[PDF] [PDF] A → B and g : B → C are functions

[PDF] Homework  Solutions Math 3283W - Math User Home Pages

11 oct 2016 · On other problems the stated solution may be complete These are the only non-surjective functions (are you convinced?), so there are 8 - 2 = 6 (3) Classify each function as injective, surjective, bijective or none of these

[PDF] Homework &# 12 Solutions - Zimmer Web Pages

= {−5+4n : n ∈ N ∪ {0}} 3 Consider functions from Z to Z Give an example of ( a) a function that is injective but not surjective;

[PDF] MATH 310 - Solutions for Homework Assignment 4 Problem 16

That fact implies that /5(x) is injective Now /5(x) > 0 for all x G R and so /5(x) cannot be a surjective function from R to R Indeed, if x > 0, then y = /6(x) = x2> 0 and so g6(y) = Vx2= x for x > 0 If x < 0, then y = /6(x) = -x2< 0 and so g6(y) = - V-y = - Vx2= -ث xث = x for x < 0

[PDF] 2 Properties of Functions 21 Injections, Surjections - FSU Math

The examples illustrate functions that are injective, surjective, and bijective injection It is not a surjection because the range is not equal to the codomain For

[PDF] 7 Functions Definition 71 A function f : A −→ B consists of: (1) the

On the other hand, there is no person who is a million inches high, so f is not surjective either 1 Page 2 The function in (2) is neither injective nor surjective as 

[PDF] CSE 20 Homework 5 Solutions - UCSD CSE

a function is only injective, you must prove that it is injective and not surjective Similarly, if you is injective Proof that fA,B is not surjective for general A and B:

[PDF] A → B and g : B → C are functions

28 oct 2011 · (a) g is not injective but g ◦ f is injective (b) f is not surjective but g ◦ f is surjective Solution The same example works for both Let A = {1}, B 

[PDF] MATH 2200 SPRING 2018 1 Show that if f : A −→ B and g : B

5 Give an example of the following functions: (a) A function f : Z>0 −→ Z that is injective but not surjective

[PDF] Functions between Sets

The definition of an injection and a surjection gives us a framework for how to go about proving that a function is or is not injective or surjective We give examples  

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MATH 052: INTRODUCTION TO PROOFS

HOMEWORK #26

Problem 3.3.7. Supposef:A!Bandg:B!Care functions.

(a) Show that ifgfis injective thenfis injective. (b) Show that ifgfis surjective thengis surjective. Solution.First, we prove (a). Suppose thatgfis injective; we show thatfis injective. To this end, letx1;x22Aand suppose thatf(x1) =f(x2). Then (gf)(x1) =g(f(x1)) =g(f(x2)) = (gf)(x2). But sincegfis injective, this implies thatx1=x2. Thereforefis injective. Next, we prove (b). Suppose thatgfis surjective. Letz2C. Then sincegfis surjective, there existsx2Asuch that (gf)(x) =g(f(x)) =z. Therefore if we lety=f(x)2B, then g(y) =z. Thusgis surjective. Problem 3.3.8. In each part of the exercise, give examples of setsA;B;Cand functionsf:A!B andg:B!Csatisfying the indicated properties. (a)gis not injective butgfis injective. (b)fis not surjective butgfis surjective. Solution.The same example works for both. LetA=f1g,B=f1;2g,C=f1g, andf:A!B byf(1) = 1 andg:B!Cbyg(1) =g(2) = 1. Thengf:A!Cis dened by (gf)(1) = 1. This map is a bijection fromA=f1gtoC=f1g, so is injective and surjective. However,gis not injective, sinceg(1) =g(2) = 1, andfis not surjective, since 262f(A) =f1g. Problem 3.3.9. Dene functionsfandgfromZtoZsuch thatfis not surjective and yetgf is surjective.

Solution.Let

f:Z!Z n7!2n and g:Z!Z n7!( n=2;ifnis even;

0;ifnis odd:

The mapfis not surjective: the image is the set of even integers. However,gfis surjective, since (gf)(n) =g(f(n)) =g(2n)) = (2n)=2 =n|so in factgfis the identity map onZ.

Solution.Let

f:Z!Z x7!( x+ 5;ifx0; x;ifx <0:Date: 28 October 2011. 1 and g:Z!Z x7!( x5;ifx0; x;ifx <0: The mapfis not surjective: the elements 1;2;3;4 are not in the image. However,gfis surjective, since (gf)(x) =g(f(x)) =g(x+ 5) =xifx0 and (gf)(x) =f(x) =xifx <0|so in fact gfis the identity map onZ. 2quotesdbs_dbs6.pdfusesText_12