{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

mat25-hw4solns

mat25-hw4solns - Math 25 Advanced Calculus UC Davis Spring...

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

Math 25: Advanced Calculus UC Davis, Spring 2011 Math 25 — Solutions to Homework Assignment #4 1. Using the Archimedean Theorem, prove each of the three statements that follow the proof of that theorem in section 1.7 of the textbook. (a) No matter how large a real number x is given, there is always a natural number n larger. Proof. Suppose that there is some x such that no natural number is larger than x . Then x is an upper bound for the set of natural numbers, which contradicts the Archimedean Property. (b) Given any positive number y , no matter how large, and any positive number x , no matter how small, there is some natural number n such that nx > y . Proof. Assume there is no such n . That is nx y for all n . Note that this is equivalent to n yx - 1 for all n , which contradicts item (a). (c) Given any positive number x , no matter how small, once can always find a fraction 1 n such that 1 n < x . Proof. Assume there is no such n . Then 1 n x for all n . Note that this is equivalent to n x - 1 for all n , which contradicts item (a). 2. The completeness axiom can be used to construct numbers whose existence we only suspected before. As an illustration of this idea, prove that the number α = sup { x R : x 2 < 2 } exists in R and satisfies α 2 = 2. (Hint: show that either of the assumptions α 2 < 2 and α 2 > 2 lead to a contradiction.) Proof. We will use the following auxiliary result: if x 2 < y 2 , then x < y where x, y > 0. To see this, note that x 2 < y 2 is equivalent to x 2 - y 2 < 0 or ( x - y )( x + y ) < 0. Since x, y > 0, we have that x + y 6 = 0, so

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}