The Instigator
cowpie1998
Pro (for)
Losing
5 Points
The Contender
Extremely-Far-Right
Con (against)
Winning
42 Points

.9 repeated is equal to 1.

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Voting Style: Open Point System: 7 Point
Started: 4/7/2011 Category: Science
Updated: 6 years ago Status: Voting Period
Viewed: 5,133 times Debate No: 15833
Debate Rounds (4)
Comments (39)
Votes (10)

 

cowpie1998

Pro

First, I will clarify my argument
I am arguing that .9 repeated (which for the sake of the argument we will call .9) is equal to one (1).
The debate will start in the second round, concluding in the fourth.
I hope to have a grueling competition!
ps. this is only my second debate, so take it easy on me
Extremely-Far-Right

Con

I thank Pro for this most interesting debate.

I agree with Pro's clarification that we shall call .9 repeated, simply .9 as for the sake of the argument.

I shall argue as Con, and support the notion that .9 isn't equal to 1.

I await Pro's argument(s) following the conclusion of this round, since it was specifically clarified that the debate will start in the second round.

Over to you Pro.

(This is also only my 6th debate)
Debate Round No. 1
cowpie1998

Pro

Thank you.
.9 is equal to 1 because .3 (repeated) is mathematically equal to a third. therefore 3 (repeated) times .3 is three thirds, and also .9.
Also, one cannot identify a space between .9 and 1, as there is no number that comes between them, therefore they are the same.
Extremely-Far-Right

Con

Thanks for the response Pro.

I shall first dissect your responses, and then show some of my own as support for my view in respect to my position taken on this topic.

You say, ".9 is equal to 1 because .3 (repeated) is mathematically equal to a third. therefore 3 (repeated) times .3 is three thirds, and also .9."

.3 (repeated) isn't actually equal to one third or a third. It is the closest number that can reach one third, and it even becomes ever more increasingly accurate to reaching one third, with the more 3's it has, but it will technically never reach it. The reason for this is the fact that you can't simply show a third using only decimals. The same applies to showing two thirds with decimals or three thirds.

You mention, "Also, one cannot identify a space between .9 and 1, as there is no number that comes between them, therefore they are the same."

This argument is incorrect due to one simple reason.
1-.9 repeated equals, 0.01 repeated.

Another reason is that one can't simply round .9 to equal 1, since rounding is antithesis and isn't correct anyways.

.9 can only equal 1 if it is rounded. And since rounding isn't accurate, that can't be a reason since you specifically stated that it is EQUAL to the number "1".

Thank you.
Debate Round No. 2
cowpie1998

Pro

Thank you.
.3 repeated is a third. Please prove that it is not. Without any logical fallacies.

1-.9 is not .01. That would equal a number larger than 1.

I don't have to round, this reasoning is only true once you have proven that .9 is not 1. Otherwise, this argument is completely null due to the fact that it is based upon a false piece of logic.

I urge the voters to vote Pro because Con cannot bring a provable piece of logic to the discussion.
Extremely-Far-Right

Con

I would like to thank Pro for his response.

Ok, since Pro doesn't think that I can bring a provable piece of logic to the discussion, lets get this debate more heated, shall we?
_________________________________________________________________________________________________

Your first argument in Round 3:
".3 repeated is a third. Please prove that it is not. Without any logical fallacies."

.3 repeated is a decimal correct? And decimals are a product of long division and/or by typing an equation into a calculator correct?

You can "solve" 1/3 by either using a "standard calculator" on your computer, or by doing it by hand (long division), or perhaps using a different calculator as well.

(If you use the "standard calculator" on your computer, and by using this definition of "standard calculator", I'm referring to the one that automatically comes pre-built into a computer and you have to do the following directions on a computer to open: Start-All Programs-Accessories-Calculator)

The point of me showing the "standard calculator" definition is to show a tool that we can both use and have knowledge of, and therefore can eliminate any confusion with regards to entering the information into the calculator.

If you "solve" the equation 1/3 (One divided by three) on your calculator or on the "standard calculator" that I showed above, you should get an answer that will look like this, 0.33333333333333333333333333333333. Or if you prefer to "solve" the equation by using long division, you will get the same answer.

You can go on and on trying to solve the problem, but will never reach the answer. Nine subtracted from ten equals one which in turn causes the problem to constantly go on forever. Therefore since the problem can never be solved, it never actually equals 1/3. The calculator just shows the number that is closest to one third (1/3), because that is what it is programmed to do and because what it is limited to, due to the information that I have previously stated.

Furthermore, if you get .3 repeated and add it together three times or multiply it three times, it equals .9 repeated. If you decide to add it three times on either a calculator or by writing it down, it will equal .9 repeated.

Here I will add .3 repeated by adding it 3 times:

0.33333333333333333333333333333333
+ 0.33333333333333333333333333333333
= 0.66666666666666666666666666666666
0.66666666666666666666666666666666
+ 0.66666666666666666666666666666666
= 0.99999999999999999999999999999999

Here I will multiply .3 repeated by 3:

0.33333333333333333333333333333333
x 3.00000000000000000000000000000000
= 0.99999999999999999999999999999999

As you can see, you get .9 repeated either way. However, if you choose the multiply it on the "standard calculator", you will notice that you do indeed get 1. This happens because the calculator rounds and the answer in return is 1.
_________________________________________________________________________________________________

Your second argument in Round 3:

"1-.9 is not .01. That would equal a number larger than 1."

(I would like to make the point that Pro has only shown two pieces of evidence to support his argument that .9 subtracted from 1 doesn't equal .01 and that it would equal a number larger than 1. I would also like to note that last round; I identified 0.01 as 0.01 repeated. I am unsure if Pro noticed this, but I assume he has because in the first round Pro said, ".9 repeated (which for the sake of the argument we will call .9)" so that would mean .1 or .01 repeated for the sake of the argument would mean .1 or .01.)

First off, I would have liked it if Pro would have clarified his argument. Does the word "that" refer to the solution of .9 subtracted from 1, or does "that" refer to the possible different solution that seems it may be numerically different than .01?

I would like Pro to clarify the message he was getting across, as one could interpret "that" many ways.

Since I have identified two possible clarifications for what "that" could actually mean, (with reference to this sentence) I will respond to both with counter arguments below:

First, if you were referring to the solution of 1 minus .9, the solution does indeed equal .1 or .01 repeated. For however many 9's you put after the nine in .9, that many zeros is put in front of the one in .1. For example, if you were to perform the following equation on your calculator; 1 minus .9 or .99 or 0.999999999999999999999999999999, the answer would be .1 or .01 or 0.000000000000000000000000000001. Since you clarified that .9 was to be meant as .9 repeated, that means that .1 or .01 is .1 repeated or .01 repeated.

Another reason is that since you chose to label .9 with infinite 9's after .9, I can select to represent the .9 as .99, or .999999 or however many 9's is needed to represent that. Therefore, that means that .9 in your argument could simply have meant .99 which subtracted from 1 equals .01.

Second, if you were referring to the possible different solution that seems it may be numerically different than .01; this response pertains to that notion. Now, you seem to say that the solution of 1 minus .9 would equal a number larger than 1 (this point of view is only coming from my second interpretation).

I hope we can agree on a couple things here. .9 is a positive number correct? So how can the solution of 1 minus .9 equal a number greater than 1? Even if you follow the assertion that .9 is equal to 1, 1 minus .9 would equal 0. Is 0 equal to 1? No, it definitely isn't. So that being said, how would the solution equal a number larger than 1?
_________________________________________________________________________________________________

Your third argument in Round 3:

"I don't have to round, this reasoning is only true once you have proved that .9 is not 1. Otherwise, this argument is completely null due to the fact that it is based upon a false piece of logic."

I'm assuming what you mean is that you don't have to round .9 repeated in order for it to be equal to 1. If that is the case then that means they are two entirely different numbers. You then go on to mention that the reasoning is only true once when I have proved that .9 isn't 1, yet that is evidence right there that .9 isn't equal to 1 since it has to be rounded in order to equal it.

I believe I have provided sufficient amount of evidence to prove that .9 isn't 1 so by your logic that means I can use rounding as a reason to prove that .9 isn't equal to 1.

So that would in effect counter your argument asserting that my argument is null due to a "false piece of logic".

_________________________________________________________________________________________________

Your fourth "statement and/or argument" in round 3:

"I urge the voters to vote Pro because Con cannot bring a provable piece of logic to the discussion."

I would have to say that this is somewhat bad conduct by Pro. You really shouldn't urge and/or persuade the voters to vote for you until it is the last round.

You also make the assertion that I cannot bring any provable piece of logic to the discussion.

I would like to point out an interesting point you bring there. Since cannot and can not mean the same thing, you are essentially saying that Con can not have the ability to bring a logical argument to this debate. So if I were to bring a logical argument to the debate than I will have successfully countered your assertion and immediately made it false.

Well, I would encourage you to at least be more proffesional about conduct.

Thank you.
Debate Round No. 3
cowpie1998

Pro

Congratulations. I am very sorry for my error. You are quite obviously the better of the two of us, and for that reason I will concede this to you. I am also sorry that I could not provide you with a better debate, but I am simply a beginner.
Extremely-Far-Right

Con

I would like to thank Pro for his response, and for this wonderful debate.

I would like to offer my understanding for Pro's case, but remember that this is a debate. Keep in mind it certainly helps an argument if one side were to bring much more evidence than their opponent to help insure victory in the debate.

Now you did mention that you were a beginner, but I would like to remind you that even beginners to the website can be excellent debaters on their first debate, as there is no way I could have known if you were an experienced debater or not. (Then again, it also depends on your definition of 'experienced')

As a side note:

(It may have helped if Pro spent more time on his responses, and try to extend his arguments a bit further.)

I thank you for the debate Pro, and best of luck to both of us!
Debate Round No. 4
39 comments have been posted on this debate. Showing 1 through 10 records.
Posted by gizmo1650 6 years ago
gizmo1650
worst place for the indent to change.
Posted by gizmo1650 6 years ago
gizmo1650
wow, I am actually going to have to vote con on this. For anyone interested, the simple algebraic proof is:
let x=0.(9) | an infinite amount of nines
x=0.9(9) | 9 followed by one less than in infinite amount of nines, or simply followed by an infinite amount of nines
10x=9.(9)
-{x= 0.(9)} | this was our definition of x
9x=9 |subtract the two equations
x=1
x=0.(9) | again, by definition
1=.(9) | transitive property
Posted by RoyLatham 6 years ago
RoyLatham
twsurber, 1 is to the right of the number you referenced. However, 0.999... = 1 exactly.

I think you are up against the problem of accepting something that is true, but not intuitive. Modern physics is loaded with such things, such as the non-intuitive fact that nothing can exceed the speed of light. Einstein made the claim and it has been well proven, but it doesn't seem to make any sense. With many such things, it is better to accept the evidence over your intuition. In this case let x = 0.999... then 10x = 9.999..., 9 x = 9, and therefore x = 1. Q.E.D.

There are many other things in math that may not be intuitive. The four-color map theorem may not seem true, but it is. Or perhaps 1/3 = 1/4 + 1/16 + ... The tricks used to sum infinite series are like that used to prove 0.999... = 1. One you get used to applying them, you get to trust the result.
Posted by twsurber 6 years ago
twsurber
Question for cowpie. I agree that 3/3 equals 1 whole. I am not seeing that .333 equals a third though, unless you are going off the infinity principle Roy spoke of.

Let's say that I have a dollar in pennies. If I count out 33 pennies I have .33, but I do not have a third. In order to have a complete third I would need 33 and 1/3 pennies.

If I had 3 groups of 99 pennies would be .99, therefore I would not have a whole dollar. I would need to have 3 groups of 33 and 1/3 pennies to make a full dollar.

To Roy & Cowpie:
I'm seriously not trying to be sarcastic, just sharing another view.

Should I accept a universally accepted mathematics principle as absolute truth if I still have a question that has not been answered satisfactorily? Do I not have the right to ask why? It's not exactly circular reasoning, but I am being asked to accept something as fact that another has presented while there still remains a legitimate question about it's validity. While .999 infinite does go on indefinitely, it can never actually achieve a full 1, but rather is merely "expressed" as 1.

I concur & accept that (.999 infinite) is "expressed" as 1 whole, in fact I would likely not say $999,999.99 but rather would just say a million dollars for simplicity. However, don't both .999 infinite and the whole number 1 each have a separate and respective position on the number line?
Posted by twsurber 6 years ago
twsurber
Roy,

I am arguing that every number has it's own place on the number line. The whole number 1 has it's place to the right of: 0.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
Posted by RoyLatham 6 years ago
RoyLatham
twsurber, I have no idea what you are trying to get at. Yes, arithmetic works. It is a fact that every mathematics text says and every mathematician understands that 0.999... = 1, and that is true whether you agree or not. Your job is to prove they are all wrong. A Nobel Prize awaits.
Posted by RoyLatham 6 years ago
RoyLatham
Gizmo, The problem 0.999... = 1 is posed with real numbers. "In the 1960s Abraham Robinson proved that the hyperreals were logically consistent if and only if the reals were. This put to rest the fear that any proof involving infinitesimals might be unsound, provided that they were manipulated according to the logical rules which Robinson delineated." http://en.wikipedia.org... In other words, if you have discovered an inconsistency in real numbers, you also shoot down the hyperreal numbers. Your claim is that real numbers are inconsistent.

Your hangup is with repeating decimals converging. You agree that the repeating decimal 0.111... [base 10] is exactly equal to 0.1 [base 9]. You are correct that changing bases cannot change the result. Therefore repeating decimals converge.

Have you taken differential calculus yet? Do you agree that it is invalid if limits do not work?
Posted by gizmo1650 6 years ago
gizmo1650
sorry, I meant |x| < 1/n for all positive n.
Posted by gizmo1650 6 years ago
gizmo1650
@Roy
I agree that their is no real number x. The nearest real number to x would be 0, also referred to as st x, or x's standard part. As 0 and x differ only infinitesimally, and infinitesimals do not exist in the real number system, do you see how .(9)=1 in the real number system, and .(9)=1-x in the hyperreal system.

As far as bases go, changing bases should not change your result.
Posted by RoyLatham 6 years ago
RoyLatham
Gizmo, The x for which |x| < 1/n for all real n is x = 0. You understand that if you say that lim 1/n is not zero, then differential calculus is invalid. So you are arguing that calculus is all a big mistake, right?

I was referring to notation systems. Real numbers have many alternate representations. For example, do you agree that 0.1111 ... [base 10] = 0.1 [base 9] ?
10 votes have been placed for this debate. Showing 1 through 10 records.
Vote Placed by iholland95 4 years ago
iholland95
cowpie1998Extremely-Far-RightTied
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Total points awarded:07 
Reasons for voting decision: No matter how many 9's you put after the decimal point doesn't change the number value. It's still .9, not 1.0
Vote Placed by Man-is-good 6 years ago
Man-is-good
cowpie1998Extremely-Far-RightTied
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Total points awarded:14 
Reasons for voting decision: Pro failed to rebut Con's case and loses a conduct point for his attitude during the third round.
Vote Placed by gizmo1650 6 years ago
gizmo1650
cowpie1998Extremely-Far-RightTied
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Total points awarded:03 
Reasons for voting decision: All math aside, pro has not show .(9) eqauls 1
Vote Placed by Ryanconqueso 6 years ago
Ryanconqueso
cowpie1998Extremely-Far-RightTied
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Total points awarded:25 
Reasons for voting decision: Really it came down to minor errors and connotations for all 7 votes.
Vote Placed by GeorgeCarlinWorshipper 6 years ago
GeorgeCarlinWorshipper
cowpie1998Extremely-Far-RightTied
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Total points awarded:23 
Reasons for voting decision: Conduct was pretty good on both sides, but Con seemed to get a bit unnecessarily snarky in R3, and Pro had marginally better spelling. That said, Pro did concede, so arguments goes to Con.
Vote Placed by boredinclass 6 years ago
boredinclass
cowpie1998Extremely-Far-RightTied
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Reasons for voting decision: Pro doesn't even try to refute cons arguments
Vote Placed by kohai 6 years ago
kohai
cowpie1998Extremely-Far-RightTied
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Total points awarded:03 
Reasons for voting decision: .9 cant = 1
Vote Placed by Floid 6 years ago
Floid
cowpie1998Extremely-Far-RightTied
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Total points awarded:07 
Reasons for voting decision: Extremely-Far-Right wins because even though he is wrong and his arguments (such as 1/3 does not equal 0.3 repeated) are in error, Cowpie1998 is incapable of pointing that out. Cowpie1998 proposes a debate where he will argue in favor of an accepted mathematical concept for which multiple proofs can be found simply by using google and fails to present these or any other relevant argument.
Vote Placed by Cliff.Stamp 6 years ago
Cliff.Stamp
cowpie1998Extremely-Far-RightTied
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Reasons for voting decision: Nice job Con for refuting the impossible.
Vote Placed by Heathen 6 years ago
Heathen
cowpie1998Extremely-Far-RightTied
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Reasons for voting decision: Con had better math skills and proved he is right.