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Chapter 0: Problem 4

The following functions all have domain \\{1,2,3,4\\} and codomain \(\\{1,2,3,4,5\\} .\) For each, determine whether it is (only) injective, (only) surjective, bijective, or neither injective nor surjective. (a) \(f=\left(\begin{array}{llll}1 & 2 & 3 & 4 \\ 1 & 2 & 5 & 4\end{array}\right)\). (b) \(f=\left(\begin{array}{llll}1 & 2 & 3 & 4 \\ 1 & 2 & 3 & 2\end{array}\right)\). (c) \(f(x)\) gives the number of letters in the English word for the number \(x .\) For example, \(f(1)=3\) since "one" contains three letters.

Short Answer

Expert verified
Function (a) is injective, (b) is neither injective nor surjective, and (c) is only injective.

Step by step solution

01

Define Injective (One-to-One)

A function is injective if each element of the codomain is mapped to by at most one element of the domain. No two different elements in the domain should map to the same element in the codomain.
02

Define Surjective (Onto)

A function is surjective if every element of the codomain is the image of at least one element from the domain. All elements in the codomain should be mapped to by the function.
03

Define Bijective

A function is bijective if it is both injective and surjective. This means that each element of the codomain is mapped to by exactly one element of the domain and vice versa.
04

Analyze function (a)

Check the mapping of function (a) to see if it's injective or surjective. Since no two domain elements map to the same codomain element and all elements in the codomain except 3 are mapped, it is only injective.
05

Analyze function (b)

Check the mapping of function (b). It is not injective because domain elements 2 and 4 map to the same codomain element 2. It is not surjective because the codomain element 4 has no pre-image in the domain. Thus, it's neither injective nor surjective.
06

Analyze function (c)

For function (c), check each word for uniqueness and coverage. One (3), Two (3), Three (5), and Four (4). It is injective because no two elements in the domain have the same number of letters. It is not surjective, as no word corresponds to a number 1 in the codomain.

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Key Concepts

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

Injective Functions
In discrete mathematics, the concept of injective functions, also known as one-to-one functions, is pivotal when analyzing the behavior of mappings between sets.

An injective function ensures that distinct elements in the domain map to distinct elements in the codomain. In simpler terms, if you have two different starting points, they must end up at two different destinations. No overlap is allowed in the codomain for an injective function.

For example, in part (a) of the exercise, the function assigned distinct values in the codomain to each element in its domain, making it injective. However, in part (b), the function was not injective because both domain elements 2 and 4 were mapped to the same codomain element 2.
Surjective Functions
A surjective function, or onto function, connects every element in the codomain with at least one element from the domain. This implies that the function covers the entire codomain; no element is left out or unreachable from the domain.

The primary question to ask when determining if a function is surjective is: 'Does every potential ending point have a starting point?鈥�

For instance, function (a) from the exercise was not surjective because the codomain element 3 was never an image of any domain element, leaving the element 3 without any corresponding input from the domain set.
Bijective Functions
A function that is both injective and surjective is honored with the title of bijective function. It's the gold standard in function relationships - a perfect paring where every unique input has a unique output, and no part of the codomain is left untouched.

This elegantly balanced scenario means that a bijective function establishes a one-to-one correspondence between each element in its domain and codomain.

An essential application of bijective functions is that they allow for inverse functions to exist, paving the way for functions to be reversed. Sadly, none of the functions in our exercise example were bijective, but understanding what makes a function bijective helps enrich our comprehension of the structure and behavior of mathematical functions.

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Most popular questions from this chapter

Let \(X=\\{n \in \mathbb{N}: 0 \leq n \leq 999\\}\) be the set of all numbers with three or fewer digits. Define the function \(f: X \rightarrow \mathbb{N}\) by \(f(a b c)=a+b+c\), where \(a, b,\) and \(c\) are the digits of the number in \(X\) (write numbers less than 100 with leading 0 's to make them three digits). For example, \(f(253)=2+5+3=10\) (a) Let \(A=\\{n \in X: 113 \leq n \leq 122\\}\). Find \(f(A)\). (b) Find \(f^{-1}(\\{1,2\\})\) (c) Find \(f^{-1}(3)\). (d) Find \(f^{-1}(28)\). (e) Is \(f\) injective? Explain. (f) Is \(f\) surjective? Explain.

You have discovered an old paper on graph theory that discusses the viscosity of a graph (which for all you know, is something completely made up by the author). A theorem in the paper claims that "if a graph satisfies condition \((V)\), then the graph is viscous." Which of the following are equivalent ways of stating this claim? Which are equivalent to the converse of the claim? (a) A graph is viscous only if it satisfies condition (V). (b) A graph is viscous if it satisfies condition (V). (c) For a graph to be viscous, it is necessary that it satisfies condition(V). (d) For a graph to be viscous, it is sufficient for it to satisfy condition(V). (e) Satisfying condition (V) is a sufficient condition for a graph to be viscous. (f) Satisfying condition ( \(\mathrm{V}\) ) is a necessary condition for a graph to be viscous. (g) Every viscous graph satisfies condition (V). (h) Only viscous graphs satisfy condition (V).

For each sentence below, decide whether it is an atomic statement, a molecular statement, or not a statement at all. (a) Customers must wear shoes. (b) The customers wore shoes. (c) The customers wore shoes and they wore socks.

For a given predicate \(P(x)\), you might believe that the statements \(\forall x P(x)\) or \(\exists x P(x)\) are either true or false. How would you decide if you were correct in each case? You have four choices: you could give an example of an element \(n\) in the domain for which \(P(n)\) is true or for which \(P(n)\) if false, or you could argue that no matter what \(n\) is, \(P(n)\) is true or is false. (a) What would you need to do to prove \(\forall x P(x)\) is true? (b) What would you need to do to prove \(\forall x P(x)\) is false? (c) What would you need to do to prove \(\exists x P(x)\) is true? (d) What would you need to do to prove \(\exists x P(x)\) is false?

Suppose \(P(x, y)\) is some binary predicate defined on a very small domain of discourse: just the integers \(1,2,3,\) and \(4 .\) For each of the 16 pairs of these numbers, \(P(x, y)\) is either true or false, according to the following table \((x\) values are rows, \(y\) values are columns). $$\begin{array}{c|cccc} & 1 & 2 & 3 & 4 \\\\\hline 1 & \mathrm{~T} & \mathrm{~F} & \mathrm{~F} & \mathrm{~F} \\ 2 & \mathrm{~F} & \mathrm{~T} & \mathrm{~T} & \mathrm{~F} \\ 3 & \mathrm{~T} & \mathrm{~T} & \mathrm{~T} & \mathrm{~T} \\ 4 & \mathrm{~F} & \mathrm{~F} & \mathrm{~F} & \mathrm{~F}\end{array}$$ For example, \(P(1,3)\) is false, as indicated by the \(\mathrm{F}\) in the first row, third column. Use the table to decide whether the following statements are true or false. (a) \(\forall x \exists y P(x, y)\) (b) \(\forall y \exists x P(x, y)\) (c) \(\exists x \forall y P(x, y)\). (d) \(\exists y \forall x P(x, y)\).
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