HW_2_Solutions

# HW_2_Solutions - MATH 413 HW 2 Solution to selected...

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Unformatted text preview: MATH 413 HW 2: Solution to selected problems. September 12, 2007 2..2.4.3-4 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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HW_2_Solutions - MATH 413 HW 2 Solution to selected...

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