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│
a-b < x < a+b Proof-: │x-a│max{ -x+a , x-a}< b <=> -x+a < b and x-a < b <=> x> a – b and x < a+b <=> a-b < x < a+b SOLUTIONS PREPARED BY SANJEEV KUMAR SHUKLA MSc Mathematics(Delhi university) Contact number-9718445143
