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 September 14th, 2009, 04:49 AM #1 Senior Member   Joined: Sep 2008 Posts: 199 Thanks: 0 bimomial expansion By assuming $0<\sqrt{5}-2<\frac{1}{4}$ , deduce that the difference between $(\sqrt{5}+2)^5$ and an integer is less than $\frac{1}{1024}$
 September 14th, 2009, 05:54 AM #2 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Try expanding $(\sqrt5+2)^5-(\sqrt5-2)^5,$ and see where that gets you...
September 15th, 2009, 02:58 AM   #3
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Re:

Quote:
 Originally Posted by mattpi Try expanding $(\sqrt5+2)^5-(\sqrt5-2)^5,$ and see where that gets you...
THanks Mattpi .. I found 1364 . So how can i continue now ?

 September 15th, 2009, 08:47 AM #4 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Right, so $(\sqrt5+2)^5-1364=(\sqrt5-2)^5.$ What do you know about $\sqrt5-2?$
September 16th, 2009, 03:21 AM   #5
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 Originally Posted by mattpi Right, so $(\sqrt5+2)^5-1364=(\sqrt5-2)^5.$ What do you know about $\sqrt5-2?$
Thanks a lot , i finally understood ..

Since $0<(\sqrt{5}-2)^5<\frac{1}{1024}$

$(\sqrt5+2)^5-1364=(\sqrt5-2)^5.$ , hence $(\sqrt5+2)^5-1364$ will be less than 1/1024

Thanks again !

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