A4 - n such that 1 =n< y Consequently y is not in the interval(0 1 =n for that n So there can be no real numbers in T 1 n =1(0 1 =n 4 Write p 2

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1. State both forms of the Archimedean property of R . This is Theorem 1.4.2: (i) Given any real number x , there is a natural number n such that n > x . (ii) Given any real number y > 0 , there is a natural number n such that 1 =n < y . 2. What do we mean when we say that the rational numbers are dense in the real numbers ? This is Theorem 1.4.3: Given two real numbers a < b , there is a rational number r such that a < r < b . 3. Prove that T 1 n =1 (0 ; 1 =n ) = ; . A real number y is in (0 ; 1 =n ) exactly when 0 < y < 1 =n . If y 2 (0 ; 1) , then y > 0 so, by the Archimedean property, there is a natural number
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Unformatted text preview: n such that 1 =n < y . Consequently, y is not in the interval (0 ; 1 =n ) for that n . So there can be no real numbers in T 1 n =1 (0 ; 1 =n ) . 4. Write p 2 as the sum of two irrational numbers . Perhaps the easiest way is p 2 = 1 2 p 2+ 1 2 p 2 . The number 1 2 p 2 is irrational because it is the product of the nonzero rational number 1 = 2 and the irrational number p 2 . Another way is p 2 = 2 p 2 + ( & 1) p 2 ....
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This note was uploaded on 05/26/2010 for the course MATH maa 4200 taught by Professor Dr. during the Spring '08 term at Miami University.

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