Real-nos - PART I. THE REAL NUMBERS This material assumes...

Info iconThis preview shows pages 1–3. 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: PART I. THE REAL NUMBERS This material assumes that you are already familiar with the real number system and the represen- tation of the real numbers as points on the real line. I.1. THE NATURAL NUMBERS AND INDUCTION Let N denote the set of natural numbers (positive integers). Axiom: If S is a nonempty subset of N , then S has a least element. That is, there is an element m S such that m n for all n S . Note: A set which has the property that each non-empty subset has a least element is said to be well-ordered . Thus, the axiom tells us that the natural numbers are well-ordered. Mathematical Induction. Let S be a subset of N . If S has the following properties: 1. 1 S , and 2. k S implies k + 1 S , then S = N . Proof: Suppose S = N . Let T = N- S . Then T = . Let m be the least element in T . Then m- 1 / T . Therefore, m- 1 S which implies that ( m- 1) + 1 = m S , a contradiction. Corollary: Let S be a subset of N such that 1. m S . 2. If k m S , then k + 1 S . Then, S = { n N : n m } . Example Prove that 1 + 2 + 2 2 + 2 3 + + 2 n- 1 = 2 n- 1 for all n N . SOLUTION Let S be the set of integers for which the statement is true. Since 2 = 1 = 2 1- 1, 1 S. Assume that the positive integer k S. Then 2 + 2 1 + + 2 k- 1 + 2 k = ( 2 + 2 1 + + 2 k- 1 ) + 2 k = 2 k- 1 + 2 k = 2 2 k- 1 = 2 k +1- 1 . Thus, k + 1 S. 1 We have shown that 1 S and that k S implies k + 1 S . It follows that S contains all the positive integers. Exercises 1.1 1. Prove that 1 + 2 + 3 + + n = n ( n + 1) 2 for all n N . 2. Prove that 1 2 + 2 2 + 3 2 + + n 2 = n ( n + 1)(2 n + 1) 6 for all n N . 3. Let r be a real number r = 1. Prove that 1 + r + r 2 + r 3 + + r n = 1- r n +1 1- r . for all n N 4. Prove that 1 + 2 n 3 n for all n N . 5. Prove that 1 1 + 1 2 + 1 3 + + 1 n n for all n N . 6. Prove that 1- 1 2 2 1- 1 3 2 1- 1 n 2 = n + 1 2 n for all n 2. 7. True or False: If S is a non-empty subset of N , then there exists an element m S such that m k for all k S . I.2. ORDERED FIELDS Let R denote the set of real numbers. The set R , together with the operations of addition (+) and multiplication ( ), satisfies the following axioms: Addition: A1. For all x, y R , x + y R (addition is a closed operation). A2. For all x, y R , x + y = y + x (addition is commutative) A3. For all x, y, z R , x + ( y + z ) = ( x + y ) + z (addition is associative). A4. There is a unique number 0 such that x +0 = 0+ x for all x R . (0 is the additive identity .) A5. For each x R , there is a unique number- x R such that x + (- x ) = 0. (- x is the additive inverse of x .) Multiplication: M1. For all x, y R , x y R (multiplication is a closed operation)....
View Full Document

Page1 / 12

Real-nos - PART I. THE REAL NUMBERS This material assumes...

This preview shows document pages 1 - 3. Sign up to view the full document.

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