I'm now fairly convinced that there is no reduced fraction a/b on (0, 1) with the same digits in the fraction and its decimal expansion. I argue as follows. If reduced a/b has a finite decimal expansion, then b = 2^k or b = 5^k. If b = 2^k, the decimal expansion has k digits (or k+1 digits if we count the zero before the decimal point, it makes little difference to the argument). b = 2^k has about log10(2) k ~= 0.7 k. Since a and b together have k digits, a has about 0.7 k digits. This means that for sufficient k, specifically k >= 3, a will have strictly more digits than b, so that a/b > 1 hence not on (0, 1). We can exhaustively rule out solutions with k < 3, so there are no solutions with b = 2^k. If b = 5^k, b has about log(5) k = 0.7 k digits. This means a has about 0.3 k digits, and a/b < 1, so a/b is on (0, 1) as required. This means that a is about 10^(0.3 k) while b is about 10^(0.7 k), so that a/b is about 10^(-0.4 k), and for large k, about 0.4k digits after the decimal point will be zeroes. But in the fraction a/b, 0.7 k of the digits are part of b = 5^k, which, for large k, appears to have equidistributed digits, so that we can expect about 9/10 of them to be nonzero, that is, about 0.63 k nonzero digits. For large k, the law of averages would dictate that b = 2^k has > 0.6 k nonzero digits, that is, more nonzero digits than the decimal expansion. I'm guessing that for some moderately large k, we can be all but certain that the fraction includes more nonzero digits than the decimal, which rules out solutions. We need only rule out the small k to get whatever degree of certainty we require that no solution exists. ----- Original Message ----- From: "Christian Boyer" <cboyer@club-internet.fr> To: "'math-fun'" <math-fun@mailman.xmission.com> Sent: Saturday, September 23, 2006 9:52 AM Subject: RE: [math-fun] Decimal question
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Here is a family of solutions -> 0 using irreducible fractions:
17537/12800 = 1.370078125 17537/128000 = 0.1370078125 17537/1280000 = 0.01370078125 17537/12800000 = 0.001370078125 ... = ...
Christian.
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