Converting Repeating Decimals to Fractions using Algebraic Method is Invalid
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dylancatlow
Voting Style:  Open  Point System:  7 Point  
Started:  2/6/2013  Category:  Science  
Updated:  3 years ago  Status:  Post Voting Period  
Viewed:  4,455 times  Debate No:  29976 
Debate Rounds (2)
Comments (37)
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I am taking the Con on this and I assert that Pro has the BoP. Seeing as how the following method is a fairly well known and well accepted method for converting repeating decimals to fractions, Con feels Pro has the BoP to show it is an invalid method. Invalid, in this debate means that this method will fail to arrive at the correct result at last for one or more cases (one case is enough to prove it is invalid). Here are two examples of this method in use (anywhere you see repeated numbers followed by "...", this indicates it repeats infinitely): 1) Show 0.999... = 1 Multiply 0.999... times 10 10 * 0.999... = 9.999... This is clearly 9 plus the original value: 10 * 0.999... = 9 + 0. 999... > subtract 0.999... from both sides 10 * 0.999...  0.999... = 9 > factor out 0.999... 0.999... * (10  1) = 9 > divide both sides by 9 to arrive at 0.999... = 9/9 = 1 Usually one makes things look simpler by defining x = 0.999..., then this argument looks quite a bit cleaner: 10x = 9 + x > 9x = 9 > x = 9/9 = 1, therefore x = 0.999... = 1 2) Show that 0.9898... = 98/99 x = 0.9898... 100x = 98.9898... = 98 + 0.9898... = 98 + x > solve 100x = 98 + x > 99x = 98 > x = 98/99 Con has given two examples of using the "algebraic" method to convert repeating decimals to fractions. Pro now has the BoP to show that one or both of these methods is invalid or to give a counterexample where this method would be invalid (note it must be invalid when applied to the problem of finding the fractional value for a repeating decimal). This debate is 2 rounds long. Pro should give the entirety of their case in Round 1 and Round 2 will be for Con to rebut any of Pro's claims and for Con to respond to Con's Round 2 rebuttals. As this debate is likely to hinge on the existence or nonexistence of a counterexample, both Pro and Con may present new examples in both rounds. The only stipulation is that these examples should be used to rebut/support statements previously made (so for instance, Con should not wait until Round 2 to present a counterexample). First off, I'd like to say that I will not be using algebra to argue in this debate. I will, however, make reference to my opponent's algebraic arguments. I must do this out of necessity for the simple fact that, yes, using algebra DOES 'prove' .999 = 1 ....however, algebra falls short. It is incorrect because it is dealing with things that it shouldn't be dealing with. Let x = .999... 10x = 9.999... (multiplying RHS and LHS by 10) 10x  x = 9.999...  x (subtracting x from both sides) 9x = 9 x = 1 .999... = 1 I'm showing a version of your argument because it's more condensed. I believe that both proofs are wrong. The problem is that they are engaging in mathematical legerdemain and using an infinite series as a real number. But I don't believe it is mathematically possible to use infinite series as a real number. You can't multiply it by anything. It is infinite. You end up with terms that cannot be mixed, like a blue mile or a red 65 miles per hour. It just doesn't make sense or describe anything real. You cannot treat multiplying a series as just multiplying the constant terms, you have to distribute each term into each other and then combine them, which is literally impossible when dealing with an infinite series. .999 repeating is not equal to 1 because .999 repeating DOES NOT EXIST exist. It's an irrational number. It's like trying to prove 1.00000000....0000001 (...... = infinity) is equal to 1. I just realized that I have been debating on the wrong topic that whole time ....oh well. My proof for the invalidity of the above is the same as I would have put for the 'real' debate. Thanks for debating with me and have a good day! 

One thing Pro gets correct is that repeating decimals are (or can be) represented by infinite series (I only say can because some fractions, if you choose to represent them as decimals, must be repeating decimals, but this does not mean your original fraction was an infinite series). First I find the following statement quite misinformed: "I don't believe it is mathematically possible to use infinite series as a real number." So there are two real (irrational) numbers that come to mind: π and e. The general definition of e involves a limit, however, using a Taylor Series one can arrive at the value of e using an infinite series (a series is a sum of a sequence of numbers): e = sum(1/n!), n = 0 to n = ∞ You can use the Taylor Series for an exponential and arrive at the value of e, by plugging in x = 1 (i.e. e^1) [1] There are also ways at approximating π using infinite series [2]. In both examples, note the careful use of the word approximate. We obviousy cannot actually sum an infinite series (numerically), so approximations mean doing as many terms as is feasible (which is determined by how large your data structure for storing a number is and how long you are willing to wait). So, what do we do when we have an infinite series? Is it always the case that you cannot find the actual sum (even if you can prove that the sum converges to a particular real number, i.e. in the case of π and e)? The answer is an emphatic NO! Sometimes, you can find the exact sum. This is done through algebra and equation solving. In particular and specific to this debate is in the case of geometric series. A geometric series is defined by the following: S = sum(r^i), from i = 0, to i = n [3] Indeed a repeating decimal is an instance of a geometric series. To find the exact sum, one must know what the common ratio is. In the first case of 0.999... the common ratio is 1/10 (or 0.1) hence why we multiply by 10 to find 0.999... = 1 and in the second case of 0.989898... the common ratio is 1/100 (or 0.01) hence why, in that case, we multiply by 100 to find 0.989898... = 98/99. So here is the only argument that Con can ascertain from Pro: "You cannot treat multiplying a series as just multiplying the constant terms, you have to distribute each term into each other and then combine them, which is literally impossible when dealing with an infinite series." Con does not understand what Pro means by "multiplying the constant terms". However, Con does agree that you have to distribute whatever you multiply the series by to each term in the series (i.e. 10 for 0.999... and 100 for 0.989898...). In fact, it is this distribution that makes the algebraic proof work. First, let's look at the general case: 1 + r + r^2 + r^3 + ... = sum(r^i), i = 0 to i = ∞ What happens when you multiply by the common ratio r: r * (1 + r + r^2 + r^3 + ...) = r + r^2 + r^3 + r^4 + ... = sum(r * r^i), i = 0 to i = ∞ Do you notice something about the terms you get when you multiply by r? These are all of the terms you would get in the previous sequence except for the first value of 1: So this sum: r + r^2 + r^3 + ... = (1 + r + r^2 + r^3 + ...)  1 > this leads to the following equation (assuming this series converges, which it can be shown to not converge unless r < 1) x = 1 + r + r^2 + ... rx = x  1 > x(1  r) = 1 > x = 1 / (1  r) < the formula for the sum of an infinte geometric series (when it converges, i.e. r < 1) [3] Perhaps this is unconvincing since, if you look at the original line: r * (1 + r + r^2 + r^3 + ...) = r + r^2 + r^3 + r^4 + ... Whoa, r^4 appeared! That wasn't there before, right? WRONG! This is an infinite series, we just didn't write out every single term on either the left or the right. If we had written one more term on the left, we would have seen r^4. Still, you might say, well there is always one extra (and indeed this is important when the sum doesn't converge, i.e. r >= 1). So let's look at the summation notation now: sum(r * r^i), i = 0 to i = ∞ = sum(r^(i + 1)), i = 0 to i = ∞ > this second sum can be reindexed: clearly if we start with i = 0, then the first exponent will be 0 + 1 = 1, then 2, then 3, etc. sum(r^i), i = 1 to i = ∞ But this is the original sum minus the first term, which is r^0 = 1 > { sum(r^i), i = 1 to i = ∞ } = { sum(r^i), i = 0 to i = ∞ }  1 > so again, we get rS = S  1 > S = 1 / (1  r) Let's see this in action with 0.999... 0.999... = { sum(.9 / 10^i), i = 0 to i = ∞ } = .9 * { sum((1/10)^i), i = 0 to i = ∞ } > 0.999... = .9 * (1 + (1/10) + (1/10)^2 + ...) > when we multiply by 10, we multiply every single term! 10 * 0.999... = 0.9 * (10 + 10/10 + 10/10^2 + ...) = 0.9 * (10 + 1 + 1/10 + 1/10^2 + ...) > well that sum, is just the original plus 9 (after you distribute .9 * 10 = 9) So 10x is just the original series, x, plus 9 > 10x = x + 9 > 9x = 9 > x = 1 There is no violation of "not distributing to each term". In fact, it's the distribution that makes this proof posisble. Now, I will try and help Pro by giving the example I think they are thinking of that might prove this method invalid. We have been talking about repeating decimals! So what about repeating numbers in general? What about the following number: 111... (notice no decimal point!) So what is the value of this number? Can we use the algebraic method to incorrectly arrive at a value? Obviously, this number is infinite and in fact would be represented by the following infinite sum: 1 + 10 + 100 + 1000 + ... = sum(10^i), i = 0 to i = ∞ Indeed, if we attempted to apply the same algebraic method we would arrive at a nonsensical result: x = 1 + 10 + 100 + 1000 + ... 10x = 10 + 100 + 1000 + ... =? x  1 > 10x = x  1 > 9x = 1 > x = 1/9 So this is obviously incorrect. But there are two things to note. 1) This is not a repeating decimal! (in the sense of what we normally think of as repeating decimals) 2) It should be clear that this number is infinite and thus does not have a value. So the one "assumption" that we make when using the algebraic method is that the value of the series converges! Said another way, there actually is a value of the repeating decimal! Is that a reasonable assumption? Yes, because, for instance, if we have the following number: 98,989,898.989 898... It is clear that this number is (at the very least) less than 98,989,898 + 1 and greater than 98,989,898: 98,989,898 < 98,989,898.989 898... < 98,989,899 So it is safe to assume that this number actually does converge. Furthermore, we can keep tightenting that bound (for instance, we could add .99 to the whole part for an upper bound and add .98 to the whole part for a lower bound). In fact, we can make it infinitely tight and thus it is safe to assume the number converges. We can make this argument for any finite repeating decimal number (certainly cannot make this claim for numbers that have no bound such as 111...). So while the algebraic method might hinge on this assumption (i.e. that such a number actually does exist), we do not need to prove it every single time we see a repeating decimal, rather we can look at the decimal and easily determine whether or not it's finite and, if so, we know the algebraic method will give us the correct value. Sources: [1] http://en.wikipedia.org...; [2] http://en.wikipedia.org...; [3] http://en.wikipedia.org...; You lost me there, Mr. Math. First off, following the logic of your original proof, 1.000...000 (elipses denotes infinity zeros) 00006 would equal 1, too. How could it not? You are making up for the fact that .999 doesn't equal 1 with just adding an infinite amount of them, thus making it impossible to disprove because there is no number larger than infinity. When one uses infinity in the way you are, things can get weird. .999 repeating doesn't equal 1 for the very same reason that 1.000...000 (elipses denotes infity zeros) + infinity 9's doesn't equal 1. The way you are using infinity, many many numbers can 'equal' 1 simply because of the nature of infinity. So far all intesnive purposes, .999 to the infinity equals 1 unless we can find a number larger than infinity, but that doesn't mean it really is. It just means that it's using infinity. Watch an example of how infinity can be used to distort: (1) Take the number .999 to the infinity (2) Subtract .099999 to the infinity (3) Guess what, .999 to the infinity is STILL .999 to the infinity (supposedly still 1) . Try subtracting the same numbers from 1 and you end up with .900...00001 (4) One of the most important things about numbers that are the same is that they can DO THE SAME THINGS EXACTLY THE SAME WAY EVERY FREAKING TIME Watch another example of how easily infinity can be used to distort: (1) take the number pi (3.141592...) (2) add up all the bits (3) 3.00 + 0.10 + 0.040...... (4) All these bits MUST equal infinity because the series doesn't end  infinite fodder for the number infinity to use to make itself infinity! (5) That's the nature of infinity. It simply does whatever you want it to. It can make numbers infinitely close to a number such that no number can be used to prove otherwise. It can make any infinitely repeating series infinity or zero. It really doesn't prove anything. So in conclusion, .999 to the infinity doesn't equal 1 because of the fallacy of infinity. 
2 votes have been placed for this debate. Showing 1 through 2 records.
Vote Placed by wolfman4711 3 years ago
KroneckerDelta  dylancatlow  Tied  

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Reasons for voting decision: The numbers system is just one way of doing things and pro proved it.
Vote Placed by Stephen_Hawkins 3 years ago
KroneckerDelta  dylancatlow  Tied  

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Reasons for voting decision: Kronecker proved the system to be right, dylan just proved it to be weird when you treat irrational numbers rationally.
1) I only lost because somehow wolfman decided you had better conduct...I don't see how that vote is justified whatsoever.
2) The only way to vote for Con is by looking at my R2 where I explained that the algebraic method fails when you have a repeating number (no decimals), i.e. an infinite series that does NOT converge. But seeing as how you (Con) didn't really jump on that and, I think that's not what a repeating decimal is anyway, I still think I should have won. Even worse, the guy who voted against me clearly didn't understand the debate at all and basically had no business voting (yet he ended up being the deciding vote).
.999 repeating = 1 because:
1.1 to the power of infinity = infinity.
or 555555.... (infinity 5's) = infinity.
Does this mean calculus has no real world applications? No of course not, we can use it to approximate real world things and it allows us to reason about things that we otherwise could not reason about (Quantum Mechanics comes to mind).
No, not necessarily (if I understand your question).
Again, for instance 0.999... = 9/10 + 9/100 + 9/1000 + 9/10000 + ...
The numbers are getting smaller and smaller, so there nothing that is getting infinitely big. I would really suggest reading up on series to get more information. This is a whole topic in and of itself in Calculus III where you do analysis of series and decide whether or not the series converges or is unbounded (or never reaches a value). For instance the following series is not unbounded but it just oscillates back and forth between 1 and 0:
1 + 1 + 1 + 1 + 1 + 1 ...
This series never converges, no matter how many terms you write out, the value will always be 1, then 0, then if you add another term 1, then add another term 0, etc.
An obvious rule for series to converge is that the sequence should go to zero in the limit as it goes to infinity. So for instance:
10 + 100 + 1000 + 10000 + ...
The sequence never goes to 0, it just keeps increasing, so this series clearly is unbounded and diverges.
On the other hand:
9/10 + 9/100 + 9/1000 + ...
This sequence clearly DOES go to 0 as you go to infinity, eventually you'll have nine divided by a gigantic number, which will be very close to 0. So this series MIGHT converge (and you can show that it does).
Here's an example of a divergent series where the sequence converges to 0:
1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + ...
Well, as you go to infinity, you'll eventually reach one over a very large number and thus the sequence DOES go to 0. HOWEVER, this series (the sum of the sequence) diverges!
I concede.
That's why an infinite amount of time is being used. What happens when infinity goes against infinity?
How can you say the coordinate infinity is bigger than the time infinity?
It's not infinitely bigger. The problem is it's infinity to begin with. The domain is unbounded so there is nothing to "fill". No matter how long you count for, you will always be able to give me one more integer.
Notice this is a little it different from the case of the reals filling in a FINITE domain. This is more similar to the question about 0.999... = 1.
In this case, you never actually reach one, but you get infinitely close, so in the limit that you actually DO take it to infinity it DOES reach 1. The "..." means you are going to infinity.
What you are talking about with trying to list all of the integers (for example) by listing them for an infinite amount of time is very similar to what I talked about in R2 with the number:
111... = sum(10^i), from i = 0 to i = infinity
No matter how many terms you write down, you are NEVER going to approach ANY number. It's just going to keep growing without bound. You're not approaching ANYTHING! UNLIKE the case of 0.999... in which case, each term you write down gets closer and closer to the value of the limit, which is 1.
My argument, was that the above case 111..., where you have an unbounded number is not really what we normally consider a repeating decimal (again, for one it doesn't HAVE a value!).
But wouldn't an infinite series have a part in it that is infinitely big of whatever you want an infinite number of times?