Instigator / Pro
Points: 2

0.999... is NOT equal to 1

Finished

The voting period has ended

After 5 votes the winner is ...
garai
Debate details
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Last update
Category
Education
Time for argument
One week
Voting system
Open voting
Voting period
One month
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Rating mode
Rated
Characters per argument
5,000
Contender / Con
Points: 5
Description
No information
Round 1
Published:
On my previous debate:
I definitely proved that 0.999 does = 1. The most compelling proof of that was:

X = 0.999...
10x = 9.999...
10x - x = 9.999... - 0.999...
9x = 9
X = 9/9 = 1

To prove that the conclusion is in fact wrong, i will repeat the same arithmetic on another number: ...999.0.

X = ...999.0
10x = ...9990.0
X - 10x = ...999.0 - ...9990.0
-9x = 9
X = 9/-9 = -1

Clearly something went wrong because even a single 9 before the decimal point is greater then 0, forget about an infinity of 9s. A negative number is an absurd conclusion, thus the argument that they are equal is reduced to absurdity.
Published:
This is interesting.
First of all, I want to thank you because you proved 0.999... = 1 for me. So there is no need to give some proof other than this.
But, in second "proof" you made mistake even before you started calculation. You said you will use same arithmetic on number ...999.0. And that is the mistake: number like this does not exist. Because if you said that number is ...999 then it is infinite number-there are infinite period of 9s after 0. and you can't just put 0 after infinite period of 9s otherwise it wouldn't be infinite... It isn't even a number. It's same as you wrote 0.999...0 - number like that does not exist.
Hope this proves my point.

Round 2
Published:
Your welcome, it wasn't something I was trying to hide.

However, it is contradictory to say that an infinite amount of 9s before the decimal is not valid, while accepting a number with infinite 9s after the decimal.
if ...999.0 does not exit, then neither does 0.999..., which invalidates the original proof of 0.999... = 1. Thank you very much. Please vote PRO! :)
Published:
It is contradictory to say that neither of our proofs are true and because of that one side or other should win. You tried to go off topic.
So, to review this shortly and clearly:
1) 0.999... IS periodic infinite number(check it anywhere, ask teacher in elementary school, if you aren't ashamed)
2) 0.999...0 (what you tried to calculate) IS NOT an infinite number like we said, because it has end-that's 0.

Don't vote for con, but VOTE for common sense. ;)
Thank you.

Round 3
Published:
2 correction:

1. Ending in zero doesnt make it any less equal to infinity. It has infinite 9s in front of it.

2. I was not trying to prove that ...999.0 = -1. I was trying to prove, as per topic, that 0.999... does NOT =1. Specifically by invalidating the method used to prove that it does and settle a long fought debate in mathematics.
Published:
1) You said, I quote: "Ending in zero doesnt make it any less equal to infinity." I am pretty sure that everyone reading this debate at the moment will agree with me that, if something has ENDING, then it is NOT INFINITE.
2) Objection to your second statement: You were trying to prove that 0.999... is not 1 BY proving that ...999.0 is =-1. So, if your method isnt valid and clear, conclusion from it cannot be any more valid or clear. :) 
 
Round 4
Published:
1. In numbers,  "ends in" refers to the smallest part of a number, the singles spot. Thus 1,000,009 "ends in" a 9. So having a zero in the last and smallest position doesnt change the infinite 9s before it. Hundreds, millions, billions, and on to infinity.

I trully do not understand the confusion here. Your argument revolving around "ends in" is essentially a play on words.

2. My opening statement, where i did the initial =-1 equation concluded with a declaration that both results are absurd. My arguments involved how similar the numbers and methods are, and I highlighted how if either one is invalid, both must be invalid.

Ill let #2 also be my closing statement.
As a bonus, i will show method to this madness with one more number:

X = ...999.999... 
10x = ...999.999...
10x - x = ...999.999... - ...999.999...
9x = 0
X = 0

Which again seems absurd... until you realize that 
...999.0 = -1
0.999... = 1
...999.999... = ...999.0 + 0.999...
...999.999... = -1 + 1 = 0

At least the insane equations work together :)
Published:
You found the truth my friend, we all live in matrix and all world around you is a simulation. Maybe this could be a topic for some other debate...
Seriously now, I wanted to give myself some logical answer to this "problem" and I did some research. And even basic math with infinity is just...well,undefined. So, for example:
infinity+ 1= infinity, but subtracting infinity from infinity, well we could call it 'not-a-number-ness', because subtracting from infinity doesnt have a definite result,so your calculations fall short because of this. You tried to subtract infinity, and I respect that try, minds like you could change a world for better, just focus on something more practical. :))
After all, many big mathematician brains have been thinking about this long before us, so if it is consensus, it is consensus because of these things I mentioned above,and many others for sure.
We really live in a world full of absurds and anomalies, but thank you for high-quality debate.
Hope to see you in some next debates.
P.S. This could be named a 'tie' but let the votes prevail. :)


Added:
--> @drafterman
Yeah – and also, we encounter infinite series that converge on 1 in the real world all the time! It’s not like something we’ve never heard of.
You take a circle and divide its circumference by its diameter – there you have it, an infinite decimal.
You invest some money in an economy if inflation is fixed in the short run (e.g., a really really sticky economy) – and the spending exists in an infinite cycle. That’s 1 + C'(Y) + [C'(Y)]^2 + [C'(Y)]^3 ... and since C'(Y) belongs to (0,1), that’s identical to 1/(1-C'(Y)) (C'(Y) is the derivative of consumption with respect to real GDP).
#56
Added:
--> @Tejretics
You are not wrong. I get frustrated at these debates because, in most of them, even the "right" side of the debate seems to lack a fundamental understanding of mathematics.
It all boils down to what the decimal representation of a number even means. Each digit represents some multiple of a power of ten. The multiple is equal to the value of the digit at that place. The power of ten is dependent on that digits location relative to the decimal point, starting with 10^0 with the digit immediately to the left of the decimal point and increasing by one as you move left and decreasing by one as you move to the right. Thus any decimal representation can be expanded into a series as follows:
abc = a(10^2) + b(10^1) + c(10^0)
Example:
192 = 1(10^2) + 9(10^1) + 2(10^0) = 1(100) + 9(10) + 2(1) = 100 + 90 + 2 = 192
In either case of 0.999... or ...999.0 we have an infinite series. In the former case, the infinite series is equal to:
9/10 + 9/100 + 9/1000 + 9/10000 ...
Each term gets smaller and smaller and this series converges on a single value. Unsurprisingly, this series converges on (and is therefore equal to) 1. For the latter case however, we have:
9 + 90 + 900 + 9000 ...
Each term gets larger and larger and this series diverges. It means that its sum has no value (not even infinity). It just doesn't make mathematical sense to say that the sum of the series even exists.
#55
Added:
--> @Nemiroff
Sorry, I’m not very well-versed in pure mathematics, but here’s my intuitive reaction to your argument (I may be wrong).
While the number 0.999... is clearly a number in the interval [0,1], the number ....999.0 seems to equal infinity if it extends indefinitely in the leftward direction. So your proof seems to rely on adding and subtracting infinities. I may be wrong here, though.
#54
Added:
--> @RationalMadman
Vote Reported: RM // Mod action: Not remove
Vote is good
#53
Added:
Man, how do I always miss these slam dunk debates.
#52
Added:
--> @RationalMadman
“So... let me get this straight:
10-0.999...=0.000...1, correct? Ok.
The "..." in 0.000...1 represents an infinite number of 0s before the one. Since there is an infinite number of zeros, there will always be a 0 after the previous one. Thus, the 1 at the end will never be reached. Since the 1 will never be reached, 0.000...1 is the same as 0.000..., which is simply 0.
Going back to our equation, we now have 10-0.999...=0
Adding 0.999... on both sides, we get 10 = 0.999...
QED”
Where you see a 10, replace it with a 1. I had big brain fart lol
#51
Added:
--> @Nemiroff
No prob!
#50
Added:
--> @PressF4Respect
Wow. Gonna take me a bit of time to sit down and digest that, but thanks, looks interesting.
Instigator
#49
Added:
--> @Nemiroff
If you want me to show you the problem with the equation in this debate using this math, let me know ;)
#48
Added:
Dang I feel like a Math Professor now lol
#47
Added:
--> @Nemiroff
Since 0.999… is a repeating decimal, it can be expressed as an infinite sum:
0.999… = 0.9 + 0.09 + 0.009 + 0.0009 + …
S∞ = The entire infinite sum → S∞ = 0.999...
A = The first term → A = 0.9
R = The constant multiple (this can be derived from dividing the second term from the first)
R = (9/100)/(9/10) ← Dividing two fractions is the same as multiplying the first fraction by the reciprocal of the second.
R = (9/100)*(10/9)
R = 90/900 ← Simplify
R = 1/10
Since |1/10| < 1, we will use the second case of the equation. We will now plug in our numbers into the equation.
0.999… = 0.9/(1 - 1/10)
0.999… = (9/10)/(10/10 - 1/10)
0.999… = (9/10)/(9/10) ← Any number divided by itself is 1
0.999… = 1
QED
#46
Added:
--> @Nemiroff
For the second case, remember that x^n will increase (gets closer to infinity) as n increases (gets closer to infinity). In this case, N = ∞. So, R^N = R^∞ = ∞. With that, we get this:
S∞ = A(1 - R^∞)/(1 - R), |R| > 1
S∞ = A(1 - ∞)/(some negative number)
S∞ = A(-∞ + 1)/(some negative number)
At this point, it is important to note that infinity plus/minus any finite number is still infinity. Similarly, negative infinity plus/minus any finite number is still negative infinity. Thus:
S∞ = A(-∞ + 1)/(some negative number)
S∞ = A(-∞)/(some negative number)
S∞ = -A(∞)/-(some positive number) ← remove the negative from the negative number
S∞ = A(∞)/(some positive number)
S∞ = ∞/(some positive number) ← Infinity/(Any finite number) = Infinity
S∞ = ∞
For this case, we can see that the infinite geometric series equation will equal to infinity. This leaves us with the third case:
S∞ = A(1 - R^∞)/(1 - R), |R| < 1
For the third case, remember that x^n will decrease (gets closer to zero) as n increases (gets closer to infinity). In this case, N = ∞. So, R^N = R^∞ = 0. With that, we get this:
S∞ = A(1 - R^∞)/(1 - R), |R| < 1
S∞ = A(1 - 0)/(1 - R)
S∞ = A(1)/(1 - R) ← Remember: A(1) = A
S∞ = A/(1 - R)
With that said, there are two possible cases for this equation:
S∞ = A(1 - R^∞)/(1 - R), |R| > 1 → S∞ = ∞
S∞ = A/(1 - R), |R| < 1
I will now use this equation to prove that 0.999… = 1.
#45
Added:
--> @Nemiroff
Now that we have the equation for the geometric series, we need to establish what an infinite geometric series is.
An infinite geometric series is one which has an infinite amount of terms in its sequence. Since N is the number of terms in the sequence, and an infinite geometric series has an infinite amount of terms in its sequence (by definition), N = ∞. Let the entire infinite geometric series be S∞.
Retrieving the equation for the geometric series, and plugging ∞ into N, we get this:
S∞ = A(1 - R^∞)/(1 - R)
At this point, it is important to note the nature of exponents. Any number (x) to the power of something (n) will get larger or smaller when n gets larger, depending on what x is. The following will be the case for positive and negative numbers, so we can put an absolute value sign (| |) around x. This will make x an absolute value number. What an absolute value sign does is that it turns a negative number into a positive one. Positive numbers will not be affected.
If |x| > 1, then x^n will increase as n increases. (eg, 2^2 = 4, 2^3 = 8, etc.)
If |x| < 1, then x^n will decrease as n increases. (eg, (½)^2 = ¼, (½)^3 = ⅛)
If |x| = 1, then x^n will remain the same.
With this in mind, we can address the equation at hand. An absolute value number increasing can be thought of as getting closer to infinity, while an absolute value number decreasing can be thought of as getting closer to zero. There are three possible cases for the equation:
S∞ = A(1 - R^∞)/(1 - R), |R| = 1
S∞ = A(1 - R^∞)/(1 - R), |R| > 1
S∞ = A(1 - R^∞)/(1 - R), |R| < 1
For the first case, we get this:
S∞ = A(1 - R^∞)/(1 - R), |R| = 1 → S∞ = A(1 - 1^∞)/(1 - 1) → S∞ = A(1 - 1^∞)/0
There are two problems with this. Firstly, we cannot divide by zero. Secondly, 1^∞ is indeterminate. Thus, we can outright eliminate this case.
[continued in next comment]
#44
Added:
--> @Nemiroff
Since any number subtracted by itself is 0, we are left with:
S - SR = A - AR^N + 0 + 0 + 0 + ...(all of the terms in here are 0)... + 0 + 0
Simplified, we get this:
S(1) - SR = A(1) - AR^N
Since any number times 1 is itself, the number can be restated as itself times 1.
We can factor out the S and the A to get this:
S(1 - R) = A(1 - R^N)
Since we are trying to find S, we divide both sides by 1-R. As a result, we get:
S = A(1 - R^N)/(1 - R)
And this is the equation for the geometric series.
[Continued in next comment]
#43
Added:
--> @Nemiroff
This goes on for the amount of terms that are in the sequence. Let's call the number of terms in the sequence "N", and the sequence as a whole as "s".(Again, what we call these by is completely arbitrary)
So, a geometric sequence can be shown as:
s = [A, AR, AR^2, ..., AR^(N-3), AR^(N-2), AR^(N-1)]
The reason that the final number in the sequence is AR^(N-1) and not AR^N is because on the 1st term, N = 0 (R^0 = 1, and A*1 is simply A).
Since a series is simply the sum of a sequence, a geometric series (let's call the variable for the entire series "S") can be shown as:
S = A + AR + AR^2 + ... + AR^(N-3) + AR^(N-2) + AR^(N-1)
This is our first equation for the proof.
If we multiply R by both sides, we get:
SR = R * [A + AR + AR^2 + ... + AR^(N-3) + AR^(N-2) + AR^(N-1)]
Remember that the exponent simply shows the number of times that something is being multiplied by. So, if we multiply something raised to the power of n (n is simply the exponent of that something) by that same something, then it is the same as adding 1 to n. With that in mind, we get:
SR = AR + AR^2 + AR^3 + ... + AR^(N-3+1) + AR^(N-2+1) + AR^(N-1+1)
= AR + AR^2 + AR^3 + ... + AR^(N-2) + AR^(N-1) + AR^N
This is our second equation.
Now, we subtract the second equation from the first (First equation is S, second equation is SR):
S - SR = [A + AR + AR^2 + ... + AR^(N-2) + AR^(N-1)] - [AR + AR^2 + AR^3 + ... + AR^(N-2) + AR^(N-1) + AR^N]
The minus sign in front of the first series can be distributed into each number within the series:
S - SR = [A + AR + AR^2 + ... + AR^(N-2) + AR^(N-1)] + [-AR - AR^2 - AR^3 - ... - AR^(N-2) - AR^(N-1) - AR^N]
We can get rid of the brackets:
S - SR = A + AR + AR^2 + ... + AR^(N-2) + AR^(N-1) - AR - AR^2 - AR^3 - ... - AR^(N-2) - AR^(N-1) - AR^N
Then we can rearrange the equation:
S - SR = A - AR^N + AR - AR + AR^2 - AR^2 + AR^3 - AR^3 + ... - ... + AR^(N-2) - AR^(N-2) + AR^(N-1) - AR^(N-1)
[continued in next comment]
#42
#5
Criterion Pro Tie Con Points
Winner 1 point
Reason:
First of all, whatever either said elsewhere, is not proof within this debate. Second, there's no need to accuse anyone of living in the matrix.
While I did not find pro's math convincing, it was more than I did not find his math convincing as opposed to finding con's counter case convincing (maybe breaking R1 into multiple paragraphs would have helped; and yes, for confusing stuff like this, it's really best to break the concepts apart).
I will say that infinite 9's to the left of the decimal, probably wouldn't change to infinite 9's followed by a 0 when multiplied within the base-10 system.
#4
Criterion Pro Tie Con Points
Winner 1 point
Reason:
This Math is really confusing my brain, and I feel like oromagi and Madman have covered the Math part really well, way better than I could have every tried. But here's my opinion on the non-Math things...
Neither of you guys had a format, or a full speech, even though you were given 5000 characters to do so. With a 5000 character limit, I would except solid arguments with a clear format and organized clash, but that wasn't present here.
Pro however contradicts himself using himself in the first round, which I find quite odd. He tries to disprove his own case, but ends up cancelling his case out. Con takes notice of this.
Due to that reason, and the other voters reasons, Con wins.
#3
Criterion Pro Tie Con Points
Winner 1 point
Reason:
PRO begins with solid evidence for the notion he has offered to disprove and then promises to disprove the notion using the same equation.
But PRO uses a new equation (10X - X =/= X - 10X), inserting a negative result and proving zilcho. Likewise, he replaces the irrational value of his equation with an infinite number which behaves quite differently.
Either side arguing for the "existence" of one irrational or infinite number is oxymoronic when by definition such numbers can't be accurately represented much less exist. e. I don't think PRO is using correct or legit notation for infinite nines but when CON tries to de- legitimize an infinite number of nines multiplied by arguing the irrational number he is not persuasive.
Ultimately, PRO loses this argument because he promises a magic reversal but we can see the chosen card peeking out of his sleeve.
#2
Criterion Pro Tie Con Points
Winner 1 point
Reason:
Pro did two things wrong in this debate, despite being on the objectively correct side. I don't know how he even made these errors as I have personally outside of this debate provided him the killer proofs that 0.999... is not equal to 1.
The first thing he did wrong was to bother using the erroneous algebra construct that people use to hocus pocus 0.999... to be equal to 1, then to replace it with 999.0 (instead of 999.999... which would indeed end up making it equal to 1000.000 and that's the whole point, that algebra construct enables erroneous conversion). This was correctly attacked by Con, albeit with strange wording, in the following Round. Con of course is erroneous because the attack Con uses is that 0 can't follow infinite 9's. What Con should have argued was that the number at the start was not at all like 0.999... since it doesn't have infinite 9's following the decimal point.
Pro does start to tackle this in the way I have introduced Pro to outside of this debate but Pro does it wrong. Instead of pointing out that if 0 can't follow infinite 9's, neither can the imaginary '9' at the end of it all, Pro decides to play around even more with the abusive algebra which Pro should be showing is corrupt because you can't reverse engineer it (the fact you can never start with 1 and end up with 0.999... means that there is clearly something erroneous about the algebra, even if you completely flip around the equations, you never end up with 0.999... at the start, this key way of negating the algebra isn't employed by Pro in the entire debate). Pro starts to keep almost joking around with his 'proof' that 0.999... is not equal to 1 with some really confusing algebraic display that tries to show you can make something else happen with algebra and infinite series that involve 9. This had nothing to do with the debate. What Con does is also very silly and minimalistic, but I know why Con did that. Con is on the side that is a lie, therefore Con's best bet at winning is to avoid going into details and depth and instead sticking to rhetoric like "a 0 can't follow infinite 9's therefore you are wrong" well, neither can a 9... But Pro doesn't hit that point home.
Neither side brought up the 1/3 - 0.333... angle, this is a much stronger case for Con to make as it's backed up by school syllabus content (which lies but is considered highly reliable due to what it is). This then means that 1/3 * 3 technically equals 0.999... and the only way for Pro to counter it is to expose that 1/3 does not equal 0.333... but instead equals 0.333...0333...0333 infinitely over and over again. Pro doesn't do this but Con doesn't bring it up.
Since this debate is structured with Pro having the burden of proof and since Con does kind of 'hit home' a point about that 0 can't follow infinite 9's and pro doesn't turn it back on Con with the '9' following the infinite 9's, Con takes the win.
#1
Criterion Pro Tie Con Points
Winner 1 point
Reason:
Pro opened with two mathematical equations that contradict each other. Con denied that the second equation was valid because he claimed ...9990 doesn't exist. Con then attempted a play on words to say that 0.999...0 doesn't exist because an infinite number can't end. Pro refuted these claims by pointing out that the existence of 0s at the beginning or end of a number does not prevent there from being an infinite number of 9s before the end or after the beginning.
Overall, neither side was able to prove their case. Pro relied on contradictory equations and Con stuck to attempts to refute Pro's claims.
The only source was Pro citing himself. Both sides had good conduct and S&G.