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Chapter 2: Problem 10

Determine whether each of these functions from \(\\{a, b, c, d\\}\) to itself is one-to-one. a) \(f(a)=b, f(b)=a, f(c)=c, f(d)=d\) b) \(f(a)=b, f(b)=b, f(c)=d, f(d)=c\) c) \(f(a)=d, f(b)=b, f(c)=c, f(d)=d\)

Short Answer

Expert verified
Function (a) and (c) are one-to-one; function (b) is not.

Step by step solution

01

Define one-to-one function

A function is one-to-one (injective) if every element of the codomain is mapped by at most one element of the domain. In simpler terms, no two different elements in the domain map to the same element in the codomain.
02

Analyze function (a)

Check the mappings: \(f(a) = b\), \(f(b) = a\), \(f(c) = c\), \(f(d) = d\). Each element in the codomain \(\{a, b, c, d\}\) is uniquely paired with a different element in the domain. Therefore, function (a) is one-to-one.
03

Analyze function (b)

Check the mappings: \(f(a) = b\), \(f(b) = b\), \(f(c) = d\), \(f(d) = c\). Here, \(f(a)\) and \(f(b)\) both map to \(b\). Since \(a\) and \(b\) map to the same element, function (b) is not one-to-one.
04

Analyze function (c)

Check the mappings: \(f(a) = d\), \(f(b) = b\), \(f(c) = c\), \(f(d) = d\). Each element in the codomain \(\{a, b, c, d\}\) is uniquely paired with a different element in the domain. Therefore, function (c) is one-to-one.

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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, has a special characteristic: each element in the codomain is mapped by at most one element in the domain.
If you imagine each pair as a unique relationship, you'll see that no two different inputs produce the same output.

This feature ensures that every output is distinct and exclusive to a single input. For instance, if we consider function (a) from the exercise:
  • \(f(a) = b\)
  • \(f(b) = a\)
  • \(f(c) = c\)
  • \(f(d) = d\)
We see that each element in the codomain \(\) is uniquely paired.
Therefore, function (a) is indeed an injective function.
function mapping
Function mapping describes how each element in the domain is paired with an element in the codomain.
In simple terms, it's a rule or process that assigns to every input exactly one output.

In mathematical notation, if we have a function \(f\) mapping elements from set \(A\) to set \(B\) (where \(A\) and \(B\) are the domain and codomain, respectively), we write \(f: A 鈫� B\). This shows which inputs are linked to which outputs.

Let's look at function (b) from the exercise:
  • \(f(a) = b\)
  • \(f(b) = b\)
  • \(f(c) = d\)
  • \(f(d) = c\)
Here, both \(a\) and \(b\) are mapped to the same output \(b\). This indicates that function (b) is not injective because one output (\(b\)) is linked to more than one input (\(a\) and \(b\)).

  • To check if a function is injective, always ensure each element in the codomain comes from one unique element in the domain.
codomain
The codomain of a function represents all possible outputs it can produce.
It's the set that contains all values to which functions map elements of the domain.

The concept is crucial in understanding function behavior.
For example, in function (c):
  • \(f(a) = d\)
  • \(f(b) = b\)
  • \(f(c) = c\)
  • \(f(d) = d\)
We can see that the codomain is \(\) and each value in the codomain is uniquely related to one value in the domain.
This illustrates that function (c) is injective.

Understanding the codomain helps to determine if all elements from the output set are utilized and if they are uniquely mapped from the domain.

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