The Instigator
Pro (for)
Winning
62 Points
The Contender
Con (against)
Losing
52 Points

# 0.9999 recurring is equal to 1

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after 27 votes the winner is...
Logic_on_rails
 Started: 1/19/2011 Category: Education Updated: 2 years ago Status: Post Voting Period Viewed: 3,042 times Debate No: 14439
Debate Rounds (3)

 Pro Intro The 0.999 recurring is to mean a continuous stream of nines, ever ending. I assume people know what I mean here. The Pro's burden is simple, to prove that 0.99 recurring is indeed equal to 1. I'm looking to see if there's any challengers out there. And before anyone gets at all tricky (sneaky, making new rules), don't get ridiculously tricky, it's also of course in base 10 that we're talking (no base 16 etc.) . Finally not every dash (-) or such is a minus sign or such. Case Let's say we have 2 numbers, A and B Firstly we times both by 10 - we get 10A and 10B Then we take 1 of the pronumeral from both 10A and 10B - 9A and 9B Now, if 9A = 9B then it is fair to simplify and say A=B. Essentially: A*10 = 10A B*10 = 10B 10A - A = 9A 10B - B = 9B If 9A=9B, A=B Keep this in mind. Now let's do it with the numbers in question. 1*10=10 (A*10) 0.9999... *10 = 9.999... (recurring) (B*10) 10-1=9 (10A-A) 9.999... - 0.999... = 9 (10B-B) 9=9 (9A=9B) Conclusion - A=B or in this case 0.99999999... (recurring) =1 Is there any person on the website who doesn't believe? Report this Argument Con .999... is not a whole number, it is infinite. You cannot subtract two infinite numbers and get a whole number, which is what you are doing. Saying [9.999... - .999 = 9] is flawed logic. You are at some point rounding .999... to 1. In calculus we are taught that an infinite number minus an infinite number will equal an infinite number. [9.999... - .999] is the same as saying [Infinity - Infinity]Report this Argument Pro Intro It probably doesn't help that I haven't done calculus, but I can still spy multiple flaws in my opponents logic. Furthermore my bottom argument conforms to the top argument, so by not attacking the top argument (proof if anyone wishes) , my opponent concedes that point, and the bottom is merely an application of the top, rendering it correct. Nevertheless... My case 1. If we take the opposite method in proving this case 1-0.9999... we see that there is a difference of 0.000... we will never see the one, resulting in a difference of 0. 2. On refuting my opponent the idea of 9.99... - 0.999.. being infinity - infinity is quite wrong. There is clearly a difference of 9. Indeed we times one number by 10 to get the second no. 3. How about adding a negative infinity to positive infinity, that should be 0. There's a list of other points, but you never need to round - 0.99... - 0.99... is 0, the are the same number. 4. Of course 0.999... isn't a whole number in that form, whole numbers are essentially integers (not quite accurate of course) What my opponent probably means is something like 0.888... - 0.777... = 0.111... and being unable to simplify 0.111... Con's case is refuted. Vote Pro.Report this Argument Con If 1 = 0.999... then since 1 - 1/infinity = 0.999..., 1/infinity must be proven to equal zero. Zero, when multiplied by anything, must equal itself (that is, zero). though 22 multiplied by 1/infinity equals 1/infinity (because 22/infinity equals 1/infinity), infinity multiplied by 1/infinity yields one. Zero multiplied by infinity yields zero, not one, so 0 does not equal 1/infinity so 1 does not equal 0.999...Report this Argument Pro Intro Con still hasn't attacked my main proof, granting it valid. And since the second try is based on the first, attacking the second is pointless. Also, I'm not sure if layout comes under any category of voting, but I'd like voters to realise the incoherence of my opponent's arguments. Without ado, let's proceed. My Case 1. The point is still unaddressed. To ever assume a difference is to assume a finite number of 9s, but there are an infinite number of 9s. 2-4 - Not countered, not will to waste time on them. New points: 1. If two numbers are different it should be possible to find another number in between the 2 numbers, but as I'm sure voters can deduce, this is not the case. 2. The number 0.9999.... can be expanded as: 0.9999... = 0.9+0.09.... ie. = 9/10 + 9/100.... = 9/10 + 9/10(1/10).... =9/10 + 9/10(1/10)^1 +9/10(1/10)^2.... In other words: 0.9999... = (9/10)(1/(1-1/10)) ie. = (9/10)(1/(9/10)) = (9/10)(10/9) =1 Voters, the resolution has been adequately proven in multiple ways and through multiple proofs. I strongly urge a Pro vote.Report this Argument Con I apologize for the vague second round, I wasn't near a computer for a while so I rushed something out when I saw I was near the time limit. .999... is an infinitesimally small number. This is part of the reason of why limits exist. If you were to draw this on a graph, the .999... would approach the number 1, but never intersect it. It would just get infinitely closer to it. The proofs that Pro has provided are correct under his interpretation of infinity. But to me 9.999... -.999... cannot compute, its like trying to divide an apple by an orange. To me they are just words now, since infinity cannot be expressed by any simple numerical means like this. It almost seems too easy, if you catch my drift. For example, if I wanted to subtract .999... from 9.999... I would have to perform subtraction for every number in 9.999... This process would continue into infinity, since the amount of 9's I would have to take into consideration are infinite. For Pro's proof to work means you have to stop subtracting at some point to get an answer of 1, which to me is not possible. What this debate has come down to is just an interpretation of infinity, which if I am correct, makes Pro's evidence and proofs null and void.Report this Argument
49 comments have been posted on this debate. Showing 11 through 20 records.
Posted by XimenBao 2 years ago
It does.
Posted by twsurber 2 years ago
2+2=5 Winston or whatever the Party tells you it equals.

I see what the AFF is arguing, but we all know that 0.00000000001 has got to be out there somewhere. If not, then 0.999 with a bar over it plus 0.0000001 should equal 1.000000001 or words to that effect.

Y'all know what I mean. Regardless, I voted neutral.
Posted by feverish 2 years ago
I'm no mathematician but Pro's argument made a lot more sense to me. One simple way of looking at this is that .3 recurring is one third and three thirds make one whole.
Posted by RoyLatham 2 years ago
Pro is correct and the proof offered is correct. A rational number like .999.. is not infinite, it is finite. Even if you don't think it is exactly equal to one, you must know is s very close to one. One is finite.

Th proof of 0.999... = 1 is in many math textbooks. Maybe reading it in math text would help those who don't understand it. It's not a debatable issue because it's a done deal. It's ike debating 2+2 = 4.
Posted by mongeese 2 years ago
That number arguably doesn't exist. But then, I guess that's what makes it a debate.
Posted by mongoose 2 years ago
.999... * 10 = .999...0

that
Posted by Grape 2 years ago
Unlike Pro in this debate, I'm going to specify a number system in which you cannot wriggle around the analytic proofs by using hyperreals. Are you still interested in debating if I restrict the discussion to real numbers?
Posted by Ore_Ele 2 years ago
I would challenge Pro on this (I'd take Con), or any one else. So long as the word "equals" is in the resolution.
Posted by FREEDO 2 years ago
RFD: Reverse vote-bombed.
Posted by XimenBao 2 years ago
although perhaps I should have said value rather than number. In any case, neither is equivalent to integer.
27 votes have been placed for this debate. Showing 1 through 10 records.