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Topic: sqrt(2), sqrt(3), sqrt(5) (Read 817 times) |
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Aryabhatta
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sqrt(2), sqrt(3), sqrt(5)
« on: Jun 17th, 2007, 11:17am » |
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Given an e > 0, show that there exist integers x,y,z (dependent on e, of course) such that: 0 < |x 2 + y 3 + z 5| < e
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« Last Edit: Jun 17th, 2007, 1:16pm by Aryabhatta » |
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Obob
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Re: sqrt(2), sqrt(3), sqrt(5)
« Reply #1 on: Jun 17th, 2007, 11:22am » |
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Hint: You can always pick z=0.
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