74hwsol4 - Solutions to Homework Assignment 4 Math 74, Fall...

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

View Full Document Right Arrow Icon
Math 74, Fall 2006 October 6, 2006 1. Definition 1. A natural number n is a perfect square if there exists an integer k such that n = k 2 . Theorem 2. If n N then n Q if and only if n is a perfect square. Proof. Clearly if n N is a perfect square, then n N Q . Suppose, on the other hand, that n N is not a perfect square. Define B = ± k N | k n Z ² . We will show that B is nonempty. Proceeding by way of contradiction, suppose that B is not empty. By the Well-Ordering Principle, B has a least element. Let b = min B . Then b n Z since b B . Now choose an integer m such that m < n m + 1 . By the Well-Ordering Principle, it is obvious that such an integer m exists, and that it is unique. Also, since n is not a perfect square, n 6 = m + 1, and so n < m + 1. Now define k = b ( n - m ) . Notice that k is an integer, since k = b n - bm is the difference of two integers. Also, our choice of b guarantees that 0 < n - m < 1. So we conclude that 0 < k < b. Since k n = bn - m ( b n ) is the difference of two integers, and is therefore an integer, we deduce that k B . But k < b contradicts the fact that k = min B . 2. (a)
Background image of page 1

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

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 01/12/2010 for the course CALC 2b taught by Professor Doolittle during the Winter '06 term at University of California, Berkeley.

Page1 / 5

74hwsol4 - Solutions to Homework Assignment 4 Math 74, Fall...

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