, Numbers ,
Sets and Inequalities
Rational Numbers
1. NUMBERS -
r : Fn where m.net , n±o
Note : the decimal representation of a rational number is
Real Numbers IT I
:
always repeating eg
: =
0,50
R V3 : 2-3 =
0,6666 . . .
=
0,5
④
Integers
V5 ±
4
Zo
-3
"
v5
/
Irrational numbers IN Too
z
{
-
-
Cannot be expressed 3
- ,
Natural numbers 1,213.415.6 . . .
}
as a ratio of
integers 2 I
¥ rn
↳ %
no
2. SETS a collection of objects ,
called elements .
Ex S={ a. b. c. d } .
Here a ES ,
but e¢S .
0 denotes the set i. e. it has elements
-
empty ,
no
set operations
a) Union b) Intersection
S T S T S T
Metta took }
SUT :{ xlx c- Sor x ET } SAT -
-
{ xlxes and ✗ ET }
define
t
3. INTERVALS
Open intervals ( a. b) { xlacxcb } Look at table of intervals
}
=
a b
• •
page A- 4
Closed intervals [ a. b ] =
{ x1aExEb }
a b
Rules : ① If uab and bcc ,
than aac
② If ahb ,
than ate Cbtc
③ If acb and c >o than accbc
{④
,
*
If asb and cco ,
than ac > be
⑤ If aab and cod ,
than atccbtd
* ⑥ If osa< b. than ta > ¥
[ ¥ 'z ) [
'
Note that -215 c- Note that -4¢ lz ,
as )
,
THUS 13×+41-12×-11=1 <= >
✗ c-
{ -6 ,
-215 }
, C- ✗ Solve the
inequality 2<-13×-5146
First solve the
inequality separately :
13×-51<6 13.x -51 22
⇐> -
6 53.x -516 <= > 3×-5>-2 or 32C -5£ -
2
⇐ > -1 432C 911 ⇐ > 32-77 or 32 £3
⇐ > -
'
13 Soc <
"
13 ⇐ > x
I 73 or 2C
It
( " "
x c- 13) C- I -0,1 ] [ 73 )
-
>
⇐ 3 ,
⇐ > x u ,
as
2<-13×-51 < 6
1- 13 [1-0,1]
173,2b¥ Can you
'
do this ?
"
⇐ > x c-
,
B) n u
⇐ > x c- ( -
'
13, I ] u [ >
13 ,
' '
13)
• • 0
'
I ' '
13 713 13
-
Proof
if xzo
{
"
loci =
-
x if oc co txt : Fi
for all x (R)
→
C- ✗ :
prove that
the lock = xz
Proof real number either
:
Suppose >c is a , xzo or xco
case 1 : ✗ 20
we have lock = x2 ( since bcl=x )
cases :
x co
1×12 =
1 xp-
= x2 ( since bcl = -
x )
therefore 1×12=70 for a " "
☐
[ ☐ end of proof
The
Triangle Inequality
If a ,b c- 112 ,
then latbl E Iat t Ibl
Always
*
smaller or
equal to
PROOF lal Etat & Ibl Elbl
:
Suppose a. b c- 1112 :
Case If b have the same then the either
, a and
sign sum on side is equal .
Case 2 If a andb have opposite signs ,
then
-
late a stat ①
and - Ibl Eb E Ibl ②
Adding ① ② that
and we
get
-
( lait lbl ) I atb E Iat t lbl -
c < x Cc
Put atb and
/
✗ "
y
=
=
then c = Iat t Ibl
lxtscifff.ro
that ← ✗ ,
teased
so - c ⇐
m
⑤ bet <C
lxl=ciHy#
and ④ loci =L iff x=±c
we have lxl Ec x = ± c
l
THUS latbl E Iat t Ibl
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