The Instigator
Con (against)
Anonymous
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The Contender
MagicAintReal
Pro (for)
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0 Points

does 0.9999999.......... equal 1

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Voting Style: Open Point System: 7 Point
Started: 7/14/2018 Category: Philosophy
Updated: 3 years ago Status: Post Voting Period
Viewed: 1,371 times Debate No: 116578
Debate Rounds (4)
Comments (24)
Votes (0)

 

Con

Pro starts
MagicAintReal

Pro

Pro's also going to provide definitions.
While I consider myself a science guy, I'll dabble with maths here and there.
I'm not British, I just like calling math "maths" sometimes.
I request that voters use the Opt-In voting standards and moderators remove crappy votes.

Definitions

0.999... - denotes the repeating decimal consisting of infinitely many 9s after the decimal point (and one 0 before it) that represents the smallest number no less than all decimal numbers 0.9, 0.99, 0.999, etc. shown to equal 1.
https://en.wikipedia.org...

equal - be the same as in number or amount.
https://en.oxforddictionaries.com...

1 (one) - the first number of the infinite sequence of natural numbers.
https://en.wikipedia.org...


*Resolution*

I'm attempting to affirm the resolution that 0.999... equals 1.
I will use 3 mathematical arguments to do so.


*Maths Argument #1*

Take the number 1 and divide it by 3.
You in fact get a repeating decimal of infinitely many 3s after the decimal point = 0.333...

If you multiply 0.333... by 2, you would get 0.666...
So it should follow that if we multiplied 0.333... by 3, we should get 0.999...

This means that dividing the number 1 by 3 and dividing the number 0.999... by 3 are exactly the same.
This is because 0.999... = 1.


*Maths Argument #2*

Any number subtracted by itself is zero.
3 - 3 = 0
2 - 2 = 0
and 1 - 1 = 0

When you subtract 0.999... from the number 1 the answer is 0.000...
Why?
Well, let's look at smaller 0.9 decimals.

1 - 0.9 = 0.1
1 - 0.99 = 0.01
1 - 0.999 = 0.001

The amount of 0s in the answer's decimal increases along with the amount of 9s in the decimal being subtracted from 1.
In fact, the only time a 1 appears in the answer is when there's an *end* to the amount of 9s in the decimal being subtracted from 1.
Well, 0.999... has *no end* to its 9s so when you subtract it from 1, you get,

1 - 0.999... = 0.000...
The zeroes in the answer's decimal are infinite and 0.000... = 0
This is because 0.999... = 1.


*Maths Argument #3*

Algebraically if we make x = 0.999..., you cannot distinguish between 1 and x.
For the following proof, x = 0.999...

Let's take x and multiply it by 10.
10x = 9.999...

So, when we multiply 0.999... by 10, we get 9.999...
Well, 9.999... is just 9 + 0.999...
So we can simplify 9.999... to 9 + x.

Therefore 10x = 9 + x
Subtract x from the right, and then from the left, and what remains is
9x = 9
Divide both sides by 9 and x = 1.

If you make x = 0.999... which we did, it's indistinguishable from making x = 1.
This is because 0.999... = 1.


*Conclusion*

For reasons #1, #2, and #3 above, 0.999... = 1 and I request that opt-in voters pay close attention to the definitions provided 1st round by Pro.

Con?
Debate Round No. 1

Con

Your math argument 1: 1 divided by 3 is not a correct value and here is why. The answer resorts to infinity? Why? Because there is no answer that satisfies 1 divided by 3. This is why when you multiply it by 3, you get 0.999999......, not one.

Your math argument 2: 1-0.9999...... does not equal 0. it equals 0.0000000....1. We call this number infinitesimally small. But it still is above zero.

Your math argument 3: Suppose we used your exact same algebra on an equation with infinitely many 9's going off to the left.
x=..........99999999999
Multiply both sides by 10
10x=............9999999990
But look at that. ........999999990 is the original x value minus the last 9
10x=x-9
9x=-9
x=-1
So this equation is telling me that .......999999999 equals -1. Do you believe that? Most people think that the first equation is fine, but then they look at the second equation and think what I did is ridiculous. But it is exactly the same mathematics so you can't just pick and choose when it is right and when it is wrong. If you believe 1=0.9999... then you also must believe that ......999999=-1. But let's make matters worse. Let do an equation with infinitely many 9's going of to the right and left. The equation looks as follows:

........9999999.99999999......=x
We multiply both sides by 10
........99999999.9999999......=10x
But look. ......999999.999999 is the original x value.
x=10x
subtract x from both sides
9x=0
x=0
So this is telling me that ....9999.99999.... equals 0. Honestly, do you believe that? The same mathematics is being applied to different circumstances and because it cannot be false in some and true and others, you must choose to either believe all cases have a meaningful answer, like -1, 1, or 0, or there isn't a meaningful answer because we have the concept of infinity in each equation.
MagicAintReal

Pro

Thanks Con.
I'm glad to see Con has no problems with the definitions.
Voters should use these definitions along with the Opt-In standards.
Let's rebuttal!

*Review*

The case I made was threefold:
#1. The fraction 1/3 is indistinguishable from the fraction 0.999.../3
#2. Subtracting 1 from 1 (1 - 1 = 0) is exactly the same as subtracting 0.999... from 1 (1 - 0.999... = 0.000...)
#3. Using 1 algebraically is indistinguishable from using 0.999... algebraically (10x = 9 + x)


*Con's Attempt*

Con asserts:
"1 divided by 3 is not a correct value...the answer resorts to infinity."

My response:
Dividing by 3 is not a correct value?
Don't tell equilateral triangles.
I mean those guys think they can just divide all 180 degrees by their 3 total angles and expect us to recognize it as a correct value?
I'll be long dead in my grave before I recognize 180 times 1/3.
I spit at the equilateral triangle...spit!

All joking aside,1/3 = 0.333... is a real, rational number just like 2/3 = 0.666... and 3/3 = 0.999... even though the numbers to the right of the decimal infinitely repeat; they are real, rational values used in standard arithmetic.

Either way, Con ignores that *dividing 1 by 3 is indistinguishable from dividing 0.999... by 3* thereby making 1 and 0.999... equal.


Con further asserts:
"1-0.9999...... does not equal 0. it equals 0.0000000....1. We call this number infinitesimally small. But it still is above zero."

My response:
Con ignored that I had indicated the only time a 1 appears in the answer to a 1 - 0.9 subtraction problem is when there's an *end* to the amount of 9s in the decimal being subtracted from 1.

Con didn't bother to attack that point and instead resorts to asserting that a 1 appears in the answer without showing an end to the amount of 9s in the decimal being subtracted from 1.

I maintain that there is no *end* to the amount of 9s to the right of the decimal point in the number 0.999..., therefore if subtracted from 1 would result in the answer 0.000... which is equal to 0.


Con supposes:
"Suppose we used your exact same algebra on an equation with infinitely many 9's going off to the left."

My response:
Ok, I'm not a math guy, I'm a science guy, but I don't think there's this reciprocal property between infinite decimals and infinite natural numbers whereby infinitely adding more numbers to the right of the decimal point behaves the same as infinitely adding numbers to the left of the decimal point.
But if we're just supposing, go ahead.

Con continues:
"x=...999.0
Multiply both sides by 10
10x=...9990.0"

My response:
Ok.
It looks like you took a positive number ...999.0 and multiplied it by 10, making it 10 times larger, equaling the number ...9990.0
Go on.

Con continues:
"...9990.0 is the original x value minus the last 9"

My response:
Ok. So here's the flaw.
It's not the original value minus a 9, it's actually 10 times greater than the original value, so subtracting 9 from ...9990.0 doesn't give you the original value of ...999.0 instead it gives you the value of ...99981.0.

From this point on, Con's attack ignores that
...9990.0 - 9 = ...99981.0

So from
10x = x - 9
the algebraic equation becomes:
10x = ...99981.0
And this does not get us to -1 or anything like it.

So Con has been refuted.


*Conclusion*

0.999... has been shown to equal 1 and Con's refutation falls short for the reasons provided this round.
Debate Round No. 2

Con

Your first argument: I never said that 0.3333333.... is not a rational number. I said it is not a correct value of 1/3 because it resorts to infinity which means there will never be an answer that satisfies the expression of 1/3. Infinity isn't a real number. It is the unsatisfied answer of an equation.

Your second argument: You said that 1-0.9999999999999999999.......... equals 0.0000000.......... There is still a 1 at the end of it. But it is being pushed back by infinitely many nines. Just because you will never find it, does not mean that it isn't there. That is why it is infinitesimally small.

Your third argument: Here is an equation that proves that .....99999 times 10 equals ....999999-9
..........9999999999(10)=.........9999999999-9
Simplify
..........99999999990=..............999999990
They are exactly the same, this proves my case. Let's rewrite the equation

x=.........9999999
10x=.........999990
I have already proven that ..........9999999999(10)=.........9999999999-9
10x=........99999999-9
Look at that. ........9999999 is the original x value.
10x=x-9
9x=-9
x=-1

We can use this same algebra using a different equation.

.....9999999.99999999......=x
We multiply both sides by 10
........99999999.9999999......=10x
But look. .......999999.9999999...... is the original x value.
x=10x
9x=0
x=0

It is the same algebra you used. You either believe it or you don't.
MagicAintReal

Pro

**Correction**
I incorrectly characterized part of Con's argument.
I had subtracted 9 from 10x, not 9 from x; my bad.
Subtracting 9 from x would give you 10x, Con is right...but

*Algebraically ...999.0 = -1 Makes Sense*

I mean, I'm not a math guy, but it seems to me that if in fact algebraically one could prove that 0.999... = 1 then it would follow that its polar opposite, ...999.0 i.e. the repeating decimal consisting of infinitely many 9s in the opposite direction, would be -1.

Heck, I'll even take it further.
-0.999... = -1
and
-...999.0 = 1

This does not change the fact that algebraically 1 and 0.999... are indistinguishable from each other even if it seemingly counter-intuitively implies that an infinite positive value like ...999.0 is negative.
In this case, both things are true, which actually makes sense.


*Responding to Con*

Con claims:
"I said [0.333...] is not a correct value of 1/3 because it resorts to infinity which means there will never be an answer that satisfies the expression of 1/3."

My response:
Con focus here!
Why does dividing 1 by 3 and dividing 0.999... by 3 give the same answer?
If you're going to use your logic from above, you therefore consider 0.999... "not a correct value" yet using this not correct value when dividing by 3 is EXACTLY THE SAME as dividing 1, a value I assume you think is correct, by 3.

Questions for Con:

1. Con do you think 1 is an incorrect value?
When you divide it by 3, you get, as you call it, an incorrect value, so the number 1 is necessarily made of 3 of those incorrect values.

2. Why are 3 incorrect values equal to a correct value, but 1 incorrect value is not?

3. I would also like Con to explain how it is that we divide ANY NUMBER by 3 without multiplying it by 0.333...
Can Con give an example of dividing by 3 where the same answer cannot be reached by multiplying by 0.333...?

Well everyone, 1/3 = 0.333... even if Con tells you it doesn't.


Con reasserts:
"You said that 1-0.999... equals 0.000...There is still a 1 at the end of it. But it is being pushed back by infinitely many nines."

My response:
Con, focus again!
I correctly stated that the only time you see a 1 in the difference between 1 and 0.999... is when there's AN END to the 9s, which, right here you just admitted there is no end to by referring to INFINITELY many nines.

Con childishly repeats:
"There is still a 1 at the end of it."

My response:
Why would there be a 1 at the end of something which you just admitted HAS NO END?
No end to the 9s-->no end to the answer-->no 1 at the end of the answer.

Con appeals to ignorance:
"Just because you will never find it, does not mean that it isn't there. That is why it is infinitesimally small."

My response:
I do hate this fallacy, and this example is most certainly an argument from ignorance given that it's appealing to not being able to find something as a reason to believe in its existence, but since we're talking about INFINITELY repeating 9s, I actually will never find the end, because INFINITY MEANS HAS NO END.

The end is not infinitesimally small; it's infinitesimally not there.


*Wrap Up*

Well I corrected myself in that Con was actually correct in saying that 10x = x - 9 with his ...999.0 example, and I realized that (...999.0 = -1) actually makes perfect sense given that algebraically 0.999... is proven to equal 1 and having the repeating 9s go the other way could be achieved by adding a negative sign to it.

I have not touched Con's ...999.999... example because it's really just total infinity and is not a rational number, so its behavior in algebra seems largely irrelevant to the question of whether or not 0.999... = 1.

Con has not shown why 1 divided by 3 and 0.999... divided by 3 are the same.
Con also has not been able to show an END to the answer of the subtraction problem 1 - 0.999...

Con?
Debate Round No. 3

Con

Honestly, do you believe that ....999999=-1 The equation is paradoxical because it would seem to prove that multiplying a positive number by 10 results in that number minus 9. That is what happens when you mess with infinity. You get answers that don't make sense. My point is not that ....9999=-1, my point is that whenever you have a number that resorts to infinity, you get strange answers. Just like how you get a strange answer in the equation that says 0.9999...=1. But if you believe that 0.99999.... equals 1, then you must also believe that ....99999=-1. But if you believe that ....9999 equals -1, then you are telling me that at the end of time 9+ 90 +900 +9000 +90,000...... equals -1. I don't think anyone believes that, I don't believe that.

Questions you asked me:

1. 1 is not an incorrect value. When you divide it by 3 however you don't get a correct value. 1 is technically made up of 3 incorrect values. That doesn't mean 1 is an incorrect value. If I multiply 2 negatives together, the answer isn't negative.

2. see 1.

3. No, I cannot give an example of when this happens. However, the only reason it happens here is because we apply infinitely many of something to the equation. THIS ISN't A CORRECT VALUE.

My point in this debate is that when you apply infinity or infinitely many of something to an equation, you get strange answers. Hence why you get the strange answer of 0.9999..=1

I have one question for you. Where along the line of 0.9999.... does it suddenly equal 1?
MagicAintReal

Pro

Thanks for the debate Con.
Con spent much of his effort on one of my proofs.
He basically left the other two untouched.
Nice! Let's end this.

*Maths Argument #1*

I had indicated that 1 divided by 3 is indistinguishable from 0.999... divided by 3 and this was confirmed and conceded by Con.

In round 2, Con concedes:
"the answer that satisfies 1 divided by 3...when you multiply it by 3, you get 0.999..."

My response:
This is a concession that dividing 1 by 3 and dividing 0.999... by 3 are indistinguishable and *this* alone shows that 0.999... is equal to 1.
The same three things that make 1 also make 0.999...
Conceded.

Con continues:
"1 is not an incorrect value...1 is technically made up of 3 incorrect values. That doesn't mean 1 is an incorrect value."

My response:
Ok, this bit here, along with Con's concession that 1/3 = 0.999.../3 has Con's logic indicating that 0.999... is NOT an incorrect value as he tried to claim before.
Con says that while 1 is made up of 3 "incorrect values," 1 is not an "incorrect value," but the number 0.999... is made up of THE EXACT SAME 3 "incorrect values" and Con somehow still claims that that 0.999... is an "incorrect value."

Even though Con never explained what an incorrect value is, by Con's own logic, 0.999... is NOT an incorrect value because it's made of 3 incorrect values IDENTICAL to the 3 incorrect values that make up the number 1.


*Maths Argument #2*

I had indicated that 1 - 0.999... = 0.000... because without and END to the 9s in the decimal being subtracted from 1, there is no END to the 0s in the answer's decimal thereby equaling 0, because 0.000... = 0.

Con concedes:
"1-0.999... does not equal 0. it equals 0.0000000....1. There is still a 1 at the end of it. But it is being pushed back by infinitely many nines."

My response:
So the response to me indicating NO END to the 0s in the answer's decimal because there's no end to the 9's in the decimal being subtracted from 1 is just Con conceding that there is NO END to the 9s?
I view this as a concession that the answer to 1 - 0.999... = 0.000... WITHOUT END.
Conceded.


*Maths Argument #3*

I mean, this is really where Con focused their efforts, and honestly, given the lack of a refute to the other maths arguments, I could just leave this one alone and still net the win, but I'm obnoxious.

I had indicated that since multiplying 0.999... by 10 was the exact same thing as adding 9 to 0.999... that algebraically 1 is indistinguishable from 0.999...

So Con indicated that doing this algebraic exercise with ...999.0, the infinite string of 9s going to the left instead of the right of the decimal. indicates that ...999.0 = -1.
Con also points out that putting the infinite string of 9s in both directions would then equal 0.

To me, this doesn't take away from the fact that algebraically 0.999... is indistinguishable from 1, in fact, though counter-intuitive and seemingly paradoxical, if 0.999... does in fact = 1, then it follows that the other direction of infinite 9s could be accomplished by making that value -1 and if you were to put ...999.0 and 0.999... i.e. (-1 and 1) you should get 0.

Con asks:
"Honestly, do you believe that ....999999=-1"

My response:
Sure.
If I say yes to this question, does it negate that 0.999... = 1?
Nope.
Both can be true, and are.

Con continues:
"the equation is paradoxical because it would seem to prove that multiplying a positive number by 10 results in that number minus 9."

My response:
Great.
This doesn't take away from the point that 1 and 0.999... are algebraically indistinguishable, it just shows that other counter-intuitive results can arise from algebra...that's it.
If anything it's a knock on algebra, not 0.999... = 1.


*Conclusion*

0.999... = 1 because of the untouched reasons #1 and #2 and for the irrelevant algebraic paradox Con attempts for #3.
Opt-in voters, take note.
Vote Pro.
Debate Round No. 4
24 comments have been posted on this debate. Showing 1 through 10 records.
Posted by qwaszxplm 3 years ago
qwaszxplm
this question really depends on your belief in the propose of math/calculation and the definition of infinity.
Posted by DeletedUser 3 years ago
DeletedUser
regardless of who you vote for, vote here. http://www.debate.org...
Posted by MagicAintReal 3 years ago
MagicAintReal
@sem2093
Why is dividing 1 by 3 the same as dividing 0.999... by 3?
1/3 = 0.333...
2/3 = 0.666...
3/3 = 0.999...

or

1/9 = 0.111...
2/9 = 0.222...
3/9 = 0.333...
4/9 = 0.444...
5/9 = 0.555...
6/9 = 0.666...
7/9 = 0.777...
8/9 = 0.888...
9/9 = 0.999...
Posted by sem2093 3 years ago
sem2093
If we allow 0.99999... to equal 1 then we'd also have to accept that 7.99999.... equals eight. Math is a tool of measurement and rounding up or down is a stray from preciseness in the answer.
Posted by canis 3 years ago
canis
Anyway. Yesterday I was infinite far. As in infinite close to eating a Cheese burger.. Eating one is 1.
Posted by canis 3 years ago
canis
0,9999... is far from equal 1...As in infinite far. And as in infinite close.
Posted by A1DS 3 years ago
A1DS
Can't use infinity in the way con has used it in their argument (algebraically), mathematically invalid. Pro's refutation of this argument is equally invalid because it uses infinity in the same algebraic manner (not that it matters because it's not a mathematically correct argument in the first place).

Pro's proof on the other hand is correct as it doesn't use infinity inappropriately (0.999... correctly identified as a rational number).
Posted by DeletedUser 3 years ago
DeletedUser
Ok
Posted by MagicAintReal 3 years ago
MagicAintReal
Everyone's a critic I guess
Posted by DeletedUser 3 years ago
DeletedUser
I am surprised that out of all of the things you could have chosen to attack in my argument, you chose to say that my 9's had to many points before them.
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