The Instigator
jmlandf
Con (against)
Losing
7 Points
The Contender
beem0r
Pro (for)
Winning
28 Points

.999... is equal to 1

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Voting Style: Open Point System: 7 Point
Started: 9/22/2008 Category: Science
Updated: 8 years ago Status: Voting Period
Viewed: 1,137 times Debate No: 5492
Debate Rounds (3)
Comments (6)
Votes (5)

 

jmlandf

Con

Con
Now I know full good and well that most of you reading this debate actually believe that .999... is equal to 1. I've seen a lot of Debates on the issue and frankly Con has generally done a horrible job supporting the position,not to say Pro is correct. There is no question in my mind that 1 and .999... recurring forever is not equal but entirely two different numbers and I will help you understand why.

.999... = 1 is a false statement.

Definitions
.999... refers to .9 with recurring nines

1 refers to the real number, 1

= means is exactly equal to

.999... is actually equal to .999... not one.

Most people that contend 1 is = to .999... usually provide the below mathematical proof.
Step 1) Let x = .999...
Step 2) 10x = 9.999... (multiplying RHS and LHS by 10)
Step 3) 10x - x = 9.999... - x (subtracting x from both sides)
Step 4) 9x = 9
Step 5) x = 1
Conclusion .999... = 1

HOWEVER if you will notice step 4 is incorrect. It refers to 9x=9. This is wrong it actually equals 9x=8.999...1 or 8.99forever with a one at the end. Lets deal with finite to illustrate. Now many will contend that since the nines never end that you can not put a one at the end, however the one actually occurs FIRST then the 9's follow and work backwards to the decimal point.

Step 1) Let x = .999 (just 3 of them, not infinite)
Step 2) 10x = 9.99 (multiplying RHS and LHS by 10)
Step 3) 10x - x = 9.99 - x (subtracting x from both sides)
Step 4) 9x = 8.991 (there is always one 9 less than x and a 1 at the end)
Step 5) x = .999
Conclusion .999 doesn't equal 1

The first logic most people refer to, and correctly so, is the concept that .999... is a theoretical number in which it is the number closest to one but still not 1. This is a wise individuals first thoughts when someone attempts to suggest that 1 is equal to .999... Attempts to prove this silly notion with mathematics, are being made and this is the point where most people sacrifice their intellect and simply accept that the Theoretical number .999... is equal to one.

This is going to get complex, but if you think about it you will understand. Now here is the flaw in the algebraic examples. They are numbering systems that are based on 10. Most of you only have conceptualized numbering systems based on 10 but it doesn't have to be based on 10. It could just as easily based on lets say 8, this is where you must really use your intellect to understand. If the numbering system was 8 lots of things remain the same, however a few important things change, namely decimals. Now lets say we have the number system of 8. 1/8 now would equal .1 and 1/4 would equal .2. 1/3 would now equal .2666.. as opposed to .333... This may sound silly but you have to understand that decimals simply represent the numbering system chosen and my opponent is using 10. Now regardless of which numbering system you use whether it is 8 or 10 one third (1/3) will be equal to (1/3) however the way the number is represented in decimal form changes. So .375 of the 8 numbering system equals .333... in the 10 numbering system.

To help you better understand numbering systems our calendar months per year is based on 12. If you wanted to represent half a year it would be 6 months. Represented using a numbering/numeral system based on 10 the decimal representation of half would be .5. A 12 numbering system would represent the decimal of .6. A 8 numbering system would have 6 months represent .4 in decimal format. All systems would still represent the 6 months as 6/12 in fraction representation. So if you can see it is simply the decimal representation of the fraction that varies according to the numbering/numerical system.

http://en.wikipedia.org...

The error with the algebraic expressions is the decimal system based on 10. The 10 system is the most widely used system in the world today, and many find it difficult to understand other systems because it is like learning math all over.
beem0r

Pro

My opponent plans on enlightening me, you, and the entire mathematical community. Sweet!

I accept the definitions. I will prove that the decimal representation 0.999... is exactly equal to 1 in the real number system.

I will do this claim and response style.

Claim: 0.999... is equal to 0.999..., not 1
Response: It is equal to both. One of the main misconceptions that leads people to come to my opponent's conclusion is that each number should have its own unique decimal expansion. This is not a property of the real numbers, so we cannot make a conclusion based on it.

Claim: In the CON R1 proof, 9x=8.999...1 or 8.99forever with a one at the end.
Response: Like my opponent said, it is impossible to have something AFTER an infinite series of numbers. To counter this, he claims that it works backwards, going from the 1 at the end to the 8 at the beginning. Note that besides being completely invalid mathematically, this does not solve any problems, since it means the 8 at the beginning is actually 'after' an infinite series of numbers, an erroneous concept.
Also, if I asked what the impact of that 1 is on the value of the number, the only valid answer in the real numbers would be 'zero,' assuming a one could exist 'after' an infinite number of decimal places.

Claim: 0.999... is the number closest to one, but not equal to it.
Response: One method of establishing equality between X and Y is "X - Y = 0". This can be written as "X = Y - 0", or in our specific case "0.999... = 1 - 0"
Now obviously, my opponent will contend that the difference is not zero, and therefore this equation is invalid. However, let us consider something.
0.999... can be written as a convergent infinite sequence as such: (0.9, 0.99, 0.999, 0.9999, …)
Now my opponent may not be ready to admit that the sequence is equal to the limit of the sequence [which is 1], but we can see one important thing from this. The difference between this number and zero shrinks every time we go to the next value in the sequence. Since the difference between this number and one is constantly getting smaller in this sequence, it does not have a fixed finite value, since going to the next iteration will always make the difference smaller. Indeed, this difference would not even be able to be written as a finite number. The only conclusion then to make, besides the conclusion that the difference equals zero, is that the difference is some positive, nonzero infinitesimal. However, by way of the Archimedean property of the real numbers, nonzero infinitesimals do not exist. Here is a link to an article explaining the Archimedean property of the real numbers:
http://en.wikipedia.org...
Thus, the only valid conclusion is that the difference between 0.999... and 1 is zero, making the numbers equal.

Claim: This problem arises because of the base 10 number system.
Response: Actually, this phenomenon exists within every number system. In base 8, 0.777... equals 1. In base 2, 0.111... equals 1. It is also true that in all of these number systems 1 = 1, but that does not preclude these other decimal representations from also equaling one.

Now, I will present another proof for which "there's a one _at the end_ of the _non-terminating_ decimal" will not become an issue.
1] 1/3 = 0.333... This is an exact equivalence.
2] 3 * 1/3 = 3 * 0.333... A basic tenet of algebra is that doing the same thing to both sides of a valid equation leaves us with a valid equation.
3] 1 = 0.999.... Using the multiplication methods of fractions and decimals respectively, we get this equation.

This avoids my opponent's (completely invalid) claims about a number "at the end" of a non-terminating decimal. I challenge my opponent to show me the supposed mathematical fallacy used in this proof. Remember, unless there is a fallacy or bad premise used in a proof, its conclusion must be true.

I will now give the floor back to my opponent. Thank you.
Debate Round No. 1
jmlandf

Con

Thank you for taking the Debate.

1. I plan on enlightening the entire debate community not the mathematical community. The resolution says .999... is equal to one, which is wrong either way. The MATHEMATIC community accepts .999... equals one IF and only IF you use a 10 numeral/number system. The resolution does not say that does it now? No, It simply says .999... equals one, no IF, AND, or BUTS'. Any logical and knowledgeable man (must possess both qualities) understands that .999... simply can not equal one on ANY other numerical/number system.

My opponent says "in Base 8, .888... equal 1"

He doesn't just say .888... equals 1 but rather starts with "in base 8". Then for the resolution to be correct he must also include in "base 10" but this is not mentioned in the resolution. This is the first reason to vote CON.

2.Decimals are contingent on the numeral system. I hope we all understand this at this point. For example 1/3 = .333... on a 10 system but 1/3 = .30 on a 9 system. Now the 9 systems 1/3 = 1/3 of the 10 system but the .333...=.30 is invalid. 1=1 on any whole number system. 1/3=1/3 expressed in fraction but decimals fail to meet the bill. A decimal is an ATTEMPT to express a fraction but as we can see with the .333... it is incapable.

Here is some confusion if you can't fathom
10 system invalid9 system
1/3=.333...1/3=.30
3*1/3=3*.333...3*1/3=3*.30
1=.999...1=.90

10 System versus 9 system in fraction works like a charm
1/3=1/3
3*1/3=3*1/3
1=1

So in summary the mathematics isn't saying that .999... is equal to one but rather the number systems decimal report of the fraction is invalid as some fractions can not be reported quantifiable via decimal equivalent.

3. My opponent attempts to disprove my claims with Archimedean property, however there is such a thing as non-Archimedian property such as p-adic numbers which support my rational of ending or begining at decimal point.

4.Am I wrong or is it true that no calculator with base 10 will express a .999... . This sort of disproves the final step of the 1/3 proof via calculator, because when you times .333... by 3 it simply equals 1 not .999..., on a calc. ? Why won't the calculator express this step?
beem0r

Pro

My opponent's first argument, one that concedes the entire point, is that 0.999... only equals 1 in the base 10 number system, and therefore you should vote CON because base 10 is not specified in the resolution. However, note that the base 10 number system is the default number system for all of humanity. It does not need to be specified. Under my opponent's flimsy reasoning, every mathematical formula must have an explicit specification of number system. This is ludicrous, considering that if no other number system is specified, base 10 is always assumed.

My opponent also misrepresented my argument, claiming that I said that 0.888... equals 1 in base 8. This is not true, and not what I said. I claimed that 0.777... equals 1 in base 8.

Thus, now that I have covered that the base 10 system is the default one, that first argument against me is invalid.

Next, my opponent attempts to tell us that decimal representations are not always perfect.

He attempts to do this by showing that 1/3 has a different decimal representation in base 9 than in base 10. In base 9, the 'decimal' is 0.3, whereas it is 0.333... in base 10. He then says that since 0.3 is not equal to 0.333..., decimals must be flawed.
False.
0.3 in base 9 = 0.333... in base 10
That is true.
Numbers have different representations in various number systems. For instance, 255 in hexidecimal [base 16] is FF, whereas it is 255 in decimal [base 10]. Indeed, the very word 'decimal' implies that we are talking about base 10, since base 10 is CALLED decimal.

My opponent then goes to show us some more 'invalidness' in using decimal representations [by that I mean representations that use a decimal point].

On one side, he has the function 3*1/3 proof I used earlier. Using valid math, this side comes to the conclusion that 0.999... = 1.
On the other side, he has the same function in base 9.
1/3 = 0.3 base 9 - this is true
3*1/3 =3*0.3 base 9 - this is also true
1 = 0.9 base 9 - this is half true

You see, for that third step, 0.9 is not a valid number in base 9. The 9 numbers in base 9 are 0, 1, 2, 3, 4, 5, 6, 7, and 8. When one digit gets to the theoretical '9' it becomes a 1 in the digit to the left. 0.9 in base 9 does not exist, it is instead written as '1'.

We can verify this when we use 'A', the single-digit representation of 10, in base 10.
1/2 = 0.5
2*1/2 = 2*0.5
1 = 0.A

That is invalid, since A does not exist in base 10. It is instead written as a 1 in the column to the left of it, or 1.0

So as you see, math with decimals is not broken. My opponent has just made a mistake in his execution of them, pretending that the digit 9 exists in base 9.

All fractions have exact decimal representations, whether they have recurring decimal places or not. For example, 1/7 = .142857... , where the portion that repeats is the entire 142857. It would have a different decimal representation in any other number system, but you would have to specify the number system if you wanted to set them equal to each other.

0.3 base 9 = 0.333...
That is a true statement, as is any statement you will come to doing valid math on decimals, fractions, or whole numbers.

Next, my opponent points out something that's truly laughable. I'll simply copy and paste what he said.
"3. My opponent attempts to disprove my claims with Archimedean property, however there is such a thing as non-Archimedian property such as p-adic numbers which support my rational of ending or begining at decimal point."
First off, the Archimedean property applies to the real numbers, which is what we are discussing. This is because, like base 10, real numbers rather than p-adic numbers are default. The 'non-Archimedean property' applies to the p-adic numbers [I was unable to find an actual property called called non-Archimedean, I assume it simply means that a system does not have the Archimedean property].
Now I probably know about as much as my opponent does about the p-adic system, which is net to nothing. However, let me show you something that really highlights how different the p-adic system is.
"Whereas two decimal expansions are close to one another if they differ by a large negative power of 10, two 10-adic expansions are close if they differ by a large positive power of 10. Thus 3333 and 4333 are close in the 10-adic metric, and 33333333 and 43333333 are even closer."

That's so far from traditional mathematics that it's almost ridiculous how obvious it is we aren't talking about p-adic numbers. We're talking about real numbers, since they are the standard in mathematics. And for real numbers, the Archimedean property applies, which proves that the difference between 0.999... and 1 is 0. A difference of 0 one of the basic ways of establishing equality.
1 - 0 = 0.999...
1 = 0.999...

Finally, to my opponent's last objection:

"4.Am I wrong or is it true that no calculator with base 10 will express a .999... . This sort of disproves the final step of the 1/3 proof via calculator, because when you times .333... by 3 it simply equals 1 not .999..., on a calc. ? Why won't the calculator express this step?"

This is for the same reason that a calculator will give you 0.66666666667 for 2/3. Calculators cannot display an infinite number of decimals. Therefore, it will round at the last digit it can display.
Rounding 0.99 to the nearest 0.1 is 1.0
Rounding 0.999 to the nearest 0.01 is 1.00
Rounding 0.9999 to the nearest 0.001 is 1.000
Rounding 0.99999 to the nearest 0.0001 is 1.0000

Get the point? It's simply a rounding process, since it can only display it to a certain finite number of decimals.

And with that, it is now time for my opponent to make his closing arguments. Thank you.
Debate Round No. 2
jmlandf

Con

jmlandf forfeited this round.
beem0r

Pro

My opponent has failed to conjure up a response to my counter-arguments.
He will likely give us some creative excuse in the comment section, but the fact stands that he did not respond to my arguments. Last round, I addressed all the issues my opponent had with my case, so we must view my case as valid as far as this debate is concerned. Not only that, but it is mathematical truth.
Thank you for reading.
Debate Round No. 3
6 comments have been posted on this debate. Showing 1 through 6 records.
Posted by TheSkeptic 8 years ago
TheSkeptic
jmlandf is confident he can smack the entire mathematical community on the butt, let's see :D.
Posted by beem0r 8 years ago
beem0r
The fact that he has debated this so many times is the most troubling part.
Posted by knick-knack 8 years ago
knick-knack
How can you debate against something that is a proven fact?
PRO wins every time.
Posted by TheSkeptic 8 years ago
TheSkeptic
I'd be glad to debate you jmlandf, but seeing as its late, if no one accepts it tomorrow I will :)
Posted by jmlandf 8 years ago
jmlandf
lets do it
Posted by TheSkeptic 8 years ago
TheSkeptic
Oh goodness, should I take it?
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Vote Placed by Nails 7 years ago
Nails
jmlandfbeem0rTied
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Cody_Franklin
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robert.fischer
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Tatarize
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