The Instigator
AKMath
Pro (for)
Winning
3 Points
The Contender
asta
Con (against)
Losing
0 Points

0.999...= 1 | Prove Me Wrong

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Post Voting Period
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after 1 vote the winner is...
AKMath
Voting Style: Open Point System: 7 Point
Started: 6/14/2018 Category: Education
Updated: 3 years ago Status: Post Voting Period
Viewed: 571 times Debate No: 115551
Debate Rounds (5)
Comments (5)
Votes (1)

 

AKMath

Pro

0.999... = 1

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

1/3 = 0.333...
2/3 = 0.666...
3/3 = 0.999...

What number comes in between 0.999... and 1?
asta

Con

If:

.999....=1
-.999...

0=.000...1

The number on the right is infinite times as big as the number on the left.
Debate Round No. 1
AKMath

Pro

-0.999... equals -1.
x = 0.000...1
10 x = 00.000...1
10x - x = 00.000...1 - 0.000...1
x = 0
So yes 0 = 0.000...1
Stop beating around the bush and start disproving my other two proofs please.
asta

Con

"2/3 = 0.666..." No. 2/3=.6...7, not .666. Therefor, 3/3=1.

"x = 0.000...1
10 x = 00.000...1" No. 10x would be 0.0...10.

"10x - x = 00.000...1 - 0.000...1
x = 0" You didn't even correcctly solve the problem here. THe problem's answer is 9x=.00...09.

So, nothing is less then something.

I await your response.
Debate Round No. 2
AKMath

Pro

As I've said before. Stop beating around the bush and forget about my x = 0.999... thing. Start disproving the other two proofs or you've pretty much surrendered and agree that 0.999... = 1.
asta

Con

Your 2 "proofs"

1)
"x = 0.999...
10x = 9.999..."

10x=9.99....0 because multiplying by 10 is (for lack of better terms) moves the number left one relative to the decimal point. This means that after the infinite run of nines, there is a 0 to fill the void.

2) A imilar thing exists for your 2nd proof.

You said, "0 = 0.000...1" The number on the right is "infinetly small and not yet 0". This quote is so famous that there is an article about it.

http://www.massline.org...

Debate Round No. 3
AKMath

Pro

Disprove these two proofs not the other one. Forget about the x = 0.999... thing. Try to disprove these two:
Proof 2:
1/3 = 0.333...
2/3 = 0.666...
3/3 = 0.999...
Proof 3:
What number with a decimal comes in between 0.999... and 1. All numbers that have a difference have a number that comes in between. So what number comes in between 0.999... and 1?
asta

Con

1) "Proof" 2 is wrong because if 1/3=.333.... then 2/3=.666....7 because it rounds at the end. 3/3 is therefore 1. The calculator says so.

2) "Proof" 3 is wrong because after infinite 9's, there could be a 5. If x is between .999.... and 1, it could be .999....5. .999....5 is more then .999.... because if you remove the repeating periods, .9995>.999.

I hope this makes sense.
Debate Round No. 4
AKMath

Pro

First of all I forgot to mention a proof. No matter who you are you can't disprove this one. If you type in 0.99999999999999999999 into a calculator, and push = it says it equals 1. And I mean literally put 20 9's. Any less and it may not work.

1) How does 0.333... + 0.333... = 0.666...7? Put 0.3333333333 + 0.3333333333 into a calculator it gives you 0.666666...
2) You can't take out the ... from 0.999. This completely changes the number. ... represents that the number goes on infinitely. The difference between 0.999... and 0.999 is so big I couldn't possibly begin to explain it. A simple explanation is that 0.999 is finite, but 0.999... is infinite. As I have mentioned before you can take out the ... because it makes the number finite. Consequently it loses the power of infinity. So you admit that there is no number between 0.999... and 1, because you had to change the problem to answer it.

So no this doesn't make sense.
asta

Con

While https://www.calculator.com... with you, It's because most calculators round in extreme situations like the one you presented. Calculators tend to round on 2^x because they have no other choice on huge exponents to this term and this is non disputed.

1) The calculator says you were right on this one. However, it does not prove that .999......=1. It merely proves that the online calculator says that .333...*2=.666.... The calculator also said that when you multiply that number by 3, you get 1, not .999..., but 1.
2) "You can't take out the ... from 0.999." I only took out the ... in order to be able to kindof solve the problem algebraically.

Even though I changed the number this kindof applies to logs. You need to do some math to figure out those. Just like algebra, I took .999..., and I wanted to see if it would be greater then .999...5. In order to do this, I would have to subtract what would be in common with them. The relevant portion would be the ... since both numbers have them. Subtract them from each sides, it can't tell you if they are equal, but they can you which is the bigger number.
Debate Round No. 5
5 comments have been posted on this debate. Showing 1 through 5 records.
Posted by asta 3 years ago
asta
I was talking about the number, not the decimal point.
Posted by Masterful 3 years ago
Masterful
Calculators are programmed to round up. I myself agree that 0.999.. does not equal 1.
Posted by Masterful 3 years ago
Masterful
I'm sure it's to the right

3.00
30.00
30.000
Posted by asta 3 years ago
asta
I didn't move the decimal place backwards. I moved the number to the left. This is what you do when you multiply.

Example:
0.30*10=
3.00.
Posted by Masterful 3 years ago
Masterful
Why did you move the decimal place backwards when multiplying by 10?
"x = 0.000...1
10 x = 00.000...1"

10x would equal 0.000...10
1 votes has been placed for this debate.
Vote Placed by RMTheSupreme 3 years ago
RMTheSupreme
AKMathastaTied
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Total points awarded:30 
Reasons for voting decision: Con defeated their case. If you have to eventually turn 2/3 into a number with 7 then you need to round up the 0.9 recurring if it is brought up in the same context. Pro successfully trolled Con and Con didn't catch onto him defeating his own case with asserting rounding up 2/3 to a number ending in 7 which is what Pro is saying we must do with 0.9 recurring.

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