being prepared.

Round 1
Pro
#1
Let's start with a classic argument:

Let x = 0.99999.... Then 10x = 9.99999... = 9 + 0.99999... = 9 + x, and so 9x = 9, and thus x = 1. From the definition of x, 0.99999... = 1.

Now to save time by handling some of the common counterarguments:

Common counterargument #1: "Algebraic manipulations like these do not always provide accurate results. For example, the 'proof' that 1 = 2 is a similar 'algebraic argument.'"

The algebraic "proof" that 1 = 2 involves a hidden division by zero. To demonstrate that every step in the algebraic argument that 0.99999... = 1 is valid, I will remove any variables from the argument: 10(0.99999...) = 9.99999... = 9 + 0.99999..., and so 9(0.99999...) = 9, and thus 0.99999... = 1.

Common counterargument #2: "10x should not equal 9.99999.... How did we magically create another 9? There should be a 0 at the end of this, as always occurs when multiplying by 10."

We didn't magically create another 9, just like when multiply, say, 34 by 10 and getting 340, we don't say that we magically created another 0. Now, generally this would be answered with "multiplying by 10 adds a 0 and not a 9!" That is what it does for integers, but let's take a look at what it does more generally. First of all, where would an additional 0 go? There is no such thing as the "infinityith place after the decimal." What multiplication does in general is shift each decimal place once to the left. Take the example with 34. In fact, 34 = 34.00000.... Upon multiplication by 10, the 3 in the tens place moves up to the hundreds place, the 4 in the units place moves up to the tens place, and so on, giving us 340.00000.... (Which is just 340.) Observe that we did not add a zero, it was the zero from the tenths place. The new zero in the tenths place came from the hundredths place, the zero in the hundredths place from the thousandths place, and so on. Similarly, when we say 10(0.99999...) = 9.99999..., this is because each nine moves up to occupy the spot to its left. If it still feels like something must be wrong, as there seems to be one more 9 than there was before, take a look at Hilbert's Hotel. As you can see, countably infinitely many nines is exactly the same this as countably infinitely many nines plus one more.
Con
#2
Watch. and. Learn.

I may accidentally plagiarise myself from other debates of this but I have altered a few things, to more precisely cut through the BS of Pro in these debates. Pro failed to do the 'third' argument which is the biggest one in favour of Pro. I am going to start very simply:

The flaw of thinking only in denary counting systems and the way to annihilate the mindfuckery that even official mathematics professors are naive enough to stand by and fall for.

The denary counting system is 1-9 where the value after 9 is 10. In this system, there are unique blind spots and unique lack of blind spots, just as with any alternative. The reason humans fell into the habit of denary systems is that we evolved 5 fingers on 2 hands to 'physically' count with so our brains easier handled it.

In flowers and plants, for instance, things go by 4s and 8s much more (like we go by 5s and what we call 'tens'). So, if we were more like plants, we'd be counting in a system where the number after 7 is 10 and the number after 17 is 20 (that 17 would be equal to our denary 15).

One blind spot and flaw of denary counting is that 1 divided by 3 doesn't exist in it. It is not 0.3-recurring, it is actually a number that doesn't exist which is 1 third higher than 0.3. The number 0.3 recurring is indeed a third of the number 0.9recurring. Yet, we do not say that 0.3 recurring suddenly becomes 0.4 ever or that it is equal to it, even though it infinitely approaches it, nor do we say it is 0.34 etc. Somehow when we have 0.9 recurring we are insane enough to say it equals 1 and that those that disagree are mathematically illiterate even though they are merely more logically literate and realise that the proofs for it are due to limitations in our flawed denary counting system which has inherent blind spots.

If we counted in a system where the number after 8 is 10 (so 'base 9') we would realise that 0.9 recurring in denary absolutely does not equal 1.0recurring in denary, it just erroenously appears that way.

The reason we'd realise this is is that 1 third of 1 in a base-9 system would be 3 and 1 third of the number we call 1 in denary would be 0.33. It is actually that simple. (basically what we call 1 is 1.1 in that system and 1/3 is 0.3).

The only reason we are incapable of presenting and comprehending this in denary is very simply that all counting systems have these blindspots except perhaps binary (the issue is just practicality as 'chunks of 9 plus 1' are much easier to deal with for daily life on Earth even if we ignore our fingers, than chunks of 1 plus 1, due to the latter requiring 5 binary 10s for every denary 10).

When we say 0.9recurring equals 1, what we mean is that the actual value that is 0.9 recurring is equal to 1.0recurring and this is blatantly false. The only reason we can't show that is because in denary the value of 1/3 doesn't exist and is instead represented as 0.3 recurring even though 0.3 recurring is never ever going to have the 'third' to make the 1 to add onto itself to become 1 instead of 0.9 recurring.

==

With that irrefutable part out of the way that could win me the debate on its own, let me explore another issue. Why is it that this flaw exists in the denary system and why do mathematicians fall victim to it and brainwash their students to buy into it?

The answer is simple. Mathematics is not always pure values, what mathematics is, is a representation of this imaginary thing we invented to comprehend the world around us. We made numbers up and it turns out that imaginary system has a lot of help in real-world application. Think of it like the way we think of 'notes' in music as the written note etc, even though that entire system on the pages is not the music at all. Similarly, the written mathematics that we adhere to religiously, is actually a representation, not always working how it should vs the real values we're referring to in our imaginary realm of mathematics.

The number referred to as 0.9999 recurring is a number that has the following difference between itself and 1:

0.0...1

This value has to exist and I will tell you why. Even if we stick only to the denary system with is blind spot there, the way that Pro justifies 0.999999... equalling 1 is the claim is that there is no last value. In fact let me quote Pro:

We didn't magically create another 9, just like when multiply, say, 34 by 10 and getting 340, we don't say that we magically created another 0. 
by this identical logic, we didn't magically create the 1 that exists irrefutably at the end of all those infinite 0s in the answer to what 1- 0.999... is equal to.

Either the difference and the 1 at the end of it are real or 0.999... isn't real and cannot be 1 which is the 'real' number. 

It's a situation where Pro has to deny one thing that harms his case, in order to support it the other way around. I call it a 'pincer burden of proof trap'. Enjoy.
Round 2
Pro
#3
Before going any further, let's clarify that the reason no one says that 0.33333... = 0.4 is because the reasoning behind 0.99999... = 1 does not imply that. It implies that 0.39999... = 0.4. Now let's focus on the two things which are absolutely necessary for Con's argument to work:

  • He claims that not every number can be represented in denary. (a.k.a. decimal or base 10)
  • He claims that 0.00...1 is a real number.
Everything else said in Con's argument is irrelevant unless the above two claims are true.

Looking at previous debates on this topic, I was alarmed to see that no one has refuted these claims, so let's finally set this straight. I assume that we can both agree we are working with the real numbers here. If you are unfamiliar with this term, it means the numbers on the number line. More formally, they are defined using Dedekind cuts. Should you click on the "Dedekind cuts" link, you will find, on the final page of the document, the least upper bound property, which states that every set of real numbers that had an upper bound has a least upper bound. A proof of this can be found here. Now I will prove the Archimedean property. The Archimedean property states that if x is a real number, then there is a natural number ("counting number") n such that x < n. Note that since I cannot type the correct inequality signs, I will we using >= to mean "greater than or equal to" and <= to mean "less than or equal to."

Suppose instead that the Archimedean property is false. Therefore there is a real number x such that x >= n for all natural numbers n. This means that the natural numbers have an upper bound. (Namely x.) It follows from the least upper bound property that there is a real number s that is the least upper bound of the natural numbers. Let n be an arbitrary natural number. Then so is n + 1, and thus s >= n + 1, as it is an upper bound for the natural numbers. Subtracting 1 from both sides, s - 1 >= n. Since n can be any natural number here, s - 1 is an upper bound for the natural numbers. But s - 1 < s, contradicting that s is the least upper bound! This contradiction implies that the assumption that the Archimedean property is false cannot be correct, and so the Archimedean property must be true.

Now I will demonstrate that 0.00...1 is not a real number. Suppose that it was. To make notation easier, I will define x = 0.00...1. Certainly this is less than 0.1, 0.01, 0.001, etc. Let n be any natural number. Then certainly n < 10^n, thus implying that 1/n > 1/10^n. Now 1/10^n is the nth number in the sequence 0.1, 0.01, 0.001, etc., so x < 1/10^n < 1/n. In conclusion, x < 1/n for every natural number n. Then 1/x > n for every natural number n. On the other hand, by the Archimedean property, 1/x < n for some natural number n. This contradiction implies that no such number 0.00...1 can exist.

As for the proof that every real number has a decimal representation, see pages 388 - 390 (pages 155 - 157 in the pdf) in this textbook: https://math.unm.edu/~crisp/courses/math401/tao.pdf.

Final remarks:

My algebraic argument that 0.99999... = 1 which I presented in the first round still stands. Also, if you are going to insist that I prove 0.33333... = 1/3, note that my argument is entirely independent of this, so as long as my argument stands, I can simply divide 0.99999... = 1 by 3 to get 0.33333... = 1/3. It seems that so many people use the "1/3 = 0.33333... so 1 = 0.99999..." argument that Con is automatically jumping on trying to refute this even though it has nothing to do with my argument. In reality, the easiest way to show that 1/3 = 0.33333... is by using 0.99999... = 1, (as I explained above) and so this argument is pointless, which is why I don't use it.

Let's be clear on what Con will need to do to refute my argument at this point: Firstly, if he cannot propose a value for 1 - 0.99999... other than 0 which is actually a real number, then I will have won the argument. Secondly, if he cannot either defend the two claims specified at the top of this argument, or find some sort of flaw in my argument that does not rely on these claims, then I will have won the argument.
Con
#4
Forfeited
Round 3
Pro
#5
I will use this as an opportunity to clarify something from the last round. I stated that Con claims that 0.00...1 is a real number. This is not exactly true. He claims that 1 - 0.99999... = 0.00...1. Given that 0.99999... is a real number, for 1 - 0.99999... to be equal to 0.00...1, 0.00...1 would have to be a real number. He seems to think that he's trapped me with the question of "is 0.00...1 a real number?" but in fact it is quite simple: 0.00...1 is not a real number, but this does not mean that 0.99999... is not a real number, it means that 1 - 0.99999... cannot be equal to 0.00...1 as Con claims. If there is any doubt remaining as to whether or not 0.99999... is a real number, observe that is can be viewed as the least upper bound (this is a real number by the least upper bound property) of the set {0.9,0.99,0.999,...}. The reason that I then asked him to propose another value for 1 - 0.99999... is because the only other reasonable answer to this difference besides 0.00...1 (which as I have already explained cannot be the answer) is 0! If 1 - 0.99999... = 0, then 1 = 0.99999..., so this one little subtraction problem that Con has attempted to use for his case is in reality quite devastating to him.
Con
#6
Everything else said in Con's argument is irrelevant unless the above two claims are true.
Wrong.

If 0.0000...1 is not a real number never is this:

9.999...9

Just because the ending is a 9 and the infinite series before is also 9s has no impact on the stupidity or irrationality of that value being plausible in a mathematical construct.

Let me stick to my pincer BoP concept and do what I do in this debate every time rather than focus on the denary stuff just now.

The parlour trick, illusion magic trick blablabla that Pro's on this debate ALWAYS pull off is they say 9.99999... infinitely approaches 1 and therefore is equal to 1

well

0.0000.... towards 0.000...1 infinitely approaches the 1 meaning either the difference is successfully reached or their original equivalent is fallacious, creating a hellish paradox with nails in the coffin to the case Pro makes. I don't get why Pro's in this debate don't see that.

The reason the denary vs base-9 thing mattered is that people say a third of 1 equals a third of 0.9999.... but 0.3333... is not a third of 1, this is a lie taught even to university level mathematicians. the number 1/3 does not exist you cannot represent it in denary counting system. In base-3 it is as simple as being 0.33 what is a third of the denary 1
Round 4
Pro
#7
If 0.0000...1 is not a real number never is this:

9.999...9
Con is right there, but it doesn't debunk a thing I said. Observe the distinction between 9.999...9 (with a final nine) and 9.999... (no last nine). 9.999... is a real number, but 9.999...9 (where I am assuming that there is an infinite string of nines here) is not! Let's take a look at the claim that 1 - 0.99999... = 0.000...1. Con talks about 1 - 0.99999... = 0.000...1 like it's irrefutable, but in reality, he hasn't backed it up, and I have now argued against it multiple times. I'll repeat what I said in my round three argument (which was really more of a clarification) here: 0.99999... is a real number, since it can be defined as the least upper bound (least upper bound property!) of the set {0.9,0.99,0.999,...}. 0.000...1 is not a real number, as I demonstrated in round 2. It follows that 1 - 0.99999... cannot be equal to 0.000...1. It is as simple as that.

The parlour trick, illusion magic trick blablabla that Pro's on this debate ALWAYS pull off is they say 9.99999... infinitely approaches 1 and therefore is equal to 1
I have not said anything about 0.99999... (I assume that is what Con meant to write here?) "approaching" 1. That would be sheer nonsense. It is a fixed number with a fixed value. It's not a sequence, or a function, or anything of the sort, so it isn't approaching anything. It is simply a number, namely it is the number 1.

0.0000.... towards 0.000...1 infinitely approaches the 1 meaning either the difference is successfully reached or their original equivalent is fallacious, creating a hellish paradox with nails in the coffin to the case Pro makes. I don't get why Pro's in this debate don't see that.
Given that I am not saying anything about anything approaching anything, this isn't really relevant, and it certainly isn't a "hellish paradox" that "nails in the coffin" on my case. It's no surprise that we get sheer nonsense here, because the idea that 0.99999... approaches 1 is itself sheer nonsense. It doesn't approach 1, it is 1.

The reason the denary vs base-9 thing mattered is that people say a third of 1 equals a third of 0.9999.... but 0.3333... is not a third of 1, this is a lie taught even to university level mathematicians. the number 1/3 does not exist you cannot represent it in denary counting system. In base-3 it is as simple as being 0.33 what is a third of the denary 1
"People" may say that, but I'm not using that argument, so I'm not sure why Con keeps arguing against it. Also, Con says "the number 1/3 does not exist you cannot represent it in denary counting system." The number itself certainly exists. It is defined by the relation 1/3 * 3 = 1. (The asterisk represents multiplication.) If you don't think that it is acceptable to simply define a number using such a relation, you may want to do some research into abstract algebra. (The Wikipedia article is a good starting point: https://en.wikipedia.org/wiki/Abstract_algebra) As to the existence of a base-10 (or "denary" as Con calls it) representation for this number, see my round 2 argument.

In conclusion:

Con didn't refute my arguments, and mostly just presented a straw man argument. He still has yet to debunk my original algebraic argument from round 1, has entirely ignored the proof that every real number has a decimal expansion linked to in round 2, and keeps trying to use 1 - 0.99999... = 0.000...1 as part of his argument, despite the fact that I have already debunked it.
Con
#8
 Observe the distinction between 9.999...9 (with a final nine) and 9.999... (no last nine)
No. This is the entire scam that the Pro's of this resolution sell you. This is what you fall for hook, line and sinker over and over again until someone like myself shows up.

If 0.999... never hits the last 9, it also can never hit the 1 needed to upgrade itself to 1.000...

Let's not forget that 1.000... is 1 followed by infinite 0s. 

also there is a major misconception my opponent has in the comments, he said what I call base-9 is base-10, this highlights a major lapse in knowledge and mathematical understanding on his part.

For 10 to equal the denary 9, you must be operating in base 9.

as in 1 2 3 4 5 7 8 10 11 12 etc
that is base-9.

In base-9 a third of the denary 1 is 0.33 because in base-9 that is 1.1 not 1.0

There is absolutely nothing that makes 9.999.... or 0.9999.... able to exist as values unless the difference between those values and one that is greater than them is also real.

If only 10 and 1 are 'real' then 9.999... and 0.999... do not exist. The small description of this debate says 'exactly equal' not 'provably equal under certain illusory and deceptive algebraic constructs'.

proof of description's wording:

If the difference is fake, if the 0.000...1 cannot exist then neither can 0.999...9

Just because it stays on 9s doesn't at all affect this, it's a bullshit illusion the Pro's of this debate always pull. An infinite series of values either can or cannot function as a value, no in between, no third option.
Round 5
Pro
#9
If 0.999... never hits the last 9, it also can never hit the 1 needed to upgrade itself to 1.000...
But it doesn't need to "upgrade" itself to 1.000.... It is 1.000.... Also, Con has proven a point for me. He keeps saying that 1 - 0.99999... = 0.00...1, but really, since there is no last 9 in 0.99999..., 0.99999... + 0.00...1 would be 0.99999...1 and not 1, even if 0.00...1 were a real number.

also there is a major misconception my opponent has in the comments, he said what I call base-9 is base-10, this highlights a major lapse in knowledge and mathematical understanding on his part.
Huh. I thought that was an honest mistake.

For 10 to equal the denary 9, you must be operating in base 9.

as in 1 2 3 4 5 7 8 10 11 12 etc
that is base-9.
It just occurred to me that I misunderstood what Con was referring to as base-9. I thought he was referring to base-10, which he calls denary, based on the context. I hadn't realized that he was referring to when he talked about base-9 in a previous argument. Base-9 didn't seem to be relevant, so I assumed he meant base-10. This is my mistake. I do however know what different bases mean. The explanation presented by Con here is correct.

In base-9 a third of the denary 1 is 0.33 because in base-9 that is 1.1 not 1.0
Oooooh. So that's why Con was using base-9. Once again, honest misunderstanding.

If only 10 and 1 are 'real' then 9.999... and 0.999... do not exist. The small description of this debate says 'exactly equal' not 'provably equal under certain illusory and deceptive algebraic constructs'.
When did I say that 10 and 1 are the only real numbers? I'm not sure what Con is talking about there. Also, I have proven that they are exactly equal, so this is a mute point.

If the difference is fake, if the 0.000...1 cannot exist then neither can 0.999...9
But 0.999...9 doesn't exist. The debate title and description both say that this is a debate on 0.99999.... There is a difference, because 0.99999... has no last 9! Every digit in 0.99999... is finitely many decimal places after the decimal point. That is not to say there are a finite number of digits, it is to say that there is no such thing as the "infinityith 9." Con keeps relying on infinitesimals, which I already showed cannot be real numbers in my argument from round 2. Take note: Any use of an "infinityith" decimal place, implies that numbers like 0.00...1 are valid! As I have shown they are not!

Just because it stays on 9s doesn't at all affect this, it's a bullshit illusion the Pro's of this debate always pull. An infinite series of values either can or cannot function as a value, no in between, no third option.
Some series do have a value, and some don't. Example: 1 - 1 + 1 - 1 + ... diverges, meaning that it doesn't settle down to any particular value. 1 + 1/2 + 1/4 + 1/8 + ... however converges. It converges to the number 2, as it get closer and closer to 2 as we add more terms, and adding enough terms, and can get as close to 2 as we like. We say that it has a limit of 2. That is the meaning of convergence as opposed to divergence in math. 0.99999... can be written as the limit of 9/10 + 9/100 + 9/1000 + 9/10000 + .... This series converges to 1, so 0.99999... has a value of 1. Note that the series approaches 1, but 0.99999... is not any partial sum of the series, but the limit of the series, meaning that it is 1.

Conclusion:

Here is where we are at. Con still hasn't debunked my original algebraic argument. He also hasn't acknowledged my argument that the difference 1 - 0.99999... cannot be equal to 0.00...1 as he claims. I linked to an argument that every number has a decimal expansion in my round 2 argument, and if you read it, you'll notice that the way said decimal expansions are constructed, no number would have a decimal expansion with an infinityith decimal place. This poses some major problems for him, but he has promptly ignored it. This same link backs up my interpretation of decimals as limits. Lastly, since 1 - 0.99999... cannot be 0.00...1 it must be some other real number, and the only reasonable alternative is 0. Con has yet to refute this. Thus, Con will need to:

  1. Refute my algebraic argument from the first round.
  2. Refute my argument that 1 - 0.99999... cannot be 0.00...1, or propose a different value for it that is not 0. (If it is 0 then that would be an automatic win for me, since if 1 - 0.99999... = 0, then it must be that 1 = 0.99999....) Moreover, he must back up this alternative value with a mathematically rigorous argument.
  3. Refute the argument that every number has a decimal expansion which can be found in a link in my round 2 argument, or change his argument to account for this.
If Con fails to do 1 and 2, my arguments will all still stand, and I will have won. If Con fails to do 3, he will have no argument, and I will have won.

Con
#10
I have frankly made my irrefutable case already.