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Unformatted text preview: MATH 413 HW 2: Solution to selected problems. September 12, 2007 2..2.4.34 If x is a real number show that there exists a sequence of rational numbers x 1 , x 2 , . . . representing x such that x n ≤ x n +1 and x n < x for all n . Proof: Suppose that n 7→ y n is a Cauchy sequence of rational numbers that represents x . By the definition of a Cauchy sequence there exists m (1) ∈ N such that  y m (1) y j  < 1 for j > m (1) and recursively we can find m ( n ) ∈ N , for n > 1, such that m ( n ) > m ( n 1) and  y m ( n ) y j  < 1 n for j > m ( n ). We define x 1 = y m (1) 2 and x n = max { x n 1 , y m ( n ) 2 n } for n > 1. Clearly n 7→ x n is a sequence of rational numbers such that x n +1 ≤ x n for all n . We leave as an exercise to verify that: (i) n 7→ x n is a Cauchy sequence, (ii) n 7→ x n represents x , and (iii) x n < x for all n . Remark: One can also prove this fact by using Theorem 2.2.5 (Den sity of Rational Numbers), but we did not use it here in the spirit of viewing everything in the “Cauchy sequences framework”. 2.2.4.7 Prove that  x y  ≥  x    y  for any real numbers x and y . Proof: By the triangle inequality (Theorem 2.2.3) we obtain  x  =  ( x y ) + y  ≤  x y  +  y  , which is equivalent to the inequality we wanted to prove....
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 Fall '07
 HUBBARD
 Math, Rational number, Limit of a sequence, Cauchy sequence, 0 2j

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