sol-01 - MATH 453 SOLUTIONS TO ASSIGNMENT 1 SEPTEMBER 8,...

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

View Full Document Right Arrow Icon
M  453 S   A  1 S  8, 2004 Exercise 4 from Section 3, page 28 (a) Reflexivity, symmetry and transitivity for follow from the corresponding properties for equality, so is an equivalence relation. For instance, to check symmetry, we proceed as ( a 0 a 1 ) ( f ( a 0 ) = f ( a 1 )) ( f ( a 1 ) = f ( a 0 )) ( a 1 a 0 ). (b) Let A * = { [ a ] | a A } be the equivalence classes under . The map f : A B induces a map f * : A * B given by f * ([ a ]) = f ( a ). We will check that f * is a bijection. Given b B , there is an a A such that f ( a ) = b since f is surjective. Hence f * ([ a ]) = b and so f * is surjective. For injectivity, suppose f * ([ a 0 ]) = f * ([ a 1 ]). By the definition of f * we get f ( a 0 ) = f ( a 1 ), or a 0 a 1 . Hence [ a 0 ] = [ a 1 ] and f * is injective. ± Exercise 5 from Section 3, page 28 (a) Again we have to check the three conditions for an equivalence relation. For reflex- ivity, note that x - x = 0, so ( x , x ) S 0 . If y - x is an integer, then x - y = - ( y - x ) is too, and symmetry follows. Finally, if y - x and z - y are integers, then so is z - x = ( z - y ) + ( y - x ). Hence ( x , z ) S 0 and we have transitivity. If ( x , y ) S , then y = x + 1, or y - x = 1, an integer. Hence ( x , y ) S 0 . This shows that S 0 S . Given x R , the equivalence class [ x ] consists of all y R such that y - x Z . In other words, y is of the form x + n where n Z . Denoting the set { x + n | n Z } by x + Z , we have [ x ] = x + Z . Furthermore, given any x R , there is an x 0 [0 , 1) and an integer n such that x = x 0 + n . Hence [ x ] = [ x 0 ] for some x 0 [0 , 1). Also, no two distinct elements of [0 , 1) are S 0 -related, so R / S 0 = { x + Z | x [0 , 1) } . ± Exercise 5 from Section 5, page 39 Before looking at the specific subsets in this problem, let us consider an arbitrary subset A = { x | x satisfies some condition P } . Suppose P does not mix the indices, i.e. it is of the form “for all i , x i satisfies some condition C ( i ) that depends only on i ”. This means that the i th coordinate x i can only take values from the subsets A i of
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.

Page1 / 4

sol-01 - MATH 453 SOLUTIONS TO ASSIGNMENT 1 SEPTEMBER 8,...

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