Math 250a Problem Set 1 Solutions

# Math 250a Problem Set 1 Solutions - Math 250 Fall 2003...

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

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.

Unformatted text preview: Math 250, Fall 2003 Selected Solutions, Assignment # 1 HH 1.5.5 d) Is the set Q rational numbers open or closed in R , the real numbers? Response; The set of rational numbers Q is neither open nor closed in the real numbers R . Rather, both Q and R- Q are dense in R- any number can be approximated arbitrarily closely by rational numbers, and by irrational numbers. To see that the rational numbers are dense, fix a denominator n . The fractions with denominator n are of the form k n for any integer k . They are evenly spaced, with a separation of distance 1 n between adjacent numbers. For a fixed denominator n , these numbers are not dense. However, every number is within distance 1 n (actually, 1 2 n- why?) of a fraction with denominator 1 n .* Since we may let n get arbitrarily large, we can approximate any number arbitrarily closely by rational numbers; that is, the rational numbers are dense in R . (A common way to approximate irrational numbers by rational numbers is by decimal expansions - every number has an infinite decimal expansion, and the finite parts of this are rational numbers. A given number lies within 10- k of its decimal expansion truncated at the k-th decimal place. The decimal fractions with k places to the right of the decimal point are exactly those numbers which can be written as a fraction with denominator 10- k .) To see that the irrational numbers are dense, observe that, if φ is an irrational number, then rφ is irrational for all rational numbers r . (Why?) Therefore, the same argument as above shows that any number can be approximated as closely as one wishes by numbers rφ . A fortiori , irrational numbers are dense. HH 1.5.19 : For a given angle θ , consider the sequence A n = cos nθ sin nθ- sin nθ cos nθ . For what values of θ does this converge? For what values of θ can you find a convergent subsequence? Response: The matrix A n produces a rotation of the plane through an angle- nθ . Thus, we may think of A n as being a point a n on the unit circle in R 2 , such that the angle between a n and the first standard basis vector e 1 is- nθ . (This point a n is given by the first column of A n , and the second column is just the rotation of a n through -90 ◦ . Clearly, a sequence of rotation matrices converges if and only if their first columns converge in the unit circle. Thus, we will discuss the behavior of the points a n . Of course, in all these considerations, the value of θ only matters modulo 2 π . That is, if we add a multiple of 2 π to θ , the matrix A n and the point a n do not change....
View Full Document

{[ snackBarMessage ]}

### Page1 / 3

Math 250a Problem Set 1 Solutions - Math 250 Fall 2003...

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

View Full Document
Ask a homework question - tutors are online