The Instigator
thegoddamnreaper
Con (against)
Losing
0 Points
The Contender
JustVotingTiedDebates
Pro (for)
Winning
6 Points

0.999... is equal to 1

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Post Voting Period
The voting period for this debate has ended.
after 2 votes the winner is...
JustVotingTiedDebates
Voting Style: Open Point System: 7 Point
Started: 7/7/2016 Category: Education
Updated: 7 months ago Status: Post Voting Period
Viewed: 560 times Debate No: 93439
Debate Rounds (3)
Comments (16)
Votes (2)

 

thegoddamnreaper

Con

This is the first round. Any one interested can join.
JustVotingTiedDebates

Pro

I accept.
Debate Round No. 1
thegoddamnreaper

Con

0.999... can never equal to 1 as there is an infinitesimally small difference between that and 1.
This, Pro might argue, is exactly the reason why it is equal to 1, as the infinitesimally small difference is negligible. As the 9's progress, the difference is getting closer and closer to 0. Since there is an infinite number of 9's, the difference might as well be 0.
This argument is ridiculous. Infinitesimals are used all the time in calculus. In fact, infinitesimals is root of calculus. So, statements such lim(x->1) (x/(x-1)) would make no sense. Infinitesimals require there to be a difference. Infinitesimals is a well defined concept and used frequently.
JustVotingTiedDebates

Pro

Simple.

0.33 = 1/3
0.66 = 2/3
0.99 = 3/3 = 1

I strongly ask voters to vote by power of debate not by their own opinions.

Thanks,
Adil,
Qatar.
Debate Round No. 2
thegoddamnreaper

Con

1/3=0.333.... and 2/3=0.6666....... Fine, I get these two.
But, how do you say, 0.33333....+0.666....=0.9999....? Addition is not defined for these numbers. You would have to define all the arithmetic operations to even say that 0.333...+ 0.666...=0.999....
To put it simply, you have not proved all the operations working on normal, rational numbers would work on this mathematical entity, this repeating decimal number.
Argument in the comments:
The "algebraic proof" goes like this.
Let x=0.9999....
10x=9.9999...
10x-x=9.999...-0.99999...
9x=9
x=1
The same problem lies with this argument. There is no proof of why the arithmetic operations would work on repeating decimal numbers.
Making an axiom out of this fails too. For that matter, you could have made an axiom straight away stating 0.9999...=1 The opponent is required to make an argument for the same without creating unnecessary axioms.
And, as Pro said, vote based on the power of the debate, not on your preconceived notions.
JustVotingTiedDebates

Pro

I have two things to say in the last round.

1. I heartily thank my opponent for giving me another proof that 0.999=1 which is:
Let x=0.9999....
10x=9.9999...
10x-x=9.999...-0.99999...
9x=9
x=1.

2. You say: "But, how do you say, 0.33333....+0.666....=0.9999....? Addition is not defined for these numbers." I never said that. Please read my argument in the second round again.

So everybody, I repeat:

There are two proofs that 0.999=1

1.
0.33 = 1/3
0.66 = 2/3
0.99 = 3/3 = 1

2.
Let x=0.9999....
10x=9.9999...
10x-x=9.999...-0.99999...
9x=9
x=1

VOTE PRO.

I strongly ask voters to vote by power of debate not by their own opinions.

Thanks,
Adil,
Qatar.
Debate Round No. 3
16 comments have been posted on this debate. Showing 1 through 10 records.
Posted by thegoddamnreaper 7 months ago
thegoddamnreaper
@Sashil This took me a while, I admit. The answer is 0. The paper boats will be floating upstream and A is downstream. How can A ever get a paper boat?
Posted by Sashil 7 months ago
Sashil
@ thegoddamnreaper Okay,so I understand you're a person who likes something challenging.
Try solving this problem then :
Three friends A, B and C - go boating in a stream and decide to play a game. B and C are at a point X at time t = 0 seconds. B is on a boat which is floating with the stream and C is on a boat which is anchored at X. Both B and C release paper boats at intervals of 5 seconds, beginning at t = 0 seconds, A starts from a point Y, downstream of X, at t = 0 seconds and starts collecting all the paper boats he encounters as he rows upstream towards X. The speed of A"s boat in still water is thrice the speed of the stream. Find the total number of paper boats collected by A, if B reaches Y at t = 132 seconds.

P.S: This doesn't involve any complex mathematics :P
Posted by Sashil 7 months ago
Sashil
@ lord_megatron There's no pattern, the question answers itself actually cause it begins with the statement: hypothetically if 1=5 :P
Posted by thegoddamnreaper 7 months ago
thegoddamnreaper
@Sashil this is a stupid question, but the answer is 1.
1=5 so 5=1
Posted by lord_megatron 7 months ago
lord_megatron
@Sashil there seems to be no rational pattern, especially due to the 4th number
Posted by Sashil 7 months ago
Sashil
infinitesimally*

In vote.
Posted by Sashil 7 months ago
Sashil
okay a fun question to both the contestants...If hypothetically,

1=5
2=25
3=125
4=1125

What is 5=??
;)
Posted by thegoddamnreaper 7 months ago
thegoddamnreaper
Oh, wait a second. Got the burn.
Posted by thegoddamnreaper 7 months ago
thegoddamnreaper
Thank you, mcMount! Vote Con.
Posted by McMount 7 months ago
McMount
This is fun to read because Con is actually posting stronger arguements on behalf of Pro than Pro himself.
2 votes have been placed for this debate. Showing 1 through 2 records.
Vote Placed by lord_megatron 7 months ago
lord_megatron
thegoddamnreaperJustVotingTiedDebatesTied
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Total points awarded:03 
Reasons for voting decision: Con accidently gave another proof to pro. Pro use mathematical analogies that con couldn't rebut. Con said there is is an infinitesimally small difference between 0.99 and 1, but pro's mathematical analogies show that it is equal. The first one was since 1/3=0.33, 2/3=0.66, 3/3=1 or 0.99 therefore they are equal, and x=0.99 10x=9.99 10x-x= 9 9x/9=1 x=1.
Vote Placed by Sashil 7 months ago
Sashil
thegoddamnreaperJustVotingTiedDebatesTied
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Total points awarded:03 
Reasons for voting decision: CON argued that there will always be an infinitesimally small difference between 1 and 0.9999 while PRO offered a mathematical proof showing that 1 is infact 0.9999... CON claimed that Addition is not defined for these numbers but since it's undefined there's no way to prove the credibility of PRO's proof. Since the proof couldn't be proven right we face a stalemate kind of situation in this debate. But I'm going to give the benifit of doubt to PRO here because CON never explained why such operations on the said numbers should be considered undefined. There aren't any sources supporting his statement nor is there any rationale provided as to why one must consider so. And btw PRO when you said "I never said that(about addition). Please read my argument in the second round again." You multiplied 1/3 by two which is necessarily adding 0.3333 + 0.3333. If CON had more cogently pursued with his argument of the operation being undefined you could have very well lost this debate.