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 February 15th, 2010, 12:24 PM #1 Member   Joined: Jul 2009 Posts: 88 Thanks: 0 the largest integer number $For \ x \in R$ , let [x] be the largest integer less than or equal to x . prove that :$[\sqrt [x] ]= [\sqrt x] , \forall x \geq 0$
 February 15th, 2010, 12:55 PM #2 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: the largest integer number Let x = a² + b + c, where a² is the largest square number less than or equal to x, b is [x]-a², and c is x-[x]. So a and b are integers, and 0 ? c < 1. Then [x] = a²+b, and you should be able to show that a ? ?(a²+b) ? ?x < a+1.
 March 1st, 2010, 02:25 AM #3 Member   Joined: Jul 2009 Posts: 88 Thanks: 0 Re: the largest integer number $a^2 \leq x \to$ $[a^2] \leq [x] \to$ $\sqrt {a^2 }\leq \sqrt{a^2+b}$ $\to a \leq \sqrt{a^2+b}$ __________________________________ $[x] \leq x \to$ $a^2 +b \leq x$ $\to \sqrt{a^2+b} \leq \sqrt{x}$ ----------------------------------------------- $\sqrt x < a+1 \ ? ?$ i really don't know !
 March 1st, 2010, 07:24 PM #4 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: the largest integer number a² ? a²+b ? a²+b+c < (a+1)² But a²+b = [x] so a² ? [x] ? x < (a+1)² and take square roots

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