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7/12/2013 5:16:19 AM
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Science Friday, Week 18: Why 0.999... = 1

Welcome to week 18 of Science Friday. I thought I would change up my lineup slightly by addressing an issue that I have personally seen arise a number of times on Bungie.net. It so happens it is mathematical in nature, so what better place than Science Friday. Today I am going to demonstrate that the repeating decimal 0.999... is, in fact, exactly equal to the number 1. There are two ways to prove this assertion. One involves using basic algebra and the other some simple calculus. I’ll prove it using both methods. If you are not familiar with the mathematics of calculus, do not fear. I will explain the details as I go on. Let’s start with the algebraic proof since it is shorter. I am going to try and avoid using typical textbook language. I’ll preserve mathematical integrity without sounding like a robot. Everyone wins. For this proof, we are going to assign 0.999... to the variable X. (In textbook language, “Let X = 0.999....”) The reason for this will become clear. So now we have X = 0.999... Since X is just a number, we can multiply it by another number and get yet another number as an answer. I am going to multiply X by 10 in this case, which can be written as 10X. Since X = 0.999..., we multiply 0.999... by 10 to find the number solution. We are all very comfortable with multiplying numbers by 10. Just add another zero to the number. However, this can be generalized even further. Move the decimal point one place to the right (which results in adding a zero for integers). Notice that just adding a zero does not work for decimals. This means 10X = 10(0.999...) = 9.99... Now we have two equations. One expresses the value of X, and the other expresses the value of 10X. X = 0.999... 10X = 9.99... What we are going to do now is subtract the first equation from the second one. Now, this step may seem a bit abstract, so let’s first do this with a concrete example. Below are two equations that we know to be true. 1 + 1 = 2 2 + 2 = 4 Let’s subtract the first equation from the first one. To do this, we first subtract 2 from 4, which equals 2. We then subtract 1 from 2, which equals 1. We do this once more (moving right to left), and we get 1 again. We are left with 1 + 1 = 2 The above equation is the difference between the two equations of 1 + 1 = 2 and 2 + 2 = 4. Notice that we got another equation that is still mathematically sound. While this does not formally prove that subtraction will yield a valid equation from two valid equations, it does provide an intuitive framework from which we can operate. So let’s return to our original equations X = 0.999... 10X = 9.99... We will follow the same procedure as we did before for the subtraction operation. 9.99... - 0.99... = 9 and 10X - X = 9X. In the end, we get 9X = 9 Let’s now solve for X by dividing both sides by 9. X = 9/9 = 1. Since we assigned X = 0.999..., and we have found that X = 1, therefore X = X or 0.999... = 1 This is a simple manipulation to prove that 0.999... = 1. We can also use basic calculus to prove the exact same thing. The crucial thing to understand is that, using calculus, we can get finite answers to infinite constructs. For example, the series (addition of many numbers) 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ... and so on never ending is equal to a finite number: 2. 0.999... is nothing more than an infinite series like 1 + 1/2 + 1/4 and so on. The series that represents 0.999... is 0.9 + 0.09 + 0.009 + 0.0009 + ... (never ending) The formula for finding the sum of an infinite [i]geometric[/i] sequence is S = a1 / (1 - r) where S is the sum, a1 is the first term in the sequence of numbers, and r is a multiplier formally called the [i]common ratio[/i]. The common ratio is simply the number you must multiply one term in the sequence to get the next number. In this case, the common ratio is 1/10 since multiplying 0.9 by 1/10 equals 0.09. This works for the entire sequence. The derivation of the summation formula above is rather straightforward, but I will omit it here to prevent making this post too long. If you want to see the derivation, please comment below and I would be happy to provide it. So, if the repeating decimal 0.999... is expressed by the infinite series 0.9 + 0.09 + 0.009 + 0.0009 + ... (never ending), then we can find the summation of the infinite series, which will be a finite number. The summation will be the value of the repeating decimal 0.999.... Let’s use the formula. S = a1 / (1 - r) S = 0.9 / (1 - 1/10) = 0.9 / 0.9 = 1 The infinite series is equal to 1, and since the infinite series equals the repeating decimal 0.999..., 0.999... = 1 by the transitive property of equality. I hope these two proofs have clarified why 0.999... = 1 is a mathematical fact. If you have any questions about the proofs or derivation of the formula I used in the second proof, please do not hesitate to ask. Otherwise, leave your general comments and feedback below. See you next week for more Science Friday!

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  • P.S. Mathematics was not built to sustain infinities. Multiplication is simply a shortcut as to not have to make the user add a set if the same numbers over and over again. You can simply move the decimal place over one for a finite number when multiplying by ten, but not for an infinite number. That's like multiplying a giraffe by ten and ending up with g.iraffe. Or giraffe0. With .9999999(...) you would need to add it onto itself nine more times, that's what multiplication is. But since those nines are going on forever, you can't quite have 1.9999998(...), 2.9999997, et cetera, until you get to the magic 9.999999(...) that you so brazenly entitle [i]10X[/i]. Just as trying to kick a lion to see how much it hates water doesn't work, math is not the right tool to use here. And since any infinity of anything isn't a number, even an infinite number, that would explain why math shouldn't be used. After all, you don't use it for your g.iraffes. "there's a time and place for everything, but not now." ~Professor Oak

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