solution manuals introduction to real analysis( Bartle and Sherbert)

SOLUTION CHAPTER 2.2 INTRODUCTION TO REAL ANALYSIS BARTLE AND SHERBERT Definition-: (Absolute value ) Absolute value of a real number ‘a’ is denoted by │a│ and is defined as -a

if a≤ 0

│a│= a if a ≥ 0 there are few properties of absolute value which can be proved very easily here I’ll just write them I won’t prove them you can ask me if you are not able to prove them you can mail me [email protected] Property 1

Property 3

│a│= max{ a, -a}

sqr(│a│) = sqr(a)=sqr(-a)

Property 2

Property 4

│a+ b│ ≤ │a│ + │b│

│a│ =

│-a│

Now we have some basic idea about absolute value, we can now solve the questions….

Geometrical meaning of│a│-: │a│ is distance of real number ‘a’ from 0. And │a- b│ is distance between ‘a’ and ‘b’ Q1(a)-: To show that Proof-: Case 1 when

│a│ =√ sqr(a)

a ≥ 0 we have on the L H S │a│= a  sqr( │a│) = sqr(a)  │a│ = √ sqr(a)

(taking square root)

Case 2 a≤ 0 in that case we have │a│= -a  sqr( │a│) = sqr(-a) = sqr(a) 

│a│ = √ sqr(a)

(taking square root)

Q1(b)-: │a/b│=│a│/│b│ Sqr (│a/b│) = sqr(a/b) = sqr(a)/sqr(b) = sqr( │a│)/ sqr( │b│) = sqr(│a│/│b│) => │a/b│ = │a│/│b│

(taking +ve square root)

Q2 -: │a+ b│ = │a│ + │b│ Iff ab ≥ 0 Proof-: First let us assume that │a+ b│ = │a│ + │b│ To show that-: ab ≥ 0 As │a+ b│ = │a│ + │b│  sqr(│a+ b│) = sqr(│a│ + │b│)  sqr(a+b) = sqr(│a│) + sqr(│b│) +2│a││b│  sqr(a)+sqr(b)+2ab= sqr(a)+sqr(b)+2│ab│

using facts that │a││b│=│ab│ and

sqr(│a│) = sqr(a)

now using cancelation laws we have that

2ab=2│ab│  ab=│ab│  ab ≥ 0 conversely assume that ab ≥ 0 to show that -: │a+ b│ = │a│ + │b│ consider L H S We have sqr(│a+ b│)=sqr(a+b)=sqr(a)+sqr(b)+2ab ( as sqr(│a│) = sqr(a) ) => sqr(│a+ b│)=sqr(│a│)+sqr(│b│)+2│ab│ ( as ab ≥ 0 therefore ab=│ab│) => sqr((│a+ b│)= sqr(│a│)+sqr(│b│)+2│a││b│  sqr(│a+ b│)= sqr(│a│ + │b│)  │a+ b│ = │a│ + │b│ (Taking +ve square root) Q3 Given x,y,z in R with x≤ z, show that │x-y│+│y-z│=│x-z│iff x≤ y≤ z proof-: firstly assume that x≤ y≤ z to show that -: │x-y│+│y-z│=│x-z│ consider L H S We have │x-y│= -x+y

( becoz x≤ y => │x-y│= -( x-y) )

Also

( becoz y≤ z =>│y-z│= -(y-z))

│y-z│= -y+z

Therefore │x-y│+│y-z│= -x+y -y+z = z- x = │z-x│ (“ ,“ x≤ z) => │x-y│+│y-z│=│x-z│

(“ ,“ │a│ =

│-a│)

Conversely assume that -:│x-y│+│y-z│=│x-z│……………(1) To show that -: x≤ y≤ z Now we have following possibilities Case1 x-y ≤ 0 and y-z ≤ 0 Then L H S of (1) becomes -x+y -y+z = z-x = │x-z│= R H S of (1) Therefore x-y ≤ 0 and y-z ≤ 0 is true => x ≤ y ≤ z Now we’ll claim no other possibility is there. Case2 x-y ≤ 0 and y-z ≥0 Then L H S of (1) becomes -x+y+y-z = 2y -x- z which is not equal to R H S of (1) Therefore this case is not possible Similarlly you can check that other two cases are not possible too. Q4 show that │x-a│