Date: 08/04/98 at 08:54:13 From: Scott Bennett Subject: Decimal representations of rational numbers I was recently shown a proof stating that 0.9 (repeating) is actually equal to one. I have come to accept this proof, but I'm wondering from a representation standpoint whether they are actually considered to be mathematically equal. The proof involves using an sum of 9/(10^n) for n = 1 to infinity, which is an expression for 0.9 repeating. I was actually able to justify this by pulling the 9 out of the sum and multiplying it by the sum of 1/(10^n) for n = 1 to infinity. This opens up a whole new can of worms, specifically that any irrational number could possibly be expressed as an infinite sum of rational numbers. And since the sum of two or more rational numbers is another rational number, there may not be any irrational numbers as the term is currently defined. My question is, are the decimal representations of rational numbers and the rational numbers themselves mathematically equal, or would they more appropriately be called equivalent?
Date: 08/04/98 at 16:59:58 From: Doctor Rob Subject: Re: Decimal representations of rational numbers You are beginning to discover some of the beauties and mysteries of mathematics! It is true that the sum of any finite number of rational numbers is again a rational number, but the sum of an infinite number of rational numbers may be irrational. After all, any decimal expansion, such as: Pi = 3.14159265358979323846... sqrt(2) = 1.414213562... and so on, are actually infinite sums of rational numbers: 3.14159... = 3 + 1/10 + 4/10^2 + 1/10^3 + 5/10^4 + 9/10^5 + ... Yes, every irrational number is the sum of an infinite number of rational numbers. That does not preclude them from existing. The rational numbers are just those whose decimal expansions are eventually periodic. All the others are irrational. In answer to your last question, the decimal representations and the rational numbers are mathematically equal. Some rational numbers have more than one decimal expansion, as you have come to accept: 1 = 1.00000000... = 0.99999999... for example. That is not an impediment. Keep asking good questions! - Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Date: 08/05/98 at 11:39:15 From: Anonymous Subject: Re: Decimal representations of rational How can you say that an infinite sum of rational numbers can be an irrational number? If you can add two rational numbers and get a rational number, then by the associative property of addition, you could add up any number of rational numbers, even an infinite number, and get a rational number. It defies logic to say that you can show irrationality by demonstrating that a number cannot be shown to be the root of Qx+P = 0. There are an infinite number of integers, and no one can know whether there are numbers out there for which this method will or will not work. Maybe our computing power just is not enough yet to find these numbers, but this does not mean they don't exist, they're just beyond our current understanding. To me, you could prove irrationality by showing that there is not a repeating pattern. But what if the repeating pattern is infinitely long? And what if the pattern is not necessarily (I don't know the right word for it) linear? Maybe something like: 1.32435465768798109111012 ... There is a repeating pattern, but it's not linear, like: 1.35246135246 ... (like I said, I don't know the right word for it, but linear seems appropriate). Thanks, this has been fun and educational, and has definitely made me think. This revelation has only come to me in the past few days through some research and just plain thinking about things, and it has shaken me up a little bit, but I think that's good now and then. Scott
Date: 08/05/98 at 16:09:16 From: Doctor Rob Subject: Re: Decimal representations of rational When you pass from a finite number of terms to an infinite number of terms, your proof breaks down, and no substitute proof is available. You may get for your sum either a rational or an irrational number. See below. An irrational number is defined as one which cannot be written in the form a/b for integers a and b with b nonzero. (You might as well assume that b > 0, since a/[-b] = [-a]/b.) Those are precisely the solutions of b*x + (-a) = 0, so any number which is not the solution of such a linear equation with integer coefficients is by the definition an irrational number. Every rational number a/b has a finite repeating sequence of digits in its decimal expansion. The proof goes like this. Use long division to divide b into a. Look at the sequence of remainders you get. All of them are smaller than b (if you have done the long division correctly!). After at most b steps, you will either get a remainder of 0 (and the decimal expansion terminates) or you will get a nonzero remainder you have already seen previously. This is an application of the Pigeonhole Principle. (If you put n+1 things in n boxes, at least one box must contain two of them.) From that point on, the division will exactly duplicate the steps following that previously seen point, including getting that same remainder yet again, after the same number of steps. If you are getting the same sequence of remainders, you will be getting the same sequence of quotient digits, so that will repeat with the same period as the remainder sequence. The above argument is independent of the size of a and b, so is valid for every integer choices for those numbers. If b is large, the period may be long, but it is finite, since its length is at most b-1. (Why can I use "b-1" here, instead of "b"? Think about it.) Here is an example: 16/37. .432432432432 ------------------ 37 ) 16.0000000000000 14 8 ---- 1 20 1 11 ---- 90 74 -- 160 148 --- 120 111 --- 90 74 -- 16 Sequence of remainders: 16, 12, 9, 16, 12, 9, 16, 12, 9, 16, ... --------- --------- --------- -------- If two remainders are equal, then when you bring down the 0, the result will also be the same, and the quotient digit will be the same, and the new remainder will be the same, so all the steps afterwards will look like a repetition of earlier steps. Another example: 93/176. .528409090909... --------------------- 176 ) 93.000000000000... 88 0 ---- 5 00 3 52 ---- 1 480 1 408 ----- 720 704 --- 160 0 --- 1600 1584 ---- 160 0 --- 1600 1584 ---- 16 Sequence of remainders: 93, 50, 148, 72, 16, 160, 16, 160, 16, ... ------- ------- ------- There is a certain pattern to the decimal digits, but they do not form a periodic sequence of digits. All those decimals cannot be numbers of the form a/b, for any a or b, by the above argument. Thus they are irrational. Here is another kind: 1/10^1 + 1/10^4 + 1/10^9 + 1/10^16 + 1/10^25 + ... + 1/10^(n^2) + ... = 0.1001000010000001000000001000000000010000000000001000... The gaps between the 1's get longer and longer, so it is not a periodic repeating decimal. That means it must be irrational. Keep thinking and asking good questions! - Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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