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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

Page1 / 4

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online