Module - 3
Relations and Functions : Cartesian Products and
Relations: Computer Recognition – Zero-One Matrices
and
Directed
Graphs,
Properties
of
Relations,
Equivalence Relations and Partitions, Partial Orders –
Hasse Diagrams.
Functions – Plain and One-to-One, Onto Functions. The
Pigeon-hole
Principle,
Inverse Functions.
Function
Composition
and
Functions
›
Let 𝐴 and 𝐵 be non empty sets. Then function (or mapping) from 𝐴 to 𝐵
such that for each 𝑎 in 𝐴, there exists unique 𝑏 in 𝐵 such that 𝑎, 𝑏 ∈ 𝑓.
Then we write 𝑏 = 𝑓(𝑎). Here 𝑏 is called image of 𝑎 and 𝑎 is called
preimage of 𝑏 under 𝑓.
›
A function from 𝐴 to 𝐵 is denoted by 𝑓: 𝐴 → 𝐵. The pictorial
representation of 𝑓 is as below
𝑓
𝑓
Is a Function
is a relation not a function
𝑓
›
Every function is relation but every relation is not a function.
∵ If 𝑅 is a relation from 𝐴 to 𝐵 then an element 𝐴 can be related to two
elements of 𝐵 under 𝑅.
But under function, an element of 𝑎 can be related only one element of 𝐵
Functions Cont..
𝑓
𝑓
›
For the function 𝑓: 𝐴 → 𝐵, 𝐴 is called domain of 𝑓 and 𝐵 is called codomain of 𝑓.
›
The subset of 𝐵 consisting of the images of all elements of 𝐴 under 𝑓 is
called range of 𝒇 and denoted by 𝑓(𝐴).
Note: 1. Every 𝑎 in 𝐴 belong to some pair 𝑎, 𝑏 ∈ 𝑓 and if 𝑎, 𝑏1 ∈ 𝑓 and 𝑎, 𝑏2 ∈ 𝑓
then 𝑏1 = 𝑏2.
2.
An element 𝑏 ∈ 𝐵 need not have a preimage in 𝐴, under 𝑓.
Functions Cont…
Note: 3. Two different elements of 𝐴 can have same image in 𝐵 under 𝑓.
4.
The statements 𝑎, 𝑏 ∈ 𝑓 if 𝑎𝑓𝑏 and 𝑏 = 𝑓(𝑎) are equivalent.
5.
If 𝑔 is a function from 𝐴 to 𝐵, then 𝑓 = 𝑔, iff 𝑓 𝑎 = 𝑔 𝑎 , ∀𝑎 ∈ 𝐴
6.
The range of 𝑓: 𝐴 → 𝐵 is
𝑓 𝐴 𝑖𝑠 𝑠𝑢𝑏𝑠𝑒𝑡 𝑜𝑓 𝐵.
7.
For 𝑓: 𝐴 → 𝐵, if 𝐴1 ⊆ 𝐴 and 𝑓 𝐴1 is defined by 𝑓 𝐴1 = {𝑓(𝑥)|𝑥 ∈ 𝐴1 )} then
𝑓(𝐴1 ) ⊆ 𝑓(𝐴). Here, 𝑓(𝐴1 ) is called the image of 𝐴1 under 𝑓.
8.
For 𝑓: 𝐴 → 𝐵, if 𝑏 ∈ 𝐵 and 𝑓 −1 𝑏 is defined by𝑓 −1 𝑏 = 𝑥 ∈ 𝐴 𝑓 𝑥 = 𝑏}
then𝑓 −1 𝑏 ⊆ 𝐴). Here, 𝑓 −1 𝑏 is called the preimage of 𝑏 under 𝑓.
9.
For 𝑓: 𝐴 → 𝐵, if 𝐵1 ⊆ 𝐵 and 𝑓 −1 (𝐵1 ) is defined by𝑓 −1 (𝐵1 ) = 𝑥 ∈ 𝐴 𝑓 𝑥 =
(𝐵1 } then𝑓 −1 𝑏 ⊆ 𝐴). Here, 𝑓 −1 𝑏 is called the preimage of 𝐵1 under 𝑓.
given
by
𝑓 𝐴 = 𝑓 𝑥
𝑥∈𝐴 }
and
Types of Functions
1.
Identity Function:- A function on set 𝐴 is an identity function, if the image
of every element of 𝐴(under f) is itself and is denoted by IA.
i.e. 𝑓: 𝐴 → 𝐴 is such that 𝑓(𝑎) = 𝑎, ∀𝑎 ∈ 𝐴. In case of identity function,
𝑓(𝐴) = 𝐴.
2.
Constant Function:- A function 𝑓: 𝐴 → 𝐵 is called constant function, if
𝑓 𝑎 = 𝑐, ∀𝑎 ∈ 𝐴.
i.e. 𝑓 is constant function if the image of every element of 𝐴 is same in B.
In this case 𝑓 𝐴 = {𝑐}.
3.
Onto Function (Surjective):- A function 𝑓: 𝐴 → 𝐵 is said to be Onto
function from 𝐴 to 𝐵, if every element 𝑏 of 𝐵 there exists an element 𝑎 of
𝐴 such that 𝑓(𝑎) = 𝑏.
i.e. 𝑓 is Onto function from 𝐴 to 𝐵, if every element 𝑏 of 𝐵 has a preimage
in 𝐴. (no elements in 𝐵 should be free).
Types of Functions
4.
One-to-One Function (Injective):- A function 𝑓: 𝐴 → 𝐵 is said to be One-toOne function from 𝐴 to 𝐵, if different elements of 𝐴 have different images
in 𝐵 under 𝑓.
i.e., If 𝑓(𝑎1) = 𝑓(𝑎2) then 𝑎1 = 𝑎2, where 𝑎1, 𝑎2 ∈ 𝐴
(Or) If 𝑎1 ≠ 𝑎2, then 𝑓(𝑎1) ≠ 𝑓(𝑎2)
5.
One-to-One Correspondence (Bijective):- A function 𝑓: 𝐴 → 𝐵 is said to be
One-to-One correspondence, if 𝒇 is both one-one ant onto function from 𝐴
to 𝐵.
A function 𝑓: 𝐴 → 𝐵 is a bijective function then every element of 𝐴 has
unique image in 𝐵 and every element of 𝐵 has a unique preimage in 𝐴.
Types of Functions
Note:
› If 𝑓: 𝐴 → 𝐵, |𝐴| = 𝑚, |𝐵| = 𝑛, then there are 𝒏𝒎 functions from 𝐴 to 𝐵
𝒏!
and if 𝑚 ≤ 𝑛, then there are
one-to-one functions from A to B.
𝒏−𝒎!
›
Let 𝐴 and 𝐵 be finite sets with |𝐴| = 𝑚, |𝐵| = 𝑛, where 𝑚 ≥ 𝑛, then
number of onto functions from 𝐴 to 𝐵 is given by the formula
𝒏
𝒐𝒏𝒕𝒐 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏𝒔 = ෍ −𝟏
𝒌 𝒏𝑪
𝒏−𝒌
(𝒏 − 𝒌)𝒎
𝒌=𝟎
›
If 𝑚 < 𝑛, then there are no onto functions from 𝐴 to 𝐵.
Problems on Functions
1.
Let A = {1, 2, 3, 4}, determine whether the following relations on A are
functions.
i.
f = {(2,3) (1,4)(2,1)(3,2)(4,4)}
ii. g = {(2,1) (3,4) (1,4)(2,1) (4,4)
i.
We see that 𝟐, 𝟑 ∈ 𝒇 and 𝟐, 𝟏 ∈ 𝒇, that is element 2 is related to
two different elements 3 and 1, under 𝑓.
∴ 𝒇 is not a function
ii.
We see that under 𝑔, every element of 𝐴 is related unique element in
𝐴
∴ 𝒈 is a function from 𝑨 → 𝑨
Range of g is 𝑔 (𝐴) = { 1, 4}
(2,1) appears twice. This has no special significance.
Problems on Functions
2.
Let A = {0, ±𝟏, ±𝟐, ±𝟑}, consider 𝒇: 𝑨 → 𝑹 (where R is the set of real
numbers) defined by 𝒇 𝒙 = 𝒙𝟑 − 𝟐𝒙𝟐 + 𝟑𝒙 + 𝟏, find the range of 𝒇.
𝒇 𝟎 =𝟏
𝒇 𝟏 = 𝟏𝟑 − 𝟐 𝟏
𝒇 𝟐 =𝟕
𝟐
+𝟑 𝟏 + 𝟏 = 𝟑
𝒇 −𝟐 = −𝟐𝟏
𝒇(−𝟑) = −𝟓𝟑
∴ Range of 𝒇 = −𝟓𝟑, −𝟐𝟏, −𝟓, 𝟏, 𝟑, 𝟕, 𝟏𝟗
𝒇 −𝟏 = −𝟓
𝒇 𝟑 = 𝟏𝟗
Problems on Functions
3.
Let 𝑨 = {𝟏, 𝟐, 𝟑, 𝟒, 𝟓, 𝟔} and 𝑩 = {𝟔, 𝟕, 𝟖, 𝟗, 𝟏𝟎}. If a function 𝒇: 𝑨 → 𝑩 is
defined by 𝒇 = { 𝟏, 𝟕) 𝟐, 𝟕 𝟑, 𝟖 𝟒, 𝟔 𝟓, 𝟗 𝟔, 𝟗 . Determine 𝒇−𝟏 (𝟔) and
𝒇−𝟏 (𝟗). If 𝑩𝟏 = {𝟕, 𝟖} and 𝑩𝟐 = {𝟖, 𝟗, 𝟏𝟎}. Find 𝒇−𝟏 (𝑩𝟏) and 𝒇−𝟏 (𝑩𝟐).
By definition: 𝒇−𝟏 𝒃 = {𝒙 ∈ 𝑨 | 𝒇(𝒙) = 𝒃}
∴ 𝒇−𝟏 𝟔 = {𝒙 ∈ 𝑨 | 𝒇(𝒙) = 𝟔} = {4}
∴ 𝒇−𝟏 𝟗 = {𝒙 ∈ 𝑨 | 𝒇(𝒙) = 𝟗} = {5, 6}
I𝐟 𝑩𝟏 ⊆ 𝑩 𝒃𝒚 𝒅𝒆𝒇𝒊𝒏𝒊𝒕𝒊𝒐𝒏 𝒇−𝟏 (𝑩𝟏 ) = {𝒙 ∈ 𝑨 | 𝒇 𝒙 = 𝑩𝟏 }
For 𝑩𝟏 = 𝟕, 𝟖 , 𝒇 𝒙 = 𝑩𝟏 , when 𝒇 𝒙 = 𝟕 and 𝒇 𝒙 = 𝟖
From the definition of f, we can say that 𝒇(𝒙) = 𝟕 when 𝒙 = 𝟏 and 𝒙 = 𝟐
And 𝒇(𝒙) = 𝟖 when 𝒙 = 𝟑.
∴ 𝒇−𝟏 𝑩𝟏 = {1, 2, 3}
Similarly 𝒇−𝟏 (𝑩𝟏 ) = {𝒙 ∈ 𝑨 | 𝒇(𝒙) ∈ 𝑩𝟐 } = {𝟑, 𝟓, 𝟔}
4.
Problems on Functions
Assume that A = {1, 2, 3…,14}. Define a relation R from A to A by R = {(x, y):
3x – y = 0, such that x, y ∈ A}. write down its range, domain, and codomain.
3x – y = 0
y = 3x
𝑋
𝑌 = 3𝑋
𝑊ℎ𝑒𝑡ℎ𝑒𝑟 𝑥, 𝑦 ∈ 𝑅
1
3
Yes
2
6
Yes
3
9
Yes
4
12
Yes
5
15
𝑁𝑜
6
18
𝑁𝑜
∴ 𝑅 = { 1,3 2,6 3,9 4,12 }
Domain of R = set of all first elements of the ordered pairs = { 1, 2, 3, 4}
Range of R = set of all second elements of the ordered pairs = {3, 6, 9 ,12}
Co domain of R = A = {1, 2, 3, . . . . . 14}
6.
Problems on Functions
The function 𝒇: 𝒁 × 𝒁 is defined by 𝒇 𝒙, 𝒚 = 𝟐𝒙 + 𝟑𝒚. Verify that 𝒇 is 𝒐𝒏𝒕𝒐
+
but not 𝒐𝒏𝒆 − 𝒐𝒏𝒆. 𝐈𝐟 𝐀 = 𝟎, 𝐧 𝐧 ∈ 𝒁 }, 𝒇𝒊𝒏𝒅 𝒇(𝑨).
𝐓𝐚𝐤𝐞 𝐚𝐧𝐲 𝒏 ∈ 𝒁, we note that
𝒏 = 𝟒𝒏 – 𝟑𝒏 = 𝟐(𝟐𝒏) + 𝟑 (−𝒏) = 𝒇(𝟐𝒏, −𝒏)
Thus, every 𝒏 ∈ 𝒁, has a preimage 𝒇(𝟐𝒏, −𝒏) ∈ 𝒁 under f. Therefore, 𝒇 is onto
function.
Next we check that 𝒇 𝟎, 𝟐 = 𝟐 × 𝟎 + 𝟑 × 𝟐 = 𝟔
and 𝒇 𝟑, 𝟎 = 𝟐 × 𝟑 + 𝟑 × 𝟎 = 𝟔
Thus, 𝒇(𝟎, 𝟐) = 𝒇(𝟑, 𝟎), but (𝟎, 𝟐) ≠ (𝟑, 𝟎). Therefore, f is not one-to-one.
For any 𝟎, 𝒏 = 𝟐 ∙ 𝟎 + 𝟑𝒏 = 𝟑𝒏
+
Therefore 𝐟 𝐀 = 𝟑𝐧 𝐧 ∈ 𝒁 } = {𝟑, 𝟔, 𝟗, 𝟏𝟐 . . . . . }
5.
Problems on Functions
Let 𝑨 = 𝟏, 𝟐, 𝟑, 𝟒 𝐚𝐧𝐝 𝑩 = {1, 2, 3, 4, 5, 6}
i.
Find how many functions are there from A to B. How many of these are 11 ? How many are onto?
ii. Find how many function B to A? How many of these are 1-1 ? How many
are onto?
Here |A| = m = 4 and |B| = n = 6
i.
The number of possible functions from A to B is 𝒏𝒎 = 𝟔𝟒 = 𝟏𝟐𝟗𝟔
The number of 1-1 function possible from A to B is
𝒏!
𝟔!
𝟔!
=
=
𝒏−𝒎!
𝟔−𝟒!
𝟐!
= 𝟑𝟔𝟎
There is no onto functions from A to B (∵ 𝒎 < 𝒏)
ii.
The number of possible functions from B to A is 𝒎𝒏 = 𝟒𝟔 = 𝟒𝟎𝟗𝟔
There is no one-one functions from A to B (∵ 𝒎 > 𝒏)
The number of onto functions from B to A
P(6,4) = σ𝟒𝒌=𝟎 −𝟏 𝒌 𝟒𝑪𝟒−𝒌 (𝟒 − 𝒌)𝟔 =𝟒𝟔 − 𝟒 × 𝟑𝟔 + 𝟔 × 𝟐𝟔 − 𝟒 =1560
Problems on Functions
7.
𝟑𝒙 − 𝟓
Let 𝒇: 𝑹 → 𝑹 defined by ቊ
−𝟑𝒙 + 𝟏
i.
Determine 𝐟 𝟎 , 𝐟 −𝟏 , 𝐟
𝒇𝒐𝒓
𝒇𝒐𝒓
𝒙>𝟎
𝒙≤𝟎
𝟓
𝟑
Find 𝒇−𝟏 𝟎 , 𝒇−𝟏 𝟏 , 𝒇−𝟏 𝟑 , 𝒇−𝟏 −𝟑 , 𝒇−𝟏 −𝟔
iii. What are 𝒇−𝟏 [−𝟓, 𝟓], and 𝒇−𝟏 −𝟔, 𝟓 ?
ii.
i.
› 𝒇 𝟎 = −𝟑 × 𝟎 + 𝟏 = 𝟏
› 𝒇 −𝟏 = −𝟑 × −𝟏 + 𝟏 = 𝟒
𝟓
𝟓
› 𝒇
= 𝟑× −𝟓=𝟎
𝟑
ii.
›
𝟑
ii.
𝒇−𝟏 𝒃 = 𝒙 ∈ 𝑹 𝒇 𝒙 = 𝒃}
𝒇−𝟏 𝟎 = 𝒙 ∈ 𝑹 𝒇 𝒙 = 𝟎}
𝟑𝒙 – 𝟓 = 𝟎
›
𝒙>𝟎
›
›
𝒙 =
𝟓
𝟑
Thus 𝒇−𝟏 𝟎 =
possible
𝟓
𝒇−𝟏 𝒃 = 𝒙 ∈ 𝑹 𝒇 𝒙 = 𝒃
𝒇−𝟏 𝟎 = 𝒙 ∈ 𝑹 𝒇 𝒙 = 𝟎}
−𝟑𝒙 + 𝟏 = 𝟎
𝟏
𝟑
𝒙≤𝟎
𝒙 = > 𝟎 not possible
Problems on Functions
7.
Let 𝒇: 𝑹 → 𝑹 defined by ቊ
𝟑𝒙 − 𝟓
−𝟑𝒙 + 𝟏
𝒇𝒐𝒓
𝒇𝒐𝒓
𝟓
𝟑
i.
Determine 𝐟 𝟎 , 𝐟 −𝟏 , 𝐟
ii.
Find 𝒇−𝟏 𝟎 , 𝒇−𝟏 𝟏 , 𝒇−𝟏 𝟑 , 𝒇−𝟏 −𝟑
What are 𝒇−𝟏 [−𝟓, 𝟓], and 𝒇−𝟏 −𝟔, 𝟓 ?
iii.
ii.
𝒙>𝟎
𝒙≤𝟎
𝒇−𝟏 𝟏 = 𝒙 ∈ 𝑹 𝒇 𝒙 = 𝟏} = 𝟎, 𝟐
𝟖 −𝟐
𝒇−𝟏 𝟑 = 𝒙 ∈ 𝑹 𝒇 𝒙 = 𝟑} = 𝟑 , 𝟑
𝒇−𝟏 −𝟑 = 𝒙 ∈ 𝑹 𝒇 𝒙 = −𝟑} =
𝟐
𝟑
iii. 𝒇−𝟏 [−𝟓, 𝟓] = 𝒙 ∈ 𝑹 𝒇 𝒙 ∈ [−𝟓, 𝟓]}
iii. 𝒇−𝟏 [−𝟓, 𝟓] = 𝒙 ∈ 𝑹 𝒇 𝒙 ∈ [−𝟓, 𝟓]}
›
›
›
›
›
›
›
›
›
›
›
›
𝐟𝐨𝐫 𝐟 𝐱 = 𝟑𝐱 − 𝟓 𝐱 > 𝟎
−𝟓 ≤ 𝟑𝐱 − 𝟓 ≤ 𝟓
Add 5 throughout
0 ≤ 𝟑𝐱 ≤ 𝟏𝟎
÷ by 3
𝟏𝟎
0 ≤ 𝐱≤ 𝟑
𝟒 𝟏𝟎
∴ 𝒇−𝟏 [−𝟓, 𝟓] = − ,
𝟑 𝟑
›
𝐟𝐨𝐫 𝐟 𝐱 = −𝟑𝐱 + 𝟏, 𝐱 ≤ 𝟎
−𝟓 ≤ −𝟑𝐱 + 𝟏 ≤ 𝟓
Add -1 throughout
−𝟔 ≤ −𝟑𝐱 ≤ 𝟒
÷ by 3
𝟒
−𝟐 ≤ −𝐱 ≤ 𝟑
× 𝒃𝒚 − 𝟏
𝟒
𝟒
𝟐 ≥ 𝐱 ≥ − 𝟑 i.e. , − 𝟑 ≤ 𝐱 ≤ 𝟐
Problems on Functions
8.
𝒙 + 𝟕 𝒊𝒇 𝒙 ≤ 𝟎
A function f: R → R is defined as follows 𝒇 𝒙 = −𝟐𝒙 + 𝟓 𝒊𝒇 𝟎 < 𝒙 < 𝟑
𝒙 − 𝟏 𝒊𝒇 𝒙 ≥ 𝟑
find the following 𝒊 𝒇−𝟏 𝟒
9.
𝒊𝒊 𝒇−𝟏 𝟖
𝒊𝒊𝒊 𝒇−𝟏 −𝟓, −𝟏
The functions 𝒇:𝑹→𝑹 and 𝒈:𝑹→𝑹 are defined by 𝒇(𝒙) = 𝟑𝒙 + 𝟕 for all
𝒙 ∈ 𝑹, and g 𝒙 = 𝒙(𝒙𝟑 − 𝟏) for all 𝒙 ∈ 𝑹. Verify that f is 𝒐𝒏𝒆 − 𝒕𝒐 −
𝒐𝒏𝒆 𝐛𝐮𝐭 𝒈 is not.
10. In each of the following cases A function 𝒇: 𝑹 → 𝑹 is given. Determine
whether 𝒇 is one-to-one or onto. If f is not 𝒐𝒏𝒕𝒐, find its range.
i. 𝒇 𝒙 = 𝟐𝒙 − 𝟑
ii. 𝒇(𝒙) = 𝒙𝟑
iii. 𝒇 𝒙 = 𝒙𝟐 + 𝒙
11. Let 𝑨 = {𝟏, 𝟐, 𝟑, 𝟒, 𝟓, 𝟔, 𝟕} and 𝑩 = {𝒘, 𝒙, 𝒚, 𝒛}. Find the number of onto
functions from 𝑨 𝒕𝒐 𝑩.
Problems on Functions
11. Let 𝑨 = {𝟏, 𝟐, 𝟑} and 𝑩 = {𝟏, 𝟐, 𝟑, 𝟒, 𝟓}. Find whether the following
functions from 𝑨 𝒕𝒐 𝑩 are (𝒂) 𝒐𝒏𝒆 − 𝒕𝒐 − 𝒐𝒏𝒆 (𝒃) 𝒐𝒏𝒕𝒐
i. 𝒇 = { 𝟏, 𝟏 , 𝟐, 𝟑 , 𝟑, 𝟒 }
ii. 𝒈 = {(𝟏, 𝟏), (𝟐, 𝟐), (𝟑, 𝟑)}
12. Let 𝑨 = 𝑹 and 𝑩 = 𝒙 𝒙 𝒊𝒔 𝒓𝒆𝒂𝒍 𝒂𝒏𝒅 𝒙 ≥ 𝟎 }. Is the function 𝒇: 𝑨 → 𝑩
defined by 𝒇(𝒂) = 𝒂𝟐 an onto function ? A one-to-one function?
Composition Functions
Composition Function:- Consider 3 non empty sets 𝐴, 𝐵, 𝐶 and the function
𝑓: 𝐴 → 𝐵 and 𝑔: 𝐵 → 𝐶. The composition (product) of these two functions are
defined as 𝑔 ◦ 𝑓: 𝐴 → 𝐶.
𝒈◦𝒇 𝒂 =𝒈 𝒇 𝒂
∀𝒂 ∈ 𝑨
𝑔◦𝑓
𝑎
𝐴
𝑓
𝑓(𝑎)
𝑔
𝐶
𝐵
Note: For 𝒇: 𝑨 → 𝑨, 𝒇𝟏 = 𝒇,
𝑻𝒉𝒖𝒔, 𝒇𝟏 = 𝒇
𝑔(𝑓(𝑎))
𝒇𝟐 = 𝒇 ◦ 𝒇,
𝒇𝒏 = 𝒇 ◦ 𝒇𝒏−𝟏
𝒇𝟑 = 𝒇 ◦ 𝒇𝟐
Problems on Composition Functions
Note: Composition Function need not be commutative 𝒇 ◦ 𝒈 ≠ 𝒈 ◦ 𝒇
1. Let 𝒇: 𝑨 → 𝑩, 𝒈: 𝑩 → 𝑪 𝒂𝒏𝒅 𝒉: 𝑪 → 𝑫 be functions. Then
𝒉 ◦ 𝒈 ◦ 𝒇 = 𝒉 ◦ (𝒈 ◦ 𝒇)
Solution:
› Since 𝑓: 𝐴 → 𝐵, 𝑔: 𝐵 → 𝐶 𝑎𝑛𝑑 ℎ: 𝐶 → 𝐷, We have both
ℎ ◦ 𝑔 ◦ 𝑓 and ℎ ◦ (𝑔 ◦ 𝑓) are functions from A to D.
›
For any x ∈ 𝐴,
ℎ ◦ 𝑔 ◦ 𝑓 𝑥 = ℎ ◦ 𝑔 𝑓(𝑥) = ℎ ◦ 𝑔 𝑦 where 𝑦 = 𝑓 𝑥
= ℎ◦𝑔 𝑦 =ℎ 𝑔 𝑦
= ℎ(𝑧) where 𝑧 = 𝑔 𝑦
(1)
[ℎ ◦ (𝑔 ◦ 𝑓)](x) = ℎ (𝑔 ◦ 𝑓)(x) = ℎ 𝑔(𝑓 𝑥 ) = ℎ 𝑔 𝑦
›
𝒉 ◦ 𝒈 ◦ 𝒇 𝒙 = 𝒉 ◦ 𝒈 ◦ 𝒇 𝒙 for every x ∈ A
›
𝒉◦𝒈 ◦𝒇 = 𝒉◦ 𝒈◦𝒇
= ℎ(𝑧)
⇒ Composition of three fn is associative
(2)
Problems on Composition Functions
2.
Consider the functions 𝒇 and 𝒈 defined by 𝒇 𝒙 = 𝒙𝟑 𝐚𝐧𝐝 𝒈 𝒙 = 𝒙𝟐 +
𝟏 ∀𝒙 ∈ 𝑹, 𝐅𝐢𝐧𝐝 𝒈 ◦ 𝒇 , 𝒇 ◦ 𝒈 , 𝒇𝟐 𝒂𝒏𝒅 𝒈𝟐.
Solution:
› 𝒈◦𝒇 𝒙 = 𝒈 𝒇 𝒙
= 𝒈 𝒙𝟑 = 𝒙𝟑
›
𝒇◦𝒈 𝒙 = 𝒇 𝒈 𝒙
= 𝒇 𝒙𝟐 + 𝟏 = 𝒙𝟐 + 𝟏
›
𝒇◦𝒇 𝒙 = 𝒇 𝒇 𝒙
= 𝒇 𝒙𝟑 = 𝒙𝟑
›
𝒈◦𝒈 𝒙 =𝒈 𝒈 𝒙
𝟑
𝟐
+ 𝟏 = 𝒙𝟔 + 𝟏 .
𝟑
= 𝐱𝟔 + 𝟏 + 𝟑𝐱𝟒 + 𝟑𝐱𝟐
= 𝒙𝟗.
= 𝒈 𝒙𝟐 + 𝟏 = 𝒙𝟐 + 𝟏
𝟐
+ 𝟏 = 𝒙𝟒 + 𝟐 + 𝟐𝒙𝟐.
Problems on Composition Functions
3. Let 𝒇, 𝒈, 𝒉 be functions from 𝒁 × 𝒁 defined by 𝒇 𝒙 = 𝒙 – 𝟏, 𝒈 𝒙 =
𝟎 𝒊𝒇 𝒙 𝒊𝒔 𝒆𝒗𝒆𝒏
𝟑𝒙, and 𝒉 𝒙 =
.
𝟏 𝒊𝒇 𝒙 𝒊𝒔 𝒐𝒅𝒅
𝐃𝐞𝐭𝐞𝐫𝐦𝐢𝐧𝐞 𝐟 ◦ 𝒈 ◦ 𝒉 𝒙 𝐚𝐧𝐝 𝒇 ◦ 𝒈 ◦ 𝒉 𝒙 𝐚𝐧𝐝 𝐯𝐞𝐫𝐢𝐟𝐲 𝐭𝐡𝐚𝐭 [𝐟
◦ 𝒈◦𝒉 ] = 𝒇◦𝒈 ◦𝒉
Solution:
› We have 𝒈 ◦ 𝒉 𝒙 = 𝒈 𝒉 𝒙
›
›
›
›
›
𝐟 ◦ 𝒈 ◦ 𝒉 𝒙 = 𝒇 𝟑𝒉 𝒙
= 𝟑𝒉 𝒙 − 𝟏 =
We have 𝒇 ◦ 𝒈 𝒙 = 𝒇 𝒈 𝒙
𝒇◦𝒈 ◦𝒉 𝒙= 𝒇◦𝒈 𝒉 𝒙
From (1) and (2)
𝐟◦ 𝒈◦𝒉 =
= 𝟑𝒉 𝒙
𝒇◦𝒈 ◦𝒉
−𝟏 𝒊𝒇 𝒙 𝒊𝒔 𝒆𝒗𝒆𝒏
.
𝟐 𝒊𝒇 𝒙 𝒊𝒔 𝒐𝒅𝒅
(1)
= 𝒇 𝟑𝒙 = 𝟑𝒙 − 𝟏
−𝟏 𝒊𝒇 𝒙 𝒊𝒔 𝒆𝒗𝒆𝒏
= 𝟑𝒉 𝒙 − 𝟏 =
. (2)
𝟐 𝒊𝒇 𝒙 𝒊𝒔 𝒐𝒅𝒅
Invertible Functions
Invertible Function:- A function 𝑓: 𝐴 → 𝐵 is said to be invertible function. If
there exists a function 𝑔: 𝐵 → 𝐴. Such that 𝑔 ◦ 𝑓 = 𝐼𝐴 and 𝑓 ◦ 𝑔 = 𝐼𝐵 , where
𝐼𝐴 is the identity function on 𝐴 and 𝐼𝐵 is the identity function on 𝐵. then 𝑔 is
called inverse of 𝑓 and is denoted by 𝑔 = 𝑓 −1 .
𝑓
𝑏
𝑎
𝑔 = 𝑓 −1
𝐴
𝐵
Note:› If 𝑓 is invertible function, then 𝑓 𝑎 = 𝑏 and 𝑎 = 𝑓 −1 𝑏 are equivalent.
› A function 𝑓: 𝐴 → 𝐵 is said to be invertible function iff it is ono-to-one and
onto.
› A function 𝑓: 𝐴 → 𝐵 and 𝑔: 𝐵 → 𝐴 are invertible functions then 𝑔 ◦ 𝑓 is
also invertible and (𝑔 ◦ 𝑓)−1 = 𝑓 −1 ◦ 𝑔−1
Problems on Invertible Functions
1. A function 𝑓: 𝐴→𝐵 is invertible iff it is one-to-one and onto
Solution:› Let 𝑓: 𝐴 → 𝐵 be a invertible function then there exist a unique function
𝑔: 𝐵 → 𝐴 such that 𝑔 ◦ 𝑓 = 𝐼𝐴 and 𝑓 ◦ 𝑔 = 𝐼𝐵 .
›
›
›
›
›
›
›
TPT, 𝑓 𝑖𝑠 𝑜𝑛𝑒 − 𝑜𝑛𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
Let 𝑓 𝑎1 = 𝑓 𝑎2
𝑔 𝑓 𝑎1 = 𝑔 𝑓 𝑎2
𝑔 ◦ 𝑓 𝑎1 = 𝑔 ◦ 𝑓 𝑎2
𝐼𝐴 𝑎1 = 𝐼𝐴 𝑎2
𝑎1 = 𝑎2
∴ 𝒇 𝒊𝒔 𝒐𝒏𝒆 − 𝒐𝒏𝒆 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏
›
›
›
›
›
TPT, 𝑓 𝑖𝑠 𝑜𝑛𝑡𝑜 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
Take any 𝑏 ∈ 𝐵, then 𝑔(𝑏) ∈ 𝐴
and 𝑏 = 𝐼𝐵 𝑏 = 𝑓 ◦ 𝑔 = 𝑓{𝑔 𝑏 }
Thus, 𝑏 is the image of an element 𝑔(𝑏) ∈ 𝐴 under 𝑓
∴ 𝒇 𝒊𝒔 𝒐𝒏𝒕𝒐 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏
Problems on Invertible Functions
1. A function 𝑓: 𝐴→𝐵 is invertible iff it is one-to-one and onto
Solution:› Conversely, suppose 𝑓 is 1 − 1 and 𝑜𝑛𝑡𝑜, ∴ 𝑖𝑡 𝑖𝑠 𝑏𝑖𝑗𝑒𝑐𝑡𝑖𝑣𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
›
∴ for every 𝑏 ∈ 𝐵 there exists a unique a ∈ 𝐴 such that 𝑓(𝑎) = 𝑏
›
Consider function 𝑔: 𝐵 → 𝐴 defined by 𝑔(𝑏) = 𝑎
›
Then 𝑓 ◦ 𝑔
›
›
𝑏 =𝑓 𝑔 𝑏
𝑔 ◦ 𝑓 (𝑎) = 𝑔 𝑓 𝑎
= 𝑓 𝑎 = 𝑏 = 𝐼𝐵 (𝑏)
= 𝑔(𝑏) = 𝑎 = 𝐼𝐴 (𝑎)
∴ 𝒇 𝒊𝒔 𝒊𝒏𝒗𝒆𝒓𝒕𝒊𝒃𝒍𝒆 𝒘𝒊𝒕𝒉 𝒈 𝒉𝒂𝒔 𝒕𝒉𝒆 𝒊𝒏𝒗𝒆𝒓𝒔𝒆
Problems on Invertible Functions
2.
If 𝒇: 𝑨 → 𝑩 and 𝒈: 𝑩 → 𝑪 are invertible functions, then prove that 𝒈 ◦ 𝒇 is
also invertible and (𝒈 ◦ 𝒇)−𝟏 = 𝒇−𝟏 ◦ 𝒈−𝟏
Solution:› Suppose 𝑓 and 𝑔 are invertible
› ∴ 𝑓 and 𝑔 are 1-1 and onto function
› 𝑔 ◦ 𝑓 is also1−1 and onto function
› 𝒈 ◦ 𝒇 𝐢𝐬 invertible
›
›
›
Next we have 𝑓: 𝐴 → 𝐵 and 𝑔: 𝐵 → 𝐶 and 𝑔 ◦ 𝑓 ∶ A → C
∴ 𝑓 −1 : 𝐵 → 𝐴, 𝑔−1 : 𝐶 → 𝐵 and 𝑙𝑒𝑡 ℎ = 𝑓 −1 ◦ 𝑔−1 : 𝐶 → 𝐴
Consider ℎ ◦ 𝑔 ◦ 𝑓 = (𝑓 −1 ◦ 𝑔−1 ) ◦ 𝑔 ◦ 𝑓 = 𝑓 −1 ◦ (𝑔−1 ◦ 𝑔) ◦ 𝑓
= 𝑓 −1 ◦ 𝐼𝐵 ◦ 𝑓 = 𝑓 −1 ◦ 𝑓 = 𝐼𝐴
(1)
›
Consider (𝑔 ◦ 𝑓) ◦ ℎ = 𝑔 ◦ 𝑓 ◦ (𝑓 −1 ◦ 𝑔−1 ) = 𝑔 ◦ (𝑓 −1 ◦ 𝑓) ◦ 𝑔
= 𝑔−1 ◦ 𝐼𝐴 ◦ 𝑔 = 𝑔−1 ◦ 𝑔 = 𝐼𝐵
(2)
›
›
From (1) and (2) h is the inverse of (𝑔 ◦ 𝑓)
i.e. ℎ = (𝑔 ◦ 𝑓) −1
›
Thus, (𝑔 ◦ 𝑓) −1 = 𝑓 −1 ◦ 𝑔−1
Problems on Invertible Functions
3.
Let A = B = C = R, and 𝒇: 𝑨 → 𝑩 and 𝒈: 𝑩 → 𝑪 be defined by 𝒇 𝒂 = 𝟐𝒂 +
𝟏
𝟏 𝐚𝐧𝐝 𝒈 𝒃 = 𝒃, ∀𝒂 ∈ 𝑨, ∀𝒃 ∈ 𝑩.
𝟑
Compute 𝒈 ◦ 𝒇 and show that 𝒈 ◦ 𝒇 is invertible. Find (𝒈 ◦ 𝒇)−𝟏
Pigeon Hole Principle
›
Statement: If ′𝑚′ pigeons occupy ′𝑛′ pigeon holes and if 𝑚 > 𝑛 then two
or more pigeons occupy the same pigeon hole.
OR
› If ′𝑚′ pigeons occupy ′𝑛′ pigeon holes and if 𝑚 > 𝑛 then atleast one
pigeon hole must contain two or more pigeons in it.
Generalized Pigeon hole principles
› If ′𝑚′ pigeons occupy ′𝑛′ pigeon holes and if 𝑚 > 𝑛 then at least one
pigeon hole must contain 𝑝 + 1 or more pigeons in it.
›
Where 𝑝 =
𝑚−1
𝑛
Problems on Pigeon Hole Principle
1.
ABC is a equilateral triangle whose sides are of length 1cm each. If we
select 5 points inside triangle . Prove that at least 2 of these points are
such that the distance between them is less than ½ cm.
Solution:A
•
•
D
B
E
F
C
Consider Equilateral triangle ABC, whose sides are
of length 1cm each.
Consider the triangle DEF formed from the
midpoints of the sides AB, AC and BC respectively.
•
DEF Triangle divides triangle ABC into 4 small
equilateral triangle whose length of each side is
1/2cm.
•
Let us treat 5 points as pigeons and 4 small
equilateral triangles as pigeon holes.
•
Then by pigeon hole principle, at least one small
triangle contains 2 or more points and distance
between such points is less than 1/2cm.
Problems on Pigeon Hole Principle
2.
A bag contains 12 pair of socks (each pair in different color). If a person
draws the socks one by one at random. Determine at most how many
draws are required to get at least one pair of matched socks.
Solution:› Let ‘n’ be the number of draws
›
For 𝐧 ≤ 𝟏𝟐, it is possible that socks drawn are at different . ∵ there are 12
colors
›
For 𝒏 = 𝟏𝟑, all socks cannot have different colors –atleast two must have
same colors.
›
Let us treat 13 as the number of pigeons and 12 colors as 12 pigeon holes.
›
∴ at most 13 draws are required to have at least 1 pair of socks of the same
color.
Problems on Pigeon Hole Principle
3.
If 5 colors are used to paint 26 doors. Prove that atleast 6 doors will have
same color.
Solution:› Consider 26 doors as pigeons and 5 colors as pigeon holes
›
By generalized Pigeon hole principle, at least 1 color must be assigned to
p+1 or more pigeons
›
i.e. 𝑝 + 1 =
›
∴ at least 6 doors will have same color
𝑚−1
𝑛
+1=
26−1
5
+ 1 =5+ 1 = 6
Problems on Pigeon Hole Principle
4.
Prove that if 30 dictionaries in a library contain total of 61,327 pages, the
at least one of the dictionaries must have at least 2045 pages.
Solution:› Consider 61327 pages as pigeons, i.e. m = 61327
›
and 30 dictionaries as pigeon holes n = 30
›
By generalized Pigeon hole principle, at least 1 dictionary must contain
p+1 or more pages
›
i.e. 𝑝 + 1 =
›
𝑇ℎ𝑖𝑠 𝑝𝑟𝑜𝑣𝑒𝑠 𝑡ℎ𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑟𝑒𝑠𝑢𝑙𝑡.
𝑚−1
𝑛
+1=
61327−1
30
+ 1 = 2045
Problems on Pigeon Hole Principle
5.
How many persons must have chosen on order that atleast five of them
will have birthdays in the same calendar of month.
›
Let 𝑚 be the number of persons. Number of months over which the
birthdays are distributed is n =12.
›
The least number of persons having birthday in the same month is
›
5=𝑝+1
›
𝑝+1=5⇒
›
𝑚 − 1 = 48 ⇒ 49
›
∴ number of persons in 49.
𝑚−1
𝑛
+1=5⇒
𝑚−1
12
=4