Induction on .999...
Date: 10/19/2000 at 08:19:40 From: Blake Thomson Subject: Further questions on .999... = 1 I have been having an argument with my physics teacher over the fact that point nine recurring (.999... or PNR) equals 1. I showed him the proof on your site and he pointed out a fact that I hadn't noticed. How can you multiply PNR by ten if you can't get to the beginning? For example, to multiply 215 * 3, you would start off on the right, multiplying 3 by 5, then 1, then finally 2. Now how can you multiply PNR by ten if the farthest right value cannot be reached? Thank you for any clarification you can give. Blake
Date: 10/19/2000 at 13:18:25 From: Doctor TWE Subject: Re: Further questions on .999... = 1 Hi - thanks for writing to Dr. Math. Check out our FAQ on the subject at: Why does 0.9999... = 1? http://mathforum.org/dr.math/faq/faq.0.9999.html This, and the links therein, explain several different ways of proving this interesting fact. I wanted to comment on your (or your teacher's) idea that "you can't get to the beginning." We multiply values from right to left only as a matter of convenience, so we don't have to "backtrack" each time we carry. It is equally valid to multiply from left to right. This is often done when we want to get a quick estimate for the answer, then get more accuracy later. The LEFTMOST digits contribute the most to the answer, so to get "in the ballpark," we can multiply them first. I'll demonstrate with a concrete example, then a general argument. Let's use your example of 215 * 3. I can multiply them as follows; first, I'll multiply the 3 by the 2 (actually, by 200) and put the answer in the hundreds place, like this: 215 * 3 --- 6 Then I'll multiply the 3 by 1 and put the answer in the tens place: 215 * 3 --- 6 3 Finally, I'll multiply the 3 by 5 and put the answer in the units place: 215 * 3 --- 6 3 15 When I add these up, I get the correct answer: 215 * 3 --- 6 3 15 --- 645 Notice that along the way, I got pretty good estimates for my final answer. After the first step my total was 600 (not a bad estimate, off by less than 10%), after the second step I had 630, and after the third step, I got 645. Going right to left, my totals after each step would be: 15 (not a very good estimate of the final answer), 45 (still not very good) and finally 645. In the general case, consider multiplying a 3-digit value ABC by some value X, where A, B and C are digits. What we really have, then, is (100*A + 10*B + 1*C) * X Conventionally, we solve this starting with the units digit as: X*C*1 + X*B*10 + X*A*100 But using the commutative property of addition, this is equal to: X*A*100 + X*B*10 + X*C*1 showing that the order doesn't matter. If I wanted to, I could start in the middle, for example: X*B*10 + X*A*100 + X*C*1 I'd just have to be careful not to miss any digits or do any ones twice. Without being able to start with the most significant digit, we could never find a value like 2*pi, because pi (like all irrationals) is an infinite non-repeating decimal. When we say 2*pi is approximately 6.2831853, we can do so because we started multiplying at the end and not the beginning. One final note: How can we be sure that the multiplications by 10 continue as we expect (i.e. they continue shifting the digits 1 place to the left) and that the subtractions of successive digits produce zeroes infinitely? These steps can be proven using a technique called mathematical induction. That's too complex to explain here, but if you search our Ask Dr. Math archives for the word "induction" (type it without the quotes), you'll find many questions and answers about it. Our searcher is at: http://mathforum.org/mathgrepform.html I hope this helps. If you have any more questions, write back. - Doctor TWE, The Math Forum http://mathforum.org/dr.math/
Date: 10/20/2000 at 08:51:29 From: Blake Thomson Subject: Induction concerning point nine recurring I recently sent you a letter regarding the proof that .999... = 1 in the FAQ. Specifically, I asked how was possible to multiply PNR by ten when you could not get to the right-hand side of it. This was explained quite neatly. However, the nice person who helped me out said it could be proven that one can multiply PNR by 10 by using a process of induction. I tried to search the archives, but the results I found there were not very satisfactory because it seemed to me that each induction question was actually problem-specific. To cut short, I would like to know the induction proof for multiplying PNR by ten. Cheers for any help you can give!
Date: 10/23/2000 at 16:02:47 From: Doctor TWE Subject: Re: Induction concerning point nine recurring Hi again Blake! Thanks for writing back! In general, we use proof by induction whenever we want a proof that involves an infinite sequence or series, in this case .999... Inductive proofs require two steps: Step 1 (the basis step): Prove it for some starting value, like n = 1. Step 2 (the inductive step): Prove that if it's true for n = k, then it is true for n = k+1. For the inductive step, we assume that it is true for n = k, and we usually use this assumption in the proof itself. A word of caution: because of their self-referential nature, inductive proofs are usually hard to follow; it's easy to get lost in the details, losing track of what n, k, and k+1 are supposed to be. In our case, we want to show that when multiplying the nth digit by 10, we get a 9 in the (n+1)st digit, no carry to the (n+2)nd digit, and a 0 in the nth digit (counting from the right.) We want no carry to the (n+2)nd digit so that we don't have to adjust the previous (greater place value) digits multiplied because of a "retroactive carry." We want a 0 in the nth digit so that it will not produce further carries when multiplying the next digit (smaller place value) by 10. Mathematically: 10 * (9*10^n) = 9 * 10^(n+1) We want to show that: a) the nth digit is 0 b) the (n+1)st digit is 9 c) the (n+2)nd digit is 0 To minimize the confusion, I'm going to use n = -1 as my basis step, and show that if it is true for n = k, then it is true for n = k-1 as my inductive step. This is the "negative" of the standard, but it will eliminate the need for using -k's and -(k+1)'s, etc. in my equations. Step 1: Show that it is the case for n = -1 10 * [9*10^(-1)] = 9 * 10^(-1+1) 10 * (9*0.1) = 9*10^0 10 * 0.9 = 9*1 9 = 9 a) the -1st digit (the tenths place) is 0: True b) the 0th digit (the units place) is 9: True c) the 1st digit (the tens place) is 0: True So we've proved the basis. Now for the inductive step. Step 2: Prove that if it's true for n = k, then it's true for n = k-1. Let's let d = the initial (k+1)st digit. This digit is 0 from conclusion (a) of the previous step. Then: 10 * (9*10^k) + d = 9 * 10^(k+1) 9 * (10*10^k) + 0 = 9 * 10^(k+1) 9 * 10^(k+1) = 9 * 10^(k+1) a) the kth digit is 0: True - 9*10^(k+1) has a 0 in the kth place. b) the (k+1)st digit is 9: True - 9*10^(k+1) has a 9 in the (k+1)st place c) the (k+2)nd digit is 0: True - 9*10^(k+1) has a 9 in the (k+2)nd place So, if 10 times the kth digit is 9*10^(k+1), then 10 times the (k-1)st digit is 9*10^k, and thus 10 times the (k-2)nd digit is 9*10^(k-1), and thus... Therefore 10 * 0.999... = 9.999... Q.E.D. Depending on the skepticism of the intended audience, you may have to use a similar inductive proof for the step: 9.999... - 0.999... -------- 9.000... For most audiences, the following will do: Let x = 0.999... then 9.999... = 9 + x so 9.999... - 0.999... = (9 + x) - x = 9 However, some audiences (like your teacher, perhaps) will not be satisfied that algebraic operations work on "infinite strings" (i.e. they won't be convinced that x - x = 0 when x is an infinite repeating decimal.) I hope this helps! If you have any more questions, write back again! - Doctor TWE, The Math Forum http://mathforum.org/dr.math/
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