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        riddles >> putnam exam (pure math) >> sqrt(2), sqrt(3), sqrt(5)
        
(Message started by: Aryabhatta on Jun 17th, 2007, 11:17am)
Title: sqrt(2), sqrt(3), sqrt(5)
Post by Aryabhatta on Jun 17th, 2007, 11:17am
Given an e > 0, show that there exist integers x,y,z (dependent on e, of course) such that:

0 < |xhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif2 + yhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3 + zhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif5| < e
Title: Re: sqrt(2), sqrt(3), sqrt(5)
Post by Obob on Jun 17th, 2007, 11:22am
Hint: [hide]You can always pick z=0.[/hide]


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