# Functions and Relations Number Systems N Z (German word Zahen),

```Functions and Relations
Number Systems
Natural Numbers N (1,2,3,-----)
Integer Numbers Z (German word Zahen),
(---,-1,0,1,2,3,---)
Rational Numbers Q ( Quotient, p/q, p and q are integers,
q≠0)
Number Systems
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Real numbers R (includes all
other numbers, decimal,
irrational numbers)
Complex numbers C
x and y are real and
i  1
x  iy
Functions and Relations
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Relations:
Let S and T be sets. A relation on S and T
is a subset of S&times;T. If R is a relation then
we write either (s,t) R, or sometimes
sRt to indicate that (s,t) is an element of
the relation. We also write s~t when the
relation being discussed is understood.
Relations
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Example:
Let S=N, T=R,
Define a relation
s  t  s 1
R by (s,t)ЄR if
For instance
(2,5)ЄR, (4,17)ЄR,
(5,10)R
Relations
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The Domain of a relation R where sЄS
such that there exists a t Є T with
(s,t)ЄR.
The Image of the relation R is a set of
tЄT, such that there exists a sЄS with
(s,t)ЄR. (sometimes called Codomain or
Range)
Relations
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Example:
Let S=N, and T=N. Define a
relation R on S and T by the
condition (s,t)ЄR if
Observe that, for any
element sЄS=N,
S=1 &raquo; 1‹ 2, s=2 &raquo; 4‹ 5
the number
satisfies
2
s t
t  s 1
2
s2  t
Relations
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Therefore every sЄS=N is in
the domain of relation.
Now let us think about the
image. The number 1ЄN=T
cannot be in the image since
there is no element sЄS=N
such that
2
s t
Relations
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However, any
element t Є T that
exceeds 1 satisfies
2
1 t
Thus the image of R
is set
t  N: t  2
Relations
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Example:
S=N, T=N, (s,t)ЄR if
Is itself a perfect
square.
For instance:
(3,4)ЄR, (4,3)ЄR,
(12,5)ЄR , (5,12)ЄR
s t
2
2
Relations
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Domain of R
We can see number
1=s Є S is not in the
domain of R, because
Is not a perfect square
Also number 2=s Є S is
not in the domain of R,
because
Is not a perfect square
The domain of R is a
set {s Є S : s›2}
1 t
2
2 t
2
2
2
Relations
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Image of R:
Also 1+2 are not in the Image of R
Then the image of R is a set
{t Є T : t ›2}
In fact both the domain and image of R
have infinitely many elements.
Relations
Definition:
If S=T=A, then S and T are the same set
R be a relation on a set A.
We say that R is an equivalence relation if
the following properties hold:
R is reflexive, symmetric, and transitive.
Relations
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R is reflexive : if xЄA then (x, x)ЄR
R is symmetric : if (x, y)ЄR then (y,x)ЄR
R is transitive : if (x, y)ЄR and (y, z)ЄR
then (x, z)ЄR,
Relations
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Equivalence Classes:
Let R be an equivalence relation on set
A=S=T. If xЄA then define
Ex={yЄA: (x, y)ЄR}
Or Es‫ת‬Et=&Oslash;, or Es=Et
Relations
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Example:
If (x , y)ЄR or x~y if x and y have at least
one biological parent.
We will check if R is an equivalence relation?
R is reflexive : if xЄA then (x, x)ЄR
R is symmetric : if (x, y)ЄR then (y, x)ЄR
R is not transitive : if (x, y)ЄR and (y, z)ЄR
then
( x , z)  R
Relations
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A*B&raquo;x
B * C &raquo; y y is related to x
C * D &raquo; z y is related to z, but
z is not related to x
Then R is not transitive also R is not
equivalence relation.
Relations
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Exercises:
1) Consider the relation on Z defined by (m,
n) ЄR if m+n is even. Prove that this is an
equivalence relation. What are the
equivalence classes?
If nЄZ then n+n=2n is even so (n,n)ЄR
Then R is reflexive
Relations
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If (m,n)ЄR, then m+n is even so n+m
is even hence (n,m)ЄR, then R is
symmetric.
If (m,n) ЄR, and (n,p)ЄR, then m+n=2r
is even and n+p=2s is even thus
m+n+p=2r+2s so m+p=2r+2s2n=2(r+s-n) is even then (m,p)ЄR
So R is transitive.
Relations
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Then R is an equivalence relation
The equivalence classes:
Are the set of even integers and the set
of odd integers.
Relations
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2) Consider the relation on Zx(Z\{0})
define by (m,n)R(m’,n’) provided that
m.n’=m’.n. Prove that this is an
equivalence relation. Can you describe
the equivalence classes?
If (m,n)Є Zx(Z\{0}) then m.n=m.n so
(m,n)R(m,n) then R is reflexive.
Relations
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If (m,n)R(m’,n’) then m.n’= m’.n so that
m’.n=m.n’, hence (m’,n’)R(m,n), so R is
symmetric.
If (m,n)R(m’,n’) and (m’,n’)R(m’’,n’’), then
m.n’=m’.n and m’.n’’=m’’.n’ hence
m.n’.m’.n’’=m’.n.m’’,n’
Cancelling m’.n’&raquo; m.n’’=m’’.n hence
(m,n)R(m’’,n’’) so R is Transitive.
Then R is an equivalence relation
Relations
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The equivalence classes:
Are ordered pairs (m,n) such that the
ratio of m to n represent the same
fraction. For instance (1,2), (3,6), and
(10,20) are in the same equivalence.
Relations
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3) Consider the relation defined on the
cartesian plane by (x,y)R(x’,y’) if y=y’.
Prove that this is an equivalence
relation. Can you describe the
equivalence classes? Can you pick a
representative (that is, an element) of
each equivalence class that will help to
exhibit what the equivalence relation is?
Relations
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If (x,y)ЄR.R then y=y so (x,y)R(x,y)
then R is reflexive.
If (x,y)R(x’,y’) then y=y’ so that y’=y
hence (x’,y’)R(x,y), then R is symmetric.
If (x,y)R(x’,y’) and (x’,y’)R(x’’,y’’) then
y=y’ and y’=y’’ so that y=y’’ hence
(x,y)R(x’’,y’’) then R is transitive.
So R is an Equivalence relation.
Relations
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The equivalence classes:
Are horizontal lines. A useful
representative for each equivalence
class is the point where the line crosses
the y axis.
Relations
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4) Let s be the set of all living people.
Let x,yЄS. Say that x is related to y if x
and y are siblings or the same person,
that is, x and y have both parents the
same. Prove that this is an equivalence
relation. What are the equivalence
classes?
Relations
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If a is a person then a is the same as a so
(aRa), then R is Reflexive.
If aRb then a and b are siblings(or the same
person) with the same parents. Hence b and
a are siblings (or the same person) with the
same parents. We conclude that bRa. Then R
is symmetric.
Relations
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If aRb and bRc then a and b are
siblings (or the same person) with the
same parents and b and c are siblings
(or the same person) with the same
parents. Thus a and c are siblings (or
the same person) then aRc. Then R is
transitive.
So R is an equivalence relation.
Relations
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The equivalence classes:
Are sets of siblings in the same family
with same parents.
Functions
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Definition:
Let S and T be sets. A function f from S to T
f: S
T where f is relation on Domain S
and Range T, such that:
1) Every sЄS is in the domain of f;
2) If (s,t)Єf and (s,u)Єf then t=u( each
element s of S is a ssociated to only one
member of T) its called (Rule or assign).
Functions
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Example:
Let S={1,2,3} and T={a,b,c} If
f= {(1,a),(2,a),(3,b)}
This function is a relation on S and T,because
f(1)=a, f(2)=a, f(3)=b, 1),
Every sЄS is in the domain of f.
Also notice that each element 1,2,3 of the
domain is “assinged” to one and only one
element of the range T.
Functions
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Example:
Let S={1,2,3} and T={a,b,c,d,e} If
F={(1,b),(2,a),(3,c),(4,a),(5,b)}
This is a function (relation), because
1) Every sЄS is in the domain of f;
2) If (s,t)Єf and (s,u)Єf then t=u.
However, not all elements of the range are
used. According to the definition of function,
this is allowed. Note that the range of f
is{a,b,c,d,e} while the image of f is {a,b,c}
Functions
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Definition:
Let f be a function with domain S and T.
We often write such a function as
f: S
T. We say that f is one to one
Or injective if whenever (s,t)Єf and
(s’,t)Єf then s=s’, We sometimes refer
to such a mapping as an injection. We
also refer to such a map as univalent.
Functions
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The conditions
1) Every sЄS is in the domain of f;
2) If (s,t)Єf and (s’,t)Єf then s=s’ (that
no two domain elements be associated
with the same range element).
Functions
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Example:
Let S=T=R and let f be of all
2
ordered pairs
( x, x
or
f :R  R
x  x2
or
as
f ( x)  x 2

): x  R
Functions
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f is not one-to-one function or injective
function because
The ordered pairs (-2,4) and (2,4) are
in f and f(-2)=f(2).
Example:
Let S=T=R and let f be the function
3
f ( x)  x
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Then f is strictly increasing as x moves
from left to right. In other words, if
s &lt; t then f(s) &lt; f(t). Hence f(s) ≠ f(t)
the function f is one-to-one or injective.
Functions
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Definition
Let f be a function with domain S and range
T. If for each tЄT there is an sЄS such that
f(s)=t then we say that f is onto or surjective.
We sometimes refer to such a mapping as a
surjection.
The function is onto or surjective when its
image equals its range.
Functions
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Example
Let
f ( x)
x
2
S=T=R. The point t=-1ЄT has the
property that there is no sЄS such
that f(s)=t. As a result, this
function f is not onto
Functions
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Example
Let S=R, T={xЄR:1≤x&lt;∞}. Let
g:S
T
2
g
(
x
)

x
1
Be given by
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Then for each tЄT the number
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g ( s)  t  s 2  1  t  s   t  1
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Makes sense and lies in S. then this function
g is surjective, (is not injective,
g(-1)=g(1)=2)
Combining Functions
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Definition:
Let f and g be
functions with same
domain S and the
same range T.
1)
2)
3)
4)
( f  g )( x )  f ( x )  g ( x )
( f  g )( x )  f ( x )  g ( x )
( f . g )( x )  f ( x ). g ( x )
( f / g )( x )  f ( x ) / g ( x ) provided that g ( x )  0
Combining Functions
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Example:
Let S=T=R. Define
f ( x)  x 3  x and
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g( x)  sin( x 2 )
Let us calculate f+g, f-g, f.g, f/g.
Now
Combining Functions
( f  g )( x)  ( x  x )  sin( x )
3
2
( f  g )( x)  ( x  x)  sin( x )
3
2
( f . g )( x)  ( x  x).[sin( x )]
3
2
x x
( f / g )( x) 
Pr ovided x   k k  {0,1,2,...)
2
sin( x )
3
Combining Functions
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Definition:
Let f: S
T be a function and let
g: T U be a function. Then we define,
for sЄS, the composite function
( g  f )( x )  g ( f ( x ))
We call g  f the composite functions g and f
Combining Functions
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Example:
Let
f : R  {x  R: x  0} be given by f ( x )  x 4  x 2  6
and g:{x  R: x  0}  R be given by g ( x )  x  4
( g  f )( s)  g ( f ( x ))  g ( x 4  x 2  6)  x 4  x 2  6  4
Also
( f  g )( s)  f ( g ( x ))  f ( x  4)  [ x  4]4  [ x  4]2  6
f  g and
g f
both make sense, will generally be different
Combining Functions
Let f : S  T function
and g: T  U function
g  f  {( s, u): s  S , u U , and t  T such that ( s, t )  f and (t , u)  g}
Combining Functions
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Definition:
Let S and T be sets. Let f: S
T. f and g
are mutually (common to both) inverse
provided that both
( f  g )( t )  t
( g  f )( s)  s
g  f
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1
for all t  T , and
for all s  S , we write
or f  g  1
We say f and g are invertible; we call g the
inverse of f and f the inverse of g
Combining Functions
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Example:
Let f: R R be given by
f ( s)  s 3  1
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And g: R
g (t )  3 t  1
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R be given by
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( f  g )(t )  [ 3 t  1]3  1  (t  1)  1  t , for all t
and
( g  f )( s) 
3
( s 3  1)  1 
Thus
g f
1
(or f  g 1 )
3
s 3  s, for all s
Combining Functions
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Example:
Let g : [1,+∞)
[8, +∞) be given
g( x)  x  3x  4
2
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Then g(1)=8, g(2)=14, g(3)=22,…
and g is strictly increasing without
bound.
Then g is one-to-one function
Combining Functions
If t  9  x 2  3x  4  9  x 2  3x  5  0
 3  9  20
 3  29
x1,2 
x
 119
.
2
2
If t  15  x 2  3x  4  15  x 2  3x  11  0
x1,2
 3  9  44
 3  53

x
 2.14
2
2
Combining Functions
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Then for each tЄT there is sЄS such
that f(s)=t then we say that f is onto
function.
We conclude that g has an inverse.
g ( x )  x 2  3x  4  t  x 2  3x  4  t  0
9  4( 4  t )
 3

2
 3   7  4t
x 
2
 3   7  4x
g 1 ( x ) 
2
x 
 3
9  16  4t
2
Combining Functions
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From last example we conclude that a
function that does have an inverse must
be one-to-one and onto function.
Types of Functions
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The polynomial function:
p( x)  a0  a1 x  a2 x  a3 x ...a k x
2
3
A polynomial function is easy to understand
because we can calculate its value at any
point x by simply plugging in x and then
k
Types of Functions
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The exponential function:
g( x)  2
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x
Is an exponential function. This
function is easy to calculate
when x is an integer. It is difficult
to calculate when x is a
noninteger.
Types of Functions
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The logarithmic function:
A logarithmic function is simply the
inverse of an exponential function.
h( x)  log 2 x is the inverse of g( x)  2
x
Functions and Relations
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Exercises:
5) Let S={a,b,c,d} and
T={1,2,3,4,5,6,7}. Which of the
following relations on SXT is a function?
Why?
a) {(a,4),(d,3),(c,3),(b,2)}
b) {(a,5),(c,4),(d,3)}
Functions and Relations
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6) Which of the following functions is
one-to-one? Which is onto?
a) f : N  N
f (m)  m  2
b) g: Z  Z
g(m)  2m  7
2
Functions and Relations
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8) Find the domain and image of
each of these relations:
a) ( x, y)  R  R: x 

y3
b)( ,  ): is a person,  is a person, and  is the father of 
Functions and Relations
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a) Domain ={xЄR:x≥0}
Image = {yЄR:y≥-3}
b) Domain = all people
Image =all male parents
Functions and Relations
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10) Discuss why there are infinitely many
triples of natural number k, l, m such that
k l  m
2
2
2
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