Ang aking kontrata: Ako, si , ay nangangakong magsisipag mag

Ang aking kontrata: Ako, si , ay nangangakong magsisipag mag

Ang aking kontrata:
Ako, si ______________, ay
nangangakong magsisipag mag-aral
hindi lang para sa aking sarili kundi
para rin sa aking pamilya, para sa
aking bayang Pilipinas at para sa
ikauunlad ng mundo.
Rings and Fields
Ring
Given a non-empty set S and two operation,
say * and on S , the mathematical system
S ,*,  is a ring if
1. S ,* is an abelian group.
2. S is closed under .
3.
4. a
(ᴑ is also a binary operation on S.)
is associative in S.
b * c    a b  *  a c  , for any a,b, c  S.
Ring
EXAMPLE:
* and
can be the
usual addition and
multiplication.
e.g. [ R, ,]
Commutative Ring
If the second operation
is commutative
in the set S , then S ,*,  is called a
commutative ring.
Verify that  , , is a commutative ring.
Example
Is the mathematical system Q, ,
a commutative ring?
Is Q closed in +?
Is + associative in Q ?
What is the identity element under +?
Are there additive inverses?
Is + commutative in Q ?
Hence, Q ,   is an Abelian group.
Q ,   is an Abelian group.
Is Q closed under  ?
Is  associative in Q ?
Is  distributive over +?
Is  commutative in Q ?
Thus, Q , , is a commutative ring.
QUESTIONS:
Is there a multiplicative identity in Q ?
Is there a multiplicative inverse for each
nonzero element of Q ?
Properties of [Q,+,×]
Q , , is a commutative ring.
There is a multiplicative identity in Q.
There is a multiplicative inverse for each
nonzero element of Q.
[Q,+,×] is a FIELD!
Field
A mathematical system F , ,  satisfying
the following conditions is a field.
1. F , ,  is a commutative ring.
2. There is an identity element for .
3. Every nonzero elements of F has
an inverse under .
Field
Q , , is the smallest field contained in R.
R, , is a field.
Summary
Closed Under
Addition
Addition is
Associative
Group
Abelian Group
Closed Under
Multiplication
Field
Commutative
Ring
Multiplication
is Associative
Multiplicative
Identity
Distributivity
Multiplicative
Inverses
Multiplication
is
Commutative
Addition is
Commutative
Additive
Identity
Additive
Inverses
Note
RINGS
FIELDS
Some Theorems
Theorems
Theorem 1.
The Cancellation Law for Addition
For a,b, c  R, if a  c  b  c, then a  b.
The proof of the first theorem will be
provided in the next slide.
The proofs for the other theorems are left
as exercises. Try proving them at home!
If a  c  b  c, then a  b.
Proof of Theorem 1.
ac bc
Given
 a  c    c   b  c    c 
a   c   c    b   c   c  
APE
Associativity
a0 b0
Existence of + inverses
a b
Existence of + identity
Theorems
Theorem 2. Zero Property
For a  R, a  0  0.
Theorem 3.
For any a,b  R,  a   b    a  b 
Corollary 4.
For any a  R,  1  a  a
Theorems
Corollary 5.
 1   1  1
Theorem 6.
For any a,b  R,  a    b   a  b
Theorem 7.
For any a,b  R,   a  b    a    b 
Theorems
Theorem 8.
For any a,b  R, there is a unique
solution to the equation a  x  b.
Remark: The unique solution for a  x  b
is x  b   a  .
Theorems
Theorem 9. The Cancellation Law
for Multiplication
For any a,b  R and any nonzero number c,
if a  c  b  c, then a  b.
Try to prove this theorem at home!
Theorems
Theorem 10.
If a,b  R such that a  b  0, then
either a  0 or b  0.
Remark:
This theorem can be used when solving
equations of the form  x  a  x  b   0.
Theorems
Theorem 11.
If a,b are nonzero real numbers,
1
 1  1 
   
a b  a   b 
Theorems
Theorem 12.
For any number b and any nonzero
number a, there is a unique solution
to the equation a  x  b.
Remark:
The solution for the equation a  x  b ?
1
x  b
a
Theorems
b
Theorem 13. For any number b,  b.
1
Theorem 14. If c is any nonzero real
c
number then  1.
c
 b   d  bd
Theorem 15.      
, a  0, c  0.
 a   c  ac
Theorems
Theorem 16. If c is any nonzero real
a c a
number,
 .
bc b
b d
Theorem 17. If  then b  d and
a a
b d
if  then b  c  a  d,
a c
a  0, c  0.
Theorems
b
1
a
Theorem 18. If  0, then
 .
b
a
b
a
d
b
da
c
d
a
Theorem 19. If  0, then



.
c
b c b
b
a
a
Theorems
b
b b
b b
Theorem 20.
   
and
 , a  0.
a
a a
 a  a
b d bd
Theorem 21.  
, a0
a a
a
b d bc  a d
Theorem 22.  
, a  0, c  0
a c
a c
WARNING!!!
b d bd
 
a c ac
Remark:
Division by 0 is undefined.
Why?
We know that ax  b if and only if x 
b
.
a
b
If a  0 then x  which means (0)x  b.
0
Case 1: Suppose b  0. Then there is no value
for b that satisfies (0)x  b since (0)x  0.
Case 2: Suppose b  0. Then (0)x  0 is true
0
but x can be any real number. So x  is indeterminate.
0
Examples
Perform the indicated operations.
1.
2 5
25
10
52
5





3 12 3  12 36 18  2 18
4
2.
3.
4.
20
5
4  25 100


1
20  5 100
25
2 3
23 5


 1
5 5
5
5
2 4 2  5  3  4 22
7
 

or 1
3 5
35
15
15
The Geometry of R
1 Trichotomy Axiom
For any a,b  R, one and only
one of the following holds:
a b
a b
ba
2 Transitivity Axiom of
Order
If a,b, c  R and if a  b and b  c,
then a  c.
3 Addition Property of
Inequality (API)
If a,b, c  R and if a  b then a  c  b  c.
4 Multiplication
Property of Inequality
(MPI)
If a,b, c  R and if a  b and 0  c
then a c  b c.
Since the set of real
numbers can be ordered,
that is, the 4 order axioms
are satisfied, then we say
R is an ordered field.
Other Notations
a  b : a  b or a  b
a  b : a  b or a  b
a  x  b : a  x and x  b
a  x  b : a  x and x  b
a  x  b : a  x and x  b
a  x  b : a  x and x  b
Example
Determine if the following statements
are ALWAYS TRUE or ALWAYS FALSE.
Justify your answers.
1.
3
If 2x  3, then x 
2
2
TRUE
2.
aa
3.
If a and b are positive and a  b
then a 2  b2 .
FALSE. Why?
TRUE
Theorems
Theorem 23.
The set of positive real numbers
is closed under addition and
multiplication.
If a  0 and b  0, then a  b  0
and a b  0.
Theorems
Theorem 24.
If a  0, then  a  0 and
if a  0, then  a  0
Theorem 25. If a  b, then  a  b.
2
Theorem 26. If a  R, either a  0
2
or a  0.
Theorems
Theorem 27. 1  0.
Theorem 28. If a,b, c  R and a  b
and c  0, then ac  bc.
Remark:
If a  b then a(0)  b(0) (a(0)  b(0) is acceptable).
Theorems
Theorem 29.
1
If a  0, then  0
a
Theorem 30.
1 1
If 0  a  b, then 0   .
b a
Theorems
Theorem 31.
a b
If a  b, then a 
 b.
2
a b
(
is the arithmetic mean of a and b)
2
Theorem 32.
If a  b, then a  ab  b.
( ab is the geometric mean of a and b)
Example
Solve the following inequalities.
1.
3x  6
1
1
 3x    6 
3
3
1
1 
 3 3  x  3 6


1
1 x  6
3
1
x  6
3
x 2
Example
Solve the following inequalities.
2.
3.
4.
5.
8  5  3x  11
2
x 0
2
x 0
2
x  1
RECALL: There is a one-to-one correspondence
between the set of real numbers and the set of
points on a line.
We choose an arbitrary point to correspond to the
number 0.
To the right of 0, we place the positive real
numbers while the negative real numbers are
placed to the left of 0.
Definition. If x1 and x2 are points on
the real number line, the distance
d(x1, x2) between them, is given by
2

(
x

x
)
d(x1, x2)
2
1
Example
If the points A and B have the given
coordinates respectively, find the distance
between them :
1. 3, 7
2 4
2. ,
3 5
3.  2,
5
Definition. Absolute Value of a Number
The absolute value of a number x is defined as
| x |
 x , if x > 0

2
x   x , if x < 0
 0, if x = 0

With this definition, it follows that the distance
d(x1, x2) = |x2 – x1|=|x1 – x2| is always non-negative
as expected.
Theorem 33. For any a  R,
-|a|  a  |a|.
Theorem 34. For any a, b  R,
|ab| = |a||b| and |a/b| = |a|/|b|, b ≠ 0
Theorem 35. The Triangle Inequality
For any a, b  R, |a + b|  |a| + |b|.
SETS IN INTERVAL NOTATION
Open Intervals
The solution set to the inequality
will be written as
This is the open interval from a to b.
a
b
( a , b)  { x |a  x  b }
SETS IN INTERVAL NOTATION
Closed Intervals
The solution set to the inequality
will be denoted as
This is the closed interval from a to b.
a
b
[ a , b]  { x |a  x  b }
SETS IN INTERVAL NOTATION
Other Notations
[ a , b)
(b, a ]
a
b
( , a )
( , a ]
( a,  )
a
[ a,  )
( ,  )  R
Remember that these intervals are SETS!
Example
Write the solution set to the following
inequalities as intervals and identify
them on a number line.
1. a  x  b
2. a  x  b
5. x  R
3. x  b
4. x  a
7. x   1, 2   2, 1
8. x   0, 3   1, 4.5 
6. x  2 or x  2
Example
Give a geometric interpretation of the
following statements and find the solution :
1. x  5
2. x – 2  3
3. x  3
4. x  3
5. x – 1  2
5,5
1,5
 3,3 
 ,3   3,  
 1,3 
Solution :
Upper Bound
Definition
The number u is called an upper
bound of a set S if x  u, for all
x  S.
Example:
Give upper bounds for the following sets.
1.
7,3,1,2,9,0
Upper bounds: 10,11,12.5,9
Real Numbers  9.
2.
 1 1 
1, , ,...
 2 3 
Upper bounds: Real Numbers  1
Example
3.
 1
5
x  x  
2
 2
5
Real Numbers 
2
Upper bounds:
4.
x x  1 or x  1
Upper bounds:
5.

1
y y  N,  y 
2

Upper bounds:
No upper bound
9

2
Real Numbers  4
Lower Bound
Definition
The number v is called a lower
bound of a set S if x  v, for all
x  S.
Example
Give lower bounds for the following sets.
1.
7,3,1,2,9,0
Lower bounds: 0, 0.5, 100
Numbers  0.
2.
 1 1 
1, , ,...
 2 3 
Lower bounds: 0, 0.5, 100
Numbers  0.
Example
3.
 1
5
x  x  
2
 2
1
Numbers 
2
Lower bounds:
4.
x x  1 or x  1
Lower bounds:
5.

1
y y  N,  y 
2

Lower bounds:
No lower bound
9

2
Numbers  1
Least Upper Bound
(lub)
Definition
An upper bound b of a set S is the
least upper bound if no upper bound
is less than b.
Greatest Lower Bound
(glb)
Definition
A lower bound a of a set S is the
greatest lower bound if no lower
bound is greater than a.
Example
Identify the lub and glb of the following.
1.
2.
7,3,1,2,9,0
lub : 9
glb : 0
 1 1 
1, , ,...
 2 3 
lub : 1
glb :
0
Example
3.
4.
 1
5
x  x  
2
 2
5
lub:
2
glb :
none
glb :
1
x x  1 or x  1
lub:
5.
glb :
1
2
none

1
y y  N,  y 
2

lub:
4
9

2
Completeness Axiom
Every subset S of R that has an
upper bound has an lub in R.
Similarly, every subset S of R that
has a lower bound has a glb in R.
To illustrate the use of the Axiom, consider the set
S = {x| x  0, x is rational, and x2 < 2}.
Finding GLB: This set has 0 as its greatest lower bound.
Finding LUB: It is bounded above. (Show this!)
Hence, by the Completeness Axiom, there must be a least
upper bound.
The least upper bound is _____ , which is not an
element of the set.
THEREFORE, S in not complete!
Time to think!
Using the Completeness Axiom, show that
Z+, the set of all positive integers has no
upper bound.
The Completeness Axiom
guarantees that our real number line
has no gaps in it,
that is,
between any two real number, there is
another real number. (The density
property is a consequence of the
Completeness Axiom)
Complete Ordered
Field
R, ,  is a complete ordered field.