Industry-relevant training in Business, Technology, and Design to help professionals and graduates upskill for real-world careers.
Fun, engaging games to boost memory, math fluency, typing speed, and English skills—perfect for learners of all ages.
Enroll to start learning
You’ve not yet enrolled in this course. Please enroll for free to listen to audio lessons, classroom podcasts and take practice test.
Listen to a student-teacher conversation explaining the topic in a relatable way.
Today we're going to discuss the different types of functions, specifically focusing on surjective, injective, and bijective functions. Can anyone tell me what a surjective function is?
Is it when every element in the codomain has at least one pre-image?
Exactly! Think of surjective functions as 'onto' functions. Now, how about injective functions?
Isn’t that when distinct elements map to distinct elements?
Correct! Now, if a function is both injective and surjective, what do we call it?
A bijective function!
Right, remember the acronym 'ISB' - Injective, Surjective, Bijective. Let's summarize: surjective means 'onto,' injective means 'one-to-one,' and bijective captures both properties.
Now let’s shift our focus to equivalence relations. Can anyone explain how an equivalence relation partitions a set?
It divides the set into disjoint subsets where each element relates to others in the same subset.
Exactly! Given a set with 30 elements partitioned into 3 equal subsets, how many ordered pairs can you form?
I think it would be 10 squared for each subset, so 300 in total.
Yes! Great job! This shows how equivalence relations are useful in counting and organizing elements.
Next, let's discuss Stirling numbers, particularly the second kind. Can anyone tell me what they measure?
They help us find the number of ways to partition a set into non-empty subsets!
Excellent! If we have a set X with 'r' elements and we want to partition it into 's' subsets, we're looking at the Stirling function S(r, s).
How can we use that to find surjective functions?
Great question! Each surjective function corresponds to a unique partition, and you're right! The total number of surjective functions can be calculated as S(r, s) multiplied by the count of permutations of the subsets. Remember to think of the sets as distinct.
Read a summary of the section's main ideas. Choose from Basic, Medium, or Detailed.
The section delves into functions with special properties, discussing the distinction between surjective and bijective functions and providing examples of equivalence relations. It introduces Stirling numbers, their significance in combinatorial mathematics, and how they relate to counting specific types of functions.
This section of Discrete Mathematics explores several core concepts essential for understanding function theory and combinatorial mathematics. The definitions and characteristics of surjective, injective, and bijective functions are outlined, emphasizing their distinctions:
Through examples, the distinction is made clear, especially regarding functions defined over finite versus infinite sets.
Another significant topic covered is equivalence relations, where a set is partitioned into subsets of equal size, emphasizing their relevance in set theory and functions. It discusses the number of ordered pairs in such equivalence relations.
The concept of Stirling numbers, particularly of the second kind, is introduced as a way to partition a set into non-empty subsets, forming a critical component in combinatorial applications. By understanding how to count the number of surjective functions using the Stirling function, students can broaden their knowledge of combinatorial structures.
Lastly, the text tackles questions to reinforce understanding, such as conditions for functions and equivalence relations, promoting a deeper comprehension of discrete mathematics.
Dive deep into the subject with an immersive audiobook experience.
Signup and Enroll to the course for listening the Audio Book
So, we begin with question number 6. In question number 6 you are asked to either prove or disprove the following. You are given a function f: B → A and you are given that the function is surjective. Then the question is, is it necessary that the function is bijective as well? It turns out that the statement is true provided your set A is a finite set. Because indeed if the set is a finite set and the function is from the same set to itself and surjective. Then we can show it is a bijective function as well. So, we will touch upon this fact sometime later in this course. But the statement need not be true if the set A is an infinite set.
In this chunk, we start with a question about functions in mathematics, particularly focusing on surjective and bijective functions. A function is called surjective if every element in the codomain has at least one pre-image in the domain. The chunk tells us that if we have a function from one set to another and it's surjective, we can often conclude that it is bijective (both injective and surjective) if the sets are finite. However, this doesn't hold true for infinite sets. For example, a function mapping natural numbers that misses elements still remains surjective despite not pairing every input uniquely. We use this to differentiate the properties of functions based on set size.
Imagine you are distributing invitations to a party. If everyone you invited (the codomain) ends up receiving at least one invitation (surjective), but some people receive invitations from multiple sources (not injective), it is not one-to-one (bijective). But if you ensure that every invitation goes to a unique person, then it is bijective. This illustrates how surjectivity ensures coverage but not uniqueness.
Signup and Enroll to the course for listening the Audio Book
So, here is a counter example. So, imagine the function f given from the set 0 to infinity to the set 0 to infinity. So, that is my set A and the function is defined as follows. The mapping 0 → 0 and the mapping of 1 → 0. So, clearly the function is not injective, not 1 to 1 and what about the mapping of the remaining elements. The mapping of element 2 is 1 the mapping of element 3 is (3 - 1) and so on. So, it is easy to verify that the function is indeed a surjective function because you pick any element y from the set 0 to infinity, the pre-image for that element y will be y + 1. Because f(y + 1) = y as per the function f we have defined here.
This chunk provides an example to illustrate that surjective functions can exist without being injective, especially when dealing with infinite sets. The function defined here takes elements from the natural numbers and maps them to the same first few outputs, which shows it's not injective. However, for every possible output in the target set, there is a corresponding input, proving it is surjective. This distinction is crucial as it highlights scenarios where more complex functions don’t fit neatly into the bijection category.
Consider a shoe factory where multiple shoe styles might end up in the same size category (like many 8s in stock). If a store requests 10 pairs of 8s (the set range), they can certainly fulfill that despite having limited styles, demonstrating the surjectiveness. Yet, if they were tasked with assigning unique styles to each pair, that would fail, showing it's not bijective.
Signup and Enroll to the course for listening the Audio Book
In question number 7, you are given an equivalence relation over a set A, where the set A has 30 elements. And since it is an equivalence relation, the relation partitions the set A into three subsets each of equal size. So, the question asks you how many ordered pairs are there in that equivalence relation? So, since the subsets 𝑆1, 𝑆2, 𝑆3 constitute a partition of the set A and it is also given that the size of each subset is same and since the number of elements in the set A is 30, we get that the size of each subset in the partition is 10. Recall, when we showed that for every equivalence relation there is a partition and for every partition there is an equivalence relation, we showed that if you are given a partition how you get the corresponding equivalence relation whose equivalence class will be giving you that partition.
This section discusses equivalence relations, which relate two elements based on a defined condition. The relation partitions the set into distinct subsets, and the question asks to analyze how many ordered pairs can exist within these partitions. Since the subsets are of equal size, we can calculate the total number of ordered pairs formed from combinations within these subsets. The concept is foundational in understanding how functions and relations are interconnected.
Think of a classroom where students (set A) are grouped into three project teams (the subsets). Each team (or subset) has 10 students. If we wanted to list out all friendships (ordered pairs) that exist among members within each team, the way we group them matters—like forming a sports team primarily from a football roster, emphasizing they all belong to a single sport (equivalence relation).
Learn essential terms and foundational ideas that form the basis of the topic.
Key Concepts
Surjective Function: A function where every codomain element has a pre-image.
Injective Function: All distinct domain elements map to distinct codomain elements.
Bijective Function: A function that is both surjective and injective.
Equivalence Relation: A relation dividing a set into disjoint subsets.
Stirling Numbers: Numbers representing the partitions of a set into non-empty subsets.
See how the concepts apply in real-world scenarios to understand their practical implications.
Example of surjective function: Mapping from natural numbers to themselves where every number maps back to a number in a one-to-many relationship.
Example of an equivalence relation: Defining an equivalence relation on a number line where numbers are considered equivalent if they share the same fractional part.
Use mnemonics, acronyms, or visual cues to help remember key information more easily.
In a function's land, to each pair we must bend, a surjective line means every output's a friend.
Imagine a party where each guest had to introduce themselves. In a surjective function, every guest (codomain) had at least one introduction (pre-image). In injective function, everyone introduced themselves uniquely without overlaps.
BIF: B for Bijective, I for Injective, F for Full coverage in codomain.
Review key concepts with flashcards.
Review the Definitions for terms.
Term: Surjective Function
Definition:
A function where every element in the codomain has at least one corresponding pre-image in the domain.
Term: Injective Function
Definition:
A function where distinct elements in the domain map to distinct elements in the codomain.
Term: Bijective Function
Definition:
A function that is both injective and surjective.
Term: Equivalence Relation
Definition:
A relation that partitions a set into disjoint subsets, where each element in a subset is related.
Term: Stirling Numbers
Definition:
Numbers that count the ways to partition a set into non-empty subsets.