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Unformatted text preview: Midterm 1 for Math 104, Section 1 Solutions 1. Prove that every real number is a limit of irrational numbers. Let x R be a real number. If x is irrational, then take the constant sequence ( x ). Then lim n x = x , so x is a limit of irrational numbers trivially. For each q Q , q + 2 /n is irrational for each n N , since if q + 2 /n = p Q , then 2 = n ( p q ) Q , contradicting that 2 is irrational. For all > 0, by the Archimedean property, there exists N such that N > 2 / . Then for n N ,  q + 2 /n q  = 2 /n 2 /N < . Thus lim n q + 2 /n = q . So q is a limit of irrational numbers. 2. Use the intermediate value theorem to show that 2 R . Consider the polynomial f ( x ) = x 2 . Then f ( x ) is continuous. If we let = max { 1 , / (1+  2 x  ) } , then for y such that  y x  < , we have  y 2 x 2  =  ( y x )( y + x )  =  y x  y + x  =  y x  y x +2 x   y x  (  y x  +  2 x  ) / (1+  2 x  ) (1+  2 x  ) = by the triangle inequality and our formula for...
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 Fall '08
 RIEMAN
 Math

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