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   sqrt(2), sqrt(3), sqrt(5)
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   Author  Topic: sqrt(2), sqrt(3), sqrt(5)  (Read 804 times)
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 < |x2 + y3 + z5| < e
« Last Edit: Jun 17th, 2007, 1:16pm by Aryabhatta » IP Logged
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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