Functions. Copyright © Peter Cappello2 Definition Let D and C be nonempty sets. A function f from D to C, for each element d D, assigns exactly 1 element

Functions

Copyright © Peter Cappello 2

Definition

• Let D and C be nonempty sets.

• A function f from D to C, for each element d D, assigns exactly 1 element c C, denoted f( d ) = c.d D c C ( f( d ) = c c’ D ( f( d ) = c’ c = c’ ) )

d D !c C f( d ) = c .

• If f is a function from D to C, we write f : D C.

• Functions are also known as:Mappings

Transformations.

• Functions pass the vertical line test.


Definition

• A function is a subset of a Cartesian product:

If f : D C then f D x C.

• If f : D C then:

D is f’s domain

C is f’s codomain.

• If f( d ) = c then:

c is the image of d

d is a pre-image of c.

• f’s range is { c | d f( d ) = c }.


Example

• Let f : Z N be f( x ) = x2.

• What is f’s domain?

• What is f’s codomain?

• What is the image of 4?

• What is the pre-image of 4?

• What is f’s range?


When are functions equal?

Let f1: D C and

f2: D C.

Since

– A function is a subset of a Cartesian product.

– A Cartesian product is a set.

when does f1 = f2 ?


Declaring a function’s domain & codomain

The Java statement

long square( int x ) { … }

The domain of square is?

Its codomain is?


• Let f : D C and S D.

• The image of S under f, denoted f( S ) is

{ c | s S, f( s ) = c }.

• If S is finite, is | S | <, =, or > | f( S ) | ?

• Let f : N N , f( n ) = n mod 5.

– What is f’s range?

– Let O = { n N | n is odd } .

– What is f( O ) ?


One-to-One (Injective) Functions

• Let f : D C.

• f is one-to-one (injective) when different domain elements

have different images:

a D b D ( a b f (a ) f( b ) ).

• Example

– Let n: { T, F } { T, F }, such that n( p ) = p.

– Is n injective?

• Is f : Z Z, f( z ) = z2 injective?

• Injective functions pass the horizontal line test.


Onto (Surjective) Functions• Let f : D C.

f is onto (surjective) when f’s range equals its

codomain:

c C d D ( f( d ) = c ).

• Example

– Let or : { T, F } { T, F } { T, F }, such that

or( p, q ) = p q.

– Is or surjective?

• Is f : Z Z, f( z ) = z2 surjective?

• Is f : Z Z, f( z ) = z mod 5 surjective?


One-to-One Correspondence (Bijection)

• Function f is a one-to-one correspondence (bijection)

when it is both:

– one-to-one (injective)

– onto (surjective).• Let f : R R, f( x ) = 2x – 7. Is f a bijection?• Let f : D C be a bijection, where D, C are finite.

– Can |D| > |C|?

– Can |D| < |C|?


Inverse Functions• Let g : D C be a bijection.

• The inverse function of g, denoted g-1, is the

function : C D such that if g( d ) = c, then g-1( c ) = d.

• If g is bijective, g-1 is a function because g is:

– onto: c C ( c is the image of some element in D )

– 1-to-1: c C (c is the image of at most 1 element in D )

– Diagram this.

• If g : D C is not a bijection, does g-1 exist?

Always? Sometimes? Never?


Composition of Functions

• Let functions g : B C and f : A B.

• The composition of g and f, denoted g f,

is defined by g f( a ) = g( f( a ) ).

a g(f ( a ))f( a )

A B C


Example

• Let f : Q Q, f( x ) = 2x + 1.

• Let g : Q Q, g( x ) = (x – 1)/2.

• What is (g f )( 17 )?

• In general, what is g-1 g ( x ) ?


Exercise

Let S U. The characteristic function

fS : U { 0, 1 } is such that

x S fS ( x ) = 1

x S fS ( x ) = 0.

Show that:

f A B (x) = fA( x )fB( x )

f A B (x) = fA( x ) + fB( x ) - f A B ( x )

1

23 4A B

Copyright © Peter Cappello 2011 15

End of Lecture


• The Java statementlong square( int x ) { … }

square’s domain is int; its codomain is long.

• Let f & g be functions from A to R.(f + g)( x ) = f( x ) + g( x ),

( fg )( x ) = f( x )g( x ).

• Let f( x ) = x2 and g( x ) = x – x2.

• What is f + g? gf?


Graphs of Functions

• Let f : A B.

• The graph of f = { (a, b) | a A and f( a ) = b }.

• Example: Let the domain of f be N. Draw:

f( x ) = x2

f( x ) = x mod 2.

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Functions. Copyright © Peter Cappello2 Definition Let D and C be nonempty sets. A function f from D to C, for each element d D, assigns exactly 1 element