Converting Repeating Decimals to Fractions using Algebraic Method is Invalid

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 = 1There 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 non-sensical result:x = 1 + 10 + 100 + 1000 + ...10x = 10 + 100 + 1000 + ... =? x - 1-->10x = x - 1 --> 9x = -1 --> x = -1/9So 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... < 98,989,899So 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...;