HW1 - + 1 n T. This contradicts the assumption that is the...

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MATH 324 Real Analysis I Douglas Weber 30 January 2010 Theorem : There exists a real number β such that β 2 = b . Proof : Let T = { t R : t 2 < b } . T is not empty because for t < min { b, 1 } , t 2 < t < b , so t T . In addition, t is an upper bound for T because for t > max { b, 1 } , t 2 > t > b . Let β = sup( T ) and let n N . We will show that β 2 = b . To do so, we will show that the assumptions β 2 < b and β 2 > b lead to contradictions. Case 1 : Suppose that β 2 < b . Let ± = b - β 2 . Note that ( β + 1 n ) 2 = β 2 + 2 β n + 1 n 2 β 2 + 2 β + 1 n . Let n be chosen so that 2 β +1 n ± . Then ( β + 1 n ) 2 < b . Therefore
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Unformatted text preview: + 1 n T. This contradicts the assumption that is the supremum of T . Case 2 : Suppose 2 &gt; b . Let = 2-b . Note that ( -1 n ) 2 = 2-2 n + 1 n 2 2-2 n . Let n be chosen so that 2 n . Then ( -1 n ) 2 &gt; b. So -1 n is an upper bound of T contradicting the assertion that is the least upper bound. So 2 = b. Q.E.D. 1...
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This note was uploaded on 02/05/2010 for the course MATH 324 taught by Professor Dang during the Spring '08 term at SUNY Geneseo.

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