Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

List all \(1-1\) and onto functions from (1,2,3) to itself.

Short Answer

Expert verified
There are 6 one-to-one and onto functions: (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1).

Step by step solution

01

Understanding One-to-One Functions

A function is one-to-one (injective) if each element in the domain maps to a unique element in the codomain. For the set \( \{1, 2, 3\} \), every element from the domain must map to a different element in the codomain.
02

Understanding Onto Functions

A function is onto (surjective) if every element in the codomain has at least one element from the domain mapping to it. Since our domain and codomain are the same, every element in \( \{1, 2, 3\} \) must map to every other element, fulfilling this condition naturally.
03

Listing Permutations

Since the domain and codomain are the same set \( \{1, 2, 3\} \), listing all bijections (one-to-one and onto functions) amounts to listing all permutations of the set. The set \( \{1, 2, 3\} \) can be permuted into 6 different sequences.
04

Enumerating the Permutations

The permutations of the set \( \{1, 2, 3\} \) are: 1. \((1, 2, 3)\) 2. \((1, 3, 2)\) 3. \((2, 1, 3)\) 4. \((2, 3, 1)\) 5. \((3, 1, 2)\) 6. \((3, 2, 1)\) These represent all possible one-to-one and onto functions from the set to itself.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Injective Functions
An injective function, also known as a one-to-one function, is a fascinating concept in mathematics where each element in the function's domain maps to a distinct and unique element in the codomain.
This means no two different input values will share the same output value.
Understanding injective functions becomes particularly interesting when we explore sets such as \( \{1, 2, 3\} \).
To determine if a function is injective, consider each element in the domain.
Each must correspond to a unique element in the codomain without any overlaps.
For instance, if you have a function \( f \) from the set \( \{1, 2, 3\} \) to itself, \( f \) is injective if it satisfies this unique touch condition for all elements.
Consider mapping like this:
  • 1 maps to 2
  • 2 maps to 3
  • 3 maps to 1
Such functions maintain this distinctive mapping choice, confirming their one-to-one nature. Recognizing and crafting injective functions helps build a fundamental skill set in mathematical analysis and problem-solving.
Surjective Functions
When a function is surjective, or onto, it tells an equally compelling story in mathematics. A surjective function ensures that every element in the codomain has at least one preimage in the domain.
This means the function covers every point in the codomain entirely and has impeccably filled the map from domain to codomain.
With sets like \( \{1, 2, 3\} \), understanding surjectivity is essential because we want each number in the range to be reachable by some input from our domain.
Suppose we take our function \( f \) such that it maps the elements as follows:
  • 1 maps to 1
  • 2 maps to 2
  • 3 maps to 3
Here, every number in the set is covered; thus each is in the result of the function, making the function onto.
Surjective functions are crucial in ensuring comprehensive mappings, which are vital in various mathematical applications.
Permutations
Permutations are all about rearranging elements. In a finite set like \( \{1, 2, 3\} \), permutations are particularly handy for counting the possible arrangements or sequences of the elements.
This concept is a cornerstone in understanding bijections—functions that are both injective and surjective.
Finding permutations means listing every possible way the set can be ordered.
For a set with three elements, such as \( \{1, 2, 3\} \), there are exactly 6 permutations:
  • \((1, 2, 3)\)
  • \((1, 3, 2)\)
  • \((2, 1, 3)\)
  • \((2, 3, 1)\)
  • \((3, 1, 2)\)
  • \((3, 2, 1)\)
Each permutation matches a unique one-to-one and onto function from the set to itself, showcasing the concept of bijections.
Permutations and bijections provide significant insights into how items can be mapped or rearranged, forming the backbone for many mathematical theories and applications.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A student has 37 days to prepare for an exam. From past experience, he knows that he will need no more than 60 hours of study, To keep from forgetting the material, he wants to study for at least one hour each day. Show that there is a sequence of successive days during which he will have studied exactly 13 hours.

(a) Show that the set of all infinite sequences of elements of the one element set \(\\{0\\}\) is finite. (b) Show that the set of all infinite sequences of elements of the two element set \(\\{0.1\\}\) has the same cardinality as \(\mathcal{P}(\mathbb{N})\). (c) Challenge: Show that the set of all infinite sequences of elements of N' has the same cardinality as \(\mathcal{P}(\mathbb{N})\).

(a) Show that the set of all finite sequences of elements of the one-element set \\{0\\} is countably infinite. (b) Show that the set of all finite sequences of elements of the two-element set (0,1] is countably infinite. (c) Challenge: Show that the set of all finite sequences of natural numbers is countably infinite. (Hint: Use a diagonal argument.)

Let \(A\) be any nonempty set, and let \(\mathcal{F}_{A}\) be the set of all functions from \(A\) to \(\mathbb{R}\). (a) Why is \(F+G \in \mathcal{F}_{A}\) for all \(F, G \in \mathcal{F}_{A}\). (b) Prove \((F+G)+H=F+(G+H)\) for all \(F, G, H \in \mathcal{F}_{A}\) (c) Let Zero e \(\mathcal{F}_{A}\) be defined by \(\operatorname{Zero}(a)=0\) for all \(a \in A\). Prove that \(Z e r o+F=F\) for all \(F \in \mathcal{F}_{A}\) (d) For \(F \in \mathcal{F}_{A}\), define \(\bar{F}\) by \(\vec{F}(a)=-F(a)\) for each \(a \in A\). Prove that \(F+\bar{F}=\) Zero \(=\bar{F}+F\) for all \(F \in \mathcal{F}_{\mathcal{A}}\)

Challenge: Show that \(|\mathbb{R}|=|\mathcal{P}(\mathbb{N})| .\) (Hint: Use the Cantor-Schröder-Bernstein Theorem. To show that \(|\mathbb{R}| \leq|\mathcal{P}(\mathbb{N})|\), you might use function \(D: \mathbb{R} \rightarrow \mathcal{P}(\mathbb{Q})\) where for \(r \in \mathbb{R}, D(r)=\\{q \in Q: q

See all solutions

Recommended explanations on Computer Science Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free