The Instigator
dev_101
Pro (for)
The Contender
jackgilbert
Con (against)

0. 999. . . Is not equal to 1

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Debate Round Forfeited
jackgilbert has forfeited round #3.
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Voting Style: Open Point System: 7 Point
Started: 8/3/2018 Category: Education
Updated: 1 week ago Status: Debating Period
Viewed: 358 times Debate No: 117299
Debate Rounds (4)
Comments (9)
Votes (0)

 

dev_101

Pro

You may use any ways, Including philosophical reasoning to argue your case however it must be related to the topic.

I will be arguing to convince people that 0. 999. . . Recurring is not equal to 1 and please do not use the argument such as 0. 333. . . Is equal to 1/3 as it falls under the same category.

-----------------------------------------------------------------------------------------------------------------

The common proof used to prove this is:

let x = 0. 999. . .
10x = 9. 999. . .
9x = 9
x = 1
1 = 0. 999. . .

The key fault lies in 10x = 9. 999. . .

In general when we multiply or divide by 10, 100, 1000, Etc. We are told to we move the decimal point left or right (respectively) however this was a simplification of the original method which requires there to be 'infinite' zeros before and after the decimal point. If we only look at the decimal point part please explain how you get 0. 9 * 1000 = 90. 0 purely by moving the decimal point without considering those 'invisible' zero

Therefore it should really have been:
let x = 0. 999. . . 9
10x = 9. 999. . . 0
9x = 8. 999. . . 1
x = 0. 999. . . 9
x = 0. 999. . . 9

The. . . Z is to keep track of last digit based on inductive paterns and the zero always exist rule

Here what you started off is what you ended with many would say. Yyy. . X makes no sense as there is infinite y so you will never reach x. If you can't reach it does not necessarily mean it does not exist. The function f(x) = 2x also has infinite values however you can still transform it, As a matter of fact, There is no way you can apply any operation to any set of numbers without using the recurrence proof or infinitesimals numbers to get 0. 999. . .
jackgilbert

Con

I have a few proofs why 0. 9999. . . =1

Algebraic proof: When refuting this you assume that there is an end to 0. 9999. . . , But there isn't. When you move the decimal point to the right of 0. 999. . . You don't eliminate any 9's because it is infinity. Thus, Infinity-1 equals infinity.
X=0. 9999. . .
10x=9. 99999. . . . .
10x=9+0. 9999999. . . . .
10x=9+X
9x=9
x=1

Proof 2:
1/3=0. 33333. . . .
2/3=0. 66666. . . . .
3/3=0. 999999
Because 3/3 equals 1
1=0. 999999. . . . .

Proof 3:
If 0. 9999. . . . Does not equal 1, We should be able to find a number between them. Let's subtract 0. 9999. . . . From 1.
1-0. 99999. . . . =0. 0000. . . . .
The dot dot dot means infinitely many 0's. So
1-0. 999999=0. 00000. . . . OR
1-0. 999. . . . =0
Debate Round No. 1
dev_101

Pro

Algebraic proof: When refuting this you assume that there is an end to 0. 9999. . . , But there isn't. When you move the decimal point to the right of 0. 999. . . You don't eliminate any 9's because it is infinity. Thus, Infinity-1 equals infinity.

This proof was already mentioned on round one as well as the counter for it stating it again does not help the case in any way, You have also failed to answer the question for this proof which was:

how you get 0. 9 * 1000 = 90. 0 purely by moving the decimal point without considering those 'invisible' zeros?

The movement of the decimal point relies on the zeros being there if not please show proof on how moving the decimal point not just for any multiple of 10, Without considering those 'INVISIBLE ZEROS' and without changing the question.


When refuting this you assume that there is an end to 0. 9999. . .

When I wrote 0. 999. . . 9 the 9 after the dots is based on the same pattern we assume the 9 goes on forever. The zero still has to exist otherwise the movement of the decimal place can NOT occur (if not please answer the question above with those requirements set). Which is why I mention the function f(x)=2x. The function has an infinite set of outputs however the function can still be transformed, In reference to this the zeros exist both on the left and right side otherwise you can not just move the decimal point as it relies on the zeros being there.

In reference to infinity-1=infinity has caused twice as many paradox as it has solved because we are oversimplifying and using operation on concepts that are not numbers.

Similar to the fact we are walking using over brain even if we feel the body is doing it by themselves in other words we can not oversimplify things and forget about the basic requirement otherwise we get results like this.

Proof 2 ignored: reason debate rule also remember fractions have different values for different bases.

Proof 3:
If 0. 9999. . . . Does not equal 1, We should be able to find a number between them. Let's subtract 0. 9999. . . . From 1.
1-0. 99999. . . . =0. 0000. . . . .
The dot dot dot means infinitely many 0's. So
1-0. 999999=0. 00000. . . . OR
1-0. 999. . . . =0

I'm going to assume you are not going to count epsilon as an answer between 0. 999. . . And 1 because it is not a real number if that is indeed your answer then you answered your own proof. If you restrict which class of numbers can be used to compare the difference does not automatically suggest that the answer is valid for all cases.

Also when you are subtracting you got zero for all the remaining numbers this is only possible if there was a first carry over where the 0 was a 10 and not 9.

It is easier to convince people that, As no real number exist between the two hence they are equal. This is because people will just think about rounding and the fact that they are extreamly close (not many people know or think about infinitesimal numbers). However, This argument should also make the following statement valid.

There exist no integer between 1 and 2 therefore 1 = 2.

According to your proof, This should be valid as I restricted the numbers used hence I got this result. Tell me how many people would be convinced by this proof and what is the main difference between this and your proof?

Also in your proof, You are only looking at a precession of 5(changes per line) and ignoring the rest because of the pattern so you get this answer in the integer proof the FULL numbers are used and are not constructed from a pattern.
jackgilbert

Con

Because there is no end to the 9's in 0. 9999. . . , There are no 0's that follow it.

If 0. 9999. . . . . . Does not equal one, Then there should be SOME NUMBER between them. 1 does not equal 2 because there is a distinct number between them. That number could be 1. 5, 1. 72, 1. 2, Etc. However, We can't find any distinct number between them. When we subtract them we get 0.

1-0. 9999. . . =0. 0000. . . . .
There is not end to the zero's so. . .
1-0. 99999. . . =0
Any number subtracted from itself equals 0. When we subtract 0. 99999. . . . From 1 we get 0. Hence why 1=0. 999. . .
Debate Round No. 2
dev_101

Pro

Sorry I just realised I mistakenly wrote 1000 instead of 100

how do you get 0. 9 * 100 = 90. 0 purely by moving the decimal point without considering those 'invisible' zeros?

This is the SECOND time you have ignored my question and once again stated the same point. In my case, I already mentioned that there are always zero left and right hence when you multiply or divide you move the number left or right respectively just like how you transform function.

"The entire number moved one “slot” to the left on the place value chart. This looks like moving the decimal point in the number to the right. "

source: https://www. Homeschoolmath. Net/teaching/d/multiply_divide_by_10_100_1000. Php

If 0. 9999. . . . . . Does not equal one, Then there should be SOME NUMBER between them. 1 does not equal 2 because there is a distinct number between them. That number could be 1. 5, 1. 72, 1. 2, Etc. However, We can't find any distinct number between them. When we subtract them we get 0.

I did mention a number epsilon
However, As this proof is usually argued with real numbers in mind is why I mention the integer case "1. 5, 1. 72, 1. 2" a not integer so they can not be used when finding an integer between 1 and 2. This is exactly the same as when people say:

"epsilon is not part of the real numbers, Therefore, It can not be considered when looking for a real number between 0. 999. . . And 1"


The 0. 000. . . 1 is actually in your subtraction proof

first simple subtraction:

1. 000
-0. 999
=====
||
\/
0 9 9 10
1. 0 0 0
- 0. 9 9 9
==========
0. 0 0 1


notice how for the two zeros on the right of the decimal to be 9 requires the left most to be a 10 to produce the 1 after subtraction hence giving the result as 0. 001

1 - 0. 999. . . Proof:

1. 000. . .
- 0. 999. . .
=======
||
\/
1. 000. . . 0
- 0. 999. . . 9
========
||
\/
0 9 9 9 10
1. 0 0 0. . . 0
- 0. 9 9 9. . . 9
===========
0 . 0 0 0. . . 1 <= The number you disagree with


In order to get the answer of 0. 0000. . . . . the number you disagree with must exist as there has to be a rightmost 0 which becomes a 10 for all the others to become 9.

if you reached 0. 0000. . . . . in any other ways please show your full method

subtraction method source: https://www. Skillsyouneed. Com/num/subtraction. Html

PS: Sorry about the formatting of the calculations but I hope it's still understandable

"If there was a race which can record infinite precession ad 1st place price is 1M <currency> and 2nd place was 100K <currency> and third 1K <currency>, In the first race two players tied with 0. 999. . . Seconds and 1. 0 seconds because of what you said then they re-raced and the person who scored 1 second last time won. Is this fair on the person who should have won first time round with 0. 999. . . ? "
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Debate Round No. 3
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Debate Round No. 4
9 comments have been posted on this debate. Showing 1 through 9 records.
Posted by dev_101 3 days ago
dev_101
As @jackgilbert forfeited round #3 how do I get the debate to resume onto the next round?
Posted by dev_101 1 week ago
dev_101
@jackgilbert The text in bold are your statements and my main statements and questions for you to answer
Posted by felixmendelssohn 1 week ago
felixmendelssohn
well @dev_101
let x = 0. 999. . . 9
10x = 9. 999. . . 0
by doing this you assumed the series of 99999 has an ending and that ending is followed by a bunch of zeros which contradict the condition that the series of 9s continues indefinitely.
your proof contradicts your assumption thus its fallacious. Plus, Euler and many brilliant mathematicians have not dispute this result, To claim that you alone hold the truth to the entire math community's misconception is a bit arrogant dont you think
Posted by dev_101 1 week ago
dev_101
@jackgilbert I did not claim it to be finite the explanation was given in the last paragraph If the notation was not clear I will re-explain in detail in the next round, Next time please read the entire argument before writing your argument as most of the things you wrote was already stated and countered.
Posted by jackgilbert 1 week ago
jackgilbert
@Dev_101- I did refute it. You assumed that infinity is a finite number and that if you subtract 1 from it, You get one less than that number. But infinity doesn't work like that. If you have infinitely many of something, Taking one away from it does not make it smaller. Just like 0. 999. . . . . - one of the 9's equals 0. 9999. . . . . .
Posted by jackgilbert 1 week ago
jackgilbert
@Dev_101- I did refute it. You assumed that infinity is a finite number and that if you subtract 1 from it, You get one less than that number. But infinity doesn't work like that. If you have infinitely many of something, Taking one away from it does not make it smaller. Just like 0. 999. . . . . - one of the 9's equals 0. 9999. . . . . .
Posted by dev_101 1 week ago
dev_101
@jackgilbert did you read what was on round 1 the first proof was already shown on round 1 as well as the counter for it, Which you did not refer to at all.
Proof 2 violates one of the rules of the debate defined at the beginning the reason for it is also defined there.
Posted by felixmendelssohn 1 week ago
felixmendelssohn
Hm, By this step
let x = 0. 999. . . 9
10x = 9. 999. . . 0
you already assume that there is a place where that 9 ends and follows by a bunch of zeros which contradict the initial assumption that the 9 continue indefinitely

what do you have to say about other proofs. For instance, Proof by geometric series?
Posted by ArguingPerson123 1 week ago
ArguingPerson123
Mathematically, Yes. I once debated this and lost. However, Technically, If you want to go by exact value, They do not equal the same thing because there will be an infinitely small difference between the two numbers. It would be 0. 00. . . 01
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