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The Nagell-Ljunggren Equation A Digression
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Decimal ExpansionsAlthough mostly I'll be talking about things in base 10, everything I have to say here applies to fractions in any base.If you have a repeating decimal, like 0.23, there's a nice series of transformations you can do. First you can pull out a factor of 23.
23 × 0.010101 … Then you can split the decimal expansion into pieces, …
23 ( 0.01 + 0.0001 + 0.000001 + … ) … write the pieces as fractions, …
23 ( 1/100 + 1/10000 + 1/1000000 + … ) … and simplify a little.
23 [ (1/100) + (1/100)2 + (1/100)3 + … ] By now I hope you can see where I'm going with this: the series is geometric. To sum it, normally I'd pull out a factor of 1/100 and use the power series
1/(1−x) = 1 + x + x2 + x3 + … , but here it's more convenient to use the old rule from algebra, that if the first term is a and the ratio between terms is r, then the sum is a/(1−r). Let's define N = 100, then a = r = 1/N, and the sum simplifies to 1/(N−1). Substituting back, we find the final answer.
0.23 = 23/99 In exactly the same way, if you have a decimal that repeats with length k, you can write it as a fraction m/n, where m is read off from the decimal expansion and n is just k 9s strung together. Or, to put it another way, n = 10k−1. If you're working in base b, there are a couple of ways of thinking about this result. Suppose we're interested in 0.23 again, but now in base 7. If you're a die-hard base 10 person, you'll probably want to write n = bk−1 and convert m into base 10 as soon as possible. So … m = 23 base 7, which is really 17, and then n = 72−1, which is 48, so the desired fraction is 17/48. If you're willing to stay in base 7, though, you don't need to rewrite the equation for n at all, because 10 already is 7. Then you can be formal and write out n = 100−1, or, as in base 10, you can just write down the answer, 23/66. Then you can convert to base 10 or not, as needed. That's all I'm going to say about base 7 right now. Now, if you have a decimal that doesn't start repeating right away, like 0.083, you can still do pretty much the same thing. First multiply by 100, to shift the digits over … that gives 8.3. Then write that as 8 3/9, or 75/9; then divide by 100 to get the real answer, 75/900. Thus, any number with a decimal expansion that eventually repeats can be written as a fraction with denominator n = 10j(10k−1), where j is the number of digits that come before the repeating sequence and k is the length of the sequence itself. And, conversely, if you have a fraction with one of those special denominators, you can almost immediately write down its decimal expansion. Now, here's the punchline. As you probably know, any fraction has a decimal expansion that eventually repeats. But, that expansion can then be written as a fraction with a special denominator! In other words, any fraction can be written as a fraction with a special denominator; in fact any denominator can be made special by multiplying by an integer. So, if you have a fraction you want to know the decimal expansion of, but you don't feel like actually working it out, what you can do instead is play “guess the special denominator”. In general this is a hard game. For example,
1/7 = 142857/999999. But, in most cases that actually arise, it's not so hard. If, for example, you want to know about 1/12, you can factor 12 as 2×2×3, then it's clear that multiplying by 5×5×3 will produce some nice nines and tens. So, 1/12 = 75/900, which as we saw above is 0.083.
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See AlsoDecimal Expansions (Section) Divisibility Footnote (Decimal Expansions) Fractions in Base 2 Fractions in Base N Negative Digits Powers of N Repeat Length @ April (2004) |