Suppose Again That Z = X + Y . Find Fz

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¶

A function is a rule that assigns each input exactly one output. We telephone call the output the image of the input. The fix of all inputs for a function is called the domain. The set of all allowable outputs is called the codomain. We would write \(f:X \to Y\) to describe a function with name \(f\text{,}\) domain \(10\) and codomain \(Y\text{.}\) This does not tell united states which function \(f\) is though. To define the function, we must describe the rule. This is oft done past giving a formula to compute the output for whatever input (although this is certainly not the simply manner to depict the rule).

For example, consider the function \(f:\N \to \N\) divers past \(f(ten) = x^2 + 3\text{.}\) Here the domain and codomain are the same set (the natural numbers). The rule is: take your input, multiply information technology by itself and add 3. This works because we can utilize this rule to every natural number (every element of the domain) and the effect is e'er a natural number (an chemical element of the codomain). Notice though that not every natural number actually is an output (there is no way to become 0, one, 2, 5, etc.). The set of natural numbers that are actually outputs is called the range of the function (in this example, the range is \(\{three, 4, 7, 12, 19, 28, \ldots\}\text{,}\) all the natural numbers that are 3 more than a perfect square).

The central thing that makes a rule actually a function is that there is exactly one output for each input. That is, it is important that the dominion be a good rule. What output exercise we assign to the input vii? At that place can simply be one answer for any item office.

The clarification of the rule tin vary profoundly. We might just give a listing of the images of each input. You could also describe the function with a tabular array or a graph or in words.

Example 0.4.1

The following are all examples of functions:

1. \(f:\Z \to \Z\) defined by \(f(n) = 3n\text{.}\) The domain and codomain are both the set of integers. However, the range is only the set up of integer multiples of 3.
2. \(g: \{1,two,three\} \to \{a,b,c\}\) defined by \(chiliad(1) = c\text{,}\) \(g(two) = a\) and \(g(3) = a\text{.}\) The domain is the gear up \(\{1,two,iii\}\text{,}\) the codomain is the set \(\{a,b,c\}\) and the range is the set \(\{a,c\}\text{.}\) Notation that \(g(ii)\) and \(g(three)\) are the same element of the codomain. This is okay since each element in the domain still has only one output.

3. \(h:\{ane,two,3\} \to \{one,2,3\}\) defined equally follows:

   

   This ways that the function \(f\) sends one to 2, 2 to i and iii to 3: but follow the arrows.

The arrow diagram used to define the function higher up can be very helpful in visualizing functions. We will often be working with functions with finite domains, and then this kind of film is ofttimes more useful than a traditional graph of a function. A graph of the role in example 3 above would look like this:

It would be absolutely WRONG to connect the dots or endeavour to fit them to some curve. There are simply three elements in the domain. A curve suggests that the domain contains an unabridged interval of real numbers. Call up, we are not in calculus any more!

Since we will then often utilize functions with small domains and codomains, permit'southward adopt some notation that is a niggling easier to work with than that of examples 2 and 3 above. All we need is some clear way of denoting the image of each element in the domain. In fact, writing a table of values would work perfectly:

\(x\)	0	one	2	3	4
\(f(x)\)	3	iii	2	four	1

We simplify this further past writing this as a matrix with each input direct over its output:

\begin{equation*} f = \begin{pmatrix}0 \amp i \amp 2\amp 3 \amp four \\ 3 \amp three \amp 2 \amp 4 \amp one\end{pmatrix} \end{equation*}

Note this is just notation and not the same sort of matrix you lot would find in a linear algebra class (information technology does not make sense to practise operations with these matrices, or row reduce them, for example).

It is important to know how to determine if a rule is or is not a part. Cartoon the arrow diagrams can aid.

Example 0.4.2

Which of the following diagrams stand for a office? Allow \(10 = \{1,2,iii,4\}\) and \(Y = \{a,b,c,d\}\text{.}\)

Solution

\(f\) is a function. Then is \(g\text{.}\) There is no problem with an element of the codomain non being the image of whatever input, and at that place is no trouble with \(a\) from the codomain being the image of both 2 and iii from the domain. Nosotros could use our 2-line notation to write these as

\brainstorm{equation*} f= \brainstorm{pmatrix} 1 \amp 2 \amp 3 \amp four \\ d \amp a \amp c \amp b \cease{pmatrix} \qquad g = \begin{pmatrix} 1 \amp 2 \amp three \amp 4 \\ d \amp a \amp a \amp b \terminate{pmatrix}. \finish{equation*}

Yet, \(h\) is NOT a office. In fact, it fails for two reasons. Starting time, the chemical element 1 from the domain has not been mapped to any element from the codomain. Second, the element 2 from the domain has been mapped to more than ane element from the codomain (\(a\) and \(c\)). Notation that either one of these bug is enough to make a rule non a function. In general, neither of the following mappings are functions:

It might also exist helpful to call back about how you lot would write the 2-line annotation for \(h\text{.}\) We would take something like:

\brainstorm{equation*} h=\begin{pmatrix} 1 \amp 2 \amp iii \amp 4 \\ \amp a,c? \amp d \amp b\cease{pmatrix}. \end{equation*}

There is null under 1 (bad) and we needed to put more than i thing under 2 (very bad). With a dominion that is actually a office, the two-line note will e'er "piece of work".

Subsection Surjections, Injections, and Bijections

¶

We at present turn to investigating special properties functions might or might not possess.

In the examples in a higher place, you may have noticed that sometimes there are elements of the codomain which are not in the range. When this sort of the matter does non happen, (that is, when everything in the codomain is in the range) we say the role is onto or that the function maps the domain onto the codomain. This terminology should brand sense: the part puts the domain (entirely) on superlative of the codomain. The fancy math term for an onto function is a surjection, and we say that an onto function is a surjective role.

In pictures:

Case 0.4.3

Which functions are surjective (i.e., onto)?

1. \(f:\Z \to \Z\) defined by \(f(n) = 3n\text{.}\)
2. \(thou: \{i,2,three\} \to \{a,b,c\}\) defined by \(m = \begin{pmatrix}1 \amp two \amp iii \\ c \amp a \amp a \stop{pmatrix}\text{.}\)

3. \(h:\{1,2,3\} \to \{1,2,iii\}\) defined every bit follows:

   

Solution

1. \(f\) is not surjective. There are elements in the codomain which are not in the range. For example, no \(n \in \Z\) gets mapped to the number 1 (the rule would say that \(\frac{ane}{3}\) would be sent to 1, but \(\frac{1}{iii}\) is not in the domain). In fact, the range of the office is \(3\Z\) (the integer multiples of three), which is not equal to \(\Z\text{.}\)
2. \(g\) is not surjective. In that location is no \(x \in \{1,2,3\}\) (the domain) for which \(g(x) = b\text{,}\) so \(b\text{,}\) which is in the codomain, is not in the range. Notice that there is an element from the codomain "missing" from the lesser row of the matrix.
3. \(h\) is surjective. Every element of the codomain is also in the range. Naught in the codomain is missed.

To be a part, a rule cannot assign a unmarried element of the domain to ii or more different elements of the codomain. Withal, we accept seen that the opposite is permissible: a function might assign the aforementioned element of the codomain to ii or more unlike elements of the domain. When this does not occur (that is, when each element of the codomain is the epitome of at about one element of the domain) then we say the function is i-to-one. Again, this terminology makes sense: nosotros are sending at most 1 element from the domain to one element from the codomain. One input to one output. The fancy math term for a i-to-one function is an injection. Nosotros call one-to-one functions injective functions.

In pictures:

Instance 0.4.4

Which functions are injective (i.due east., i-to-one)?

1. \(f:\Z \to \Z\) defined by \(f(n) = 3n\text{.}\)
2. \(g: \{i,2,three\} \to \{a,b,c\}\) defined by \(g = \begin{pmatrix}1 \amp ii \amp 3 \\ c \amp a \amp a \end{pmatrix}\text{.}\)

3. \(h:\{ane,two,3\} \to \{ane,2,3\}\) divers as follows:

   

Solution

1. \(f\) is injective. Each element in the codomain is assigned to at about 1 element from the domain. If \(10\) is a multiple of three, then only \(x/iii\) is mapped to \(x\text{.}\) If \(x\) is non a multiple of three, so in that location is no input corresponding to the output \(x\text{.}\)
2. \(m\) is non injective. Both inputs \(2\) and \(iii\) are assigned the output \(a\text{.}\) Notice that there is an element from the codomain that appears more than once on the bottom row of the matrix.
3. \(h\) is injective. Each output is just an output once.

From the examples above, it should exist clear that there are functions which are surjective, injective, both, or neither. In the case when a function is both one-to-1 and onto (an injection and surjection), we say the function is a bijection, or that the part is a bijective part.

Subsection Changed Image

¶

When discussing functions, we take notation for talking about an element of the domain (say \(x\)) and its corresponding element in the codomain (we write \(f(ten)\text{,}\) which is the image of \(x\)). It would too be nice to start with some element of the codomain (say \(y\)) and talk most which element or elements (if any) from the domain it is the image of. We could write "those \(x\) in the domain such that \(f(10) = y\text{,}\)" but this is a lot of writing. Here is some notation to brand our lives easier.

Suppose \(f:X \to Y\) is a function. For \(y \in Y\) (an element of the codomain), nosotros write \(f\inv(y)\) to represent the set of all elements in the domain \(X\) which get sent to \(y\text{.}\) That is, \(f\inv(y) = \{10 \in 10 \st f(x) = y\}\text{.}\) We say that \(f\inv(y)\) is the complete inverse image of \(y\) under \(f\text{.}\)

WARNING: \(f\inv(y)\) is not an changed office! Inverse functions only exist for bijections, but \(f\inv(y)\) is defined for any role \(f\text{.}\) The point: \(f\inv(y)\) is a prepare, non an chemical element of the domain.

Example 0.4.five

Consider the part \(f:\{i,2,3,4,5,6\} \to \{a,b,c,d\}\) given by

\begin{equation*} f = \begin{pmatrix}ane \amp two \amp 3 \amp 4 \amp 5 \amp 6 \\ a \amp a \amp b \amp c \amp c \amp c\end{pmatrix}. \stop{equation*}

Discover the complete inverse image of each element in the codomain.

Solution

Recall, nosotros are looking for sets.

\begin{equation*} f\inv(a) = \{1,2\} \end{equation*} \begin{equation*} f\inv(b) = \{three\} \end{equation*} \begin{equation*} f\inv(c) = \{4,5,6\} \terminate{equation*} \begin{equation*} f\inv(d) = \emptyset. \end{equation*}

Example 0.4.six

Consider the function \(g:\Z \to \Z\) defined by \(thou(n) = n^2 + 1\text{.}\) Find \(g\inv(1)\text{,}\) \(one thousand\inv(two)\text{,}\) \(g\inv(iii)\) and \(g\inv(10)\text{.}\)

Solution

To find \(g\inv(1)\text{,}\) we demand to observe all integers \(n\) such that \(n^ii + 1 = 1\text{.}\) Clearly only 0 works, so \(chiliad\inv(1) = \{0\}\) (note that even though at that place is only one element, we still write it as a set with one element in information technology).

To find \(g\inv(2)\text{,}\) we need to discover all \(n\) such that \(n^2 + 1 = two\text{.}\) We run into \(g\inv(2) = \{-1,1\}\text{.}\)

If \(north^2 + 1 = 3\text{,}\) then we are looking for an \(north\) such that \(n^ii = 2\text{.}\) There are no such integers so \(g\inv(3) = \emptyset\text{.}\)

Finally, \(g\inv(10) = \{-3, 3\}\) because \(1000(-iii) = x\) and \(chiliad(iii) = 10\text{.}\)

Since \(f\inv(y)\) is a set, it makes sense to enquire for \(\menu{f\inv(y)}\text{,}\) the number of elements in the domain which map to \(y\text{.}\)

Case 0.4.7

Find a role \(f:\{1,2,iii,four,5\} \to \Due north\) such that \(\card{f\inv(7)} = 5\text{.}\)

Solution

There is merely one such role. We demand five elements of the domain to map to the number \(seven \in \North\text{.}\) Since there are just five elements in the domain, all of them must map to seven. So

\begin{equation*} f = \begin{pmatrix}1 \amp 2 \amp 3 \amp 4 \amp 5 \\ vii \amp 7 \amp seven \amp 7 \amp seven\end{pmatrix}. \end{equation*}

Function Definitions

• A office is a rule that assigns each element of a ready, called the domain, to exactly one chemical element of a second set, called the codomain.

• Notation: \(f:X \to Y\) is our mode of saying that the function is chosen \(f\text{,}\) the domain is the set up \(X\text{,}\) and the codomain is the set \(Y\text{.}\)

• To specify the rule for a role with pocket-sized domain, use two-line notation by writing a matrix with each output direct beneath its corresponding input, as in:

  \begin{equation*} f = \begin{pmatrix}1 \amp 2 \amp iii \amp iv \\ 2 \amp one \amp three \amp i \end{pmatrix}. \end{equation*}

• \(f(x) = y\) ways the element \(ten\) of the domain (input) is assigned to the element \(y\) of the codomain. Nosotros say \(y\) is an output. Alternatively, nosotros telephone call \(y\) the image of \(x\) nether \(f\).

• The range is a subset of the codomain. It is the fix of all elements which are assigned to at least 1 element of the domain by the role. That is, the range is the ready of all outputs.

• A office is injective (an injection or one-to-i) if every chemical element of the codomain is the output for at nearly i chemical element from the domain.

• A function is surjective (a surjection or onto) if every element of the codomain is the output of at to the lowest degree one element of the domain.

• A bijection is a function which is both an injection and surjection. In other words, if every element of the codomain is the output of exactly 1 chemical element of the domain.

• The image of an element \(x\) in the domain is the chemical element \(y\) in the codomain that \(x\) is mapped to. That is, the image of \(x\) under \(f\) is \(f(x)\text{.}\)

• The consummate changed image of an element \(y\) in the codomain, written \(f\inv(y)\text{,}\) is the prepare of all elements in the domain which are assigned to \(y\) by the function.

Subsection Exercises

¶

one

Write out all functions \(f: \{1,2,iii\} \to \{a,b\}\) (using two-line notation). How many are there? How many are injective? How many are surjective? How many are both?

Solution

There are 8 different functions. In two-line notation these are:

\begin{equation*} f = \begin{pmatrix} 1 \amp two \amp 3 \\ a \amp a\amp a \stop{pmatrix} \quad f = \begin{pmatrix} 1 \amp 2 \amp 3 \\ b \amp b \amp b \end{pmatrix} \end{equation*} \brainstorm{equation*} f = \begin{pmatrix} ane \amp 2 \amp iii \\ a \amp a\amp b \end{pmatrix} \quad f = \begin{pmatrix} ane \amp 2 \amp three \\ a \amp b \amp a \end{pmatrix} \quad f = \begin{pmatrix} 1 \amp ii \amp 3 \\ b \amp a\amp a \stop{pmatrix} \end{equation*} \begin{equation*} \quad f = \begin{pmatrix} 1 \amp 2 \amp 3 \\ b \amp b \amp a \finish{pmatrix} \quad f = \brainstorm{pmatrix} ane \amp two \amp 3 \\ b \amp a\amp b \end{pmatrix} \quad f = \begin{pmatrix} ane \amp ii \amp 3 \\ a \amp b \amp b \cease{pmatrix} \end{equation*}

None of the functions are injective. Exactly six of the functions are surjective. No functions are both (since no functions here are injective).

2

Write out all functions \(f: \{1,ii\} \to \{a,b,c\}\) (in 2-line note). How many are there? How many are injective? How many are surjective? How many are both?

Solution

There are nine functions: you have a choice of three outputs for \(f(i)\text{,}\) and for each, you lot take iii choices for the output \(f(2)\text{.}\) Of these functions, 6 are injective, 0 are surjective, and 0 are both:

\brainstorm{equation*} f = \twoline{1 \amp 2}{a\amp a} \quad f = \twoline{ane \amp 2}{b \amp b} \quad f = \twoline{1 \amp 2}{c \amp c} \end{equation*} \brainstorm{equation*} f = \twoline{i \amp 2}{a\amp b} \quad f = \twoline{1 \amp two}{a \amp c} \quad f = \twoline{1 \amp 2}{b \amp c} \finish{equation*} \begin{equation*} f = \twoline{1 \amp 2}{b \amp a} \quad f = \twoline{1 \amp ii}{c \amp a} \quad f = \twoline{1 \amp ii}{c \amp b} \end{equation*}

3

Consider the function \(f:\{1,2,iii,4,5\} \to \{1,2,3,4\}\) given by the table below:

\(x\)	1	2	3	4	5
\(f(x)\)	3	two	4	1	2

1. Is \(f\) injective? Explain.

2. Is \(f\) surjective? Explain.

3. Write the function using two-line note.

4

Consider the function \(f:\{1,2,three,four\} \to \{i,2,3,4\}\) given by the graph below.

1. Is \(f\) injective? Explain.

2. Is \(f\) surjective? Explain.

3. Write the part using two-line notation.

v

For each part given below, make up one's mind whether or not the role is injective and whether or not the function is surjective.

1. \(f:\North \to \N\) given past \(f(n) = northward+four\text{.}\)
2. \(f:\Z \to \Z\) given by \(f(n) = n+4\text{.}\)
3. \(f:\Z \to \Z\) given past \(f(n) = 5n - viii\text{.}\)
4. \(f:\Z \to \Z\) given by \(f(n) = \begin{cases}n/two \amp \text{ if } due north \text{ is fifty-fifty} \\ (n+1)/two \amp \text{ if } northward \text{ is odd} . \stop{cases}\)

Solution

1. \(f\) is injective, but not surjective (since 0, for example, is never an output).
2. \(f\) is injective and surjective. Unlike in the previous question, every integers is an output (of the integer four less than it).
3. \(f\) is injective, but not surjective (x is not viii less than a multiple of five, for instance).
4. \(f\) is non injective, only is surjective. Every integer is an output (of twice itself, for example) but some integers are outputs of more one input: \(f(v) = 3 = f(6)\text{.}\)

6

Allow \(A = \{1,2,iii,\ldots,10\}\text{.}\) Consider the part \(f:\pow(A) \to \Due north\) given by \(f(B) = |B|\text{.}\) That is, \(f\) takes a subset of \(A\) as an input and outputs the cardinality of that set.

1. Is \(f\) injective? Prove your answer.

2. Is \(f\) surjective? Prove your respond.

3. Find \(f\inv(i)\text{.}\)

4. Detect \(f\inv(0)\text{.}\)

5. Detect \(f\inv(12)\text{.}\)

Solution

1. \(f\) is not injective. To testify this, we must simply find two different elements of the domain which map to the aforementioned element of the codomain. Since \(f(\{1\}) = 1\) and \(f(\{2\}) = 1\text{,}\) we see that \(f\) is not injective.
2. \(f\) is not surjective. The largest subset of \(A\) is \(A\) itself, and \(|A| = 10\text{.}\) And then no natural number greater than 10 will e'er exist an output.
3. \(f\inv(1) = \{\{1\}, \{two\}, \{3\}, \ldots \{x\}\}\) (the set of all the singleton subsets of \(A\)).
4. \(f\inv(0) = \{\emptyset\}\text{.}\) Note, information technology would be wrong to write \(f\inv(0) = \emptyset\) - that would claim that there is no input which has 0 every bit an output.
5. \(f\inv(12) = \emptyset\text{,}\) since at that place are no subsets of \(A\) with cardinality 12.

7

Allow \(A = \{northward \in \N \st 0 \le n \le 999\}\) be the set of all numbers with three or fewer digits. Define the function \(f:A \to \N\) by \(f(abc) = a+b+c\text{,}\) where \(a\text{,}\) \(b\text{,}\) and \(c\) are the digits of the number in \(A\text{.}\) For case, \(f(253) = 2 + 5 + 3 = 10\text{.}\)

1. Find \(f\inv(3)\text{.}\)

2. Find \(f\inv(28)\text{.}\)

3. Is \(f\) injective. Explain.

4. Is \(f\) surjective. Explain.

Solution

1. \(f\inv(3) = \{003, 030, 300, 012, 021, 102, 201, 120, 210, 111\}\)
2. \(f\inv(28) = \emptyset\) (since the largest sum of 3 digits is \(ix+9+9 = 27\))

3. Part (a) proves that \(f\) is non injective. The output 3 is assigned to ten dissimilar inputs.

4. Office (b) proves that \(f\) is non surjective. There is an element of the codomain (28) which is not assigned to any inputs.

8

Let \(f:Ten \to Y\) be some function. Suppose \(3 \in Y\text{.}\) What tin can you say nearly \(f\inv(three)\) if you know,

1. \(f\) is injective? Explain.
2. \(f\) is surjective? Explain.
3. \(f\) is bijective? Explicate.

Solution

1. \(|f\inv(3)| \le ane\text{.}\) In other words, either \(f\inv(iii)\) is the emptyset or is a prepare containing exactly one element. Injective functions cannot have two elements from the domain both map to 3.
2. \(|f\inv(three)| \ge 1\text{.}\) In other words, \(f\inv(iii)\) is a set containing at least ane elements, possibly more. Surjective functions must have something map to three.
3. \(|f\inv(3)| = 1\text{.}\) There is exactly one element from \(Ten\) which gets mapped to 3, so \(f\inv(3)\) is the set containing that one element.

9

Find a set up \(Ten\) and a function \(f:10 \to \N\) then that \(f\inv(0) \loving cup f\inv(1) = X\text{.}\)

Solution

\(10\) can really be whatever set, equally long every bit \(f(x) = 0\) or \(f(x) = 1\) for every \(x \in X\text{.}\) For case, \(10 = \Due north\) and \(f(north) = 0\) works.

10

What can you deduce most the sets \(Ten\) and \(Y\) if you know …

1. at that place is an injective function \(f:X \to Y\text{?}\) Explain.

2. there is a surjective part \(f:X \to Y\text{?}\) Explain.

3. there is a bijectitve office \(f:X \to Y\text{?}\) Explicate.

eleven

Suppose \(f:X \to Y\) is a function. Which of the following are possible? Explain.

1. \(f\) is injective merely not surjective.
2. \(f\) is surjective but not injective.
3. \(|X| = |Y|\) and \(f\) is injective simply not surjective.
4. \(|X| = |Y|\) and \(f\) is surjective but not injective.
5. \(|Ten| = |Y|\text{,}\) \(10\) and \(Y\) are finite, and \(f\) is injective only not surjective.
6. \(|X| = |Y|\text{,}\) \(X\) and \(Y\) are finite, and \(f\) is surjective but not injective.

12

Let \(f:Ten \to Y\) and \(thousand:Y \to Z\) be functions. We can define the composition of \(f\) and \(thou\) to be the function \(thou\circ f:10 \to Z\) which the image of each \(x \in Ten\) is \(g(f(10))\text{.}\) That is, plug \(ten\) into \(f\text{,}\) then plug the effect into \(g\) (just like composition in algebra and calculus).

1. If \(f\) and \(1000\) are both injective, must \(one thousand\circ f\) be injective? Explicate.
2. If \(f\) and \(thousand\) are both surjective, must \(1000\circ f\) be surjective? Explain.
3. Suppose \(g\circ f\) is injective. What, if anything, can you say about \(f\) and \(g\text{?}\) Explicate.
4. Suppose \(k\circ f\) is surjective. What, if anything, can yous say well-nigh \(f\) and \(1000\text{?}\) Explain.

Hint

Work with some examples. What if \(f = \twoline{1\amp 2 \amp three}{a \amp a \amp b}\) and \(chiliad = \twoline{a\amp b \amp c}{5 \amp 6 \amp seven}\text{?}\)

13

Consider the function \(f:\Z \to \Z\) given by \(f(n) = \begin{cases}due north+one \amp \text{ if }n\text{ is even} \\ n-3 \amp \text{ if }n\text{ is odd} . \end{cases}\)

1. Is \(f\) injective? Prove your reply.

2. Is \(f\) surjective? Prove your answer.

Solution

1. \(f\) is injective.

   Proof

   Allow \(x\) and \(y\) be elements of the domain \(\Z\text{.}\) Assume \(f(x) = f(y)\text{.}\) If \(x\) and \(y\) are both even, then \(f(x) = x+1\) and \(f(y) = y+1\text{.}\) Since \(f(x) = f(y)\text{,}\) we have \(x + 1 = y + i\) which implies that \(x = y\text{.}\) Similarly, if \(x\) and \(y\) are both odd, then \(x - 3 = y-three\) so again \(x = y\text{.}\) The simply other possibility is that \(x\) is even an \(y\) is odd (or visa-versa). Simply then \(x + 1\) would be odd and \(y - 3\) would be fifty-fifty, so information technology cannot be that \(f(x) = f(y)\text{.}\) Therefore if \(f(x) = f(y)\) we then accept \(ten = y\text{,}\) which proves that \(f\) is injective.

2. \(f\) is surjective.

   Proof

   Let \(y\) be an chemical element of the codomain \(\Z\text{.}\) We will show there is an element \(n\) of the domain (\(\Z\)) such that \(f(northward) = y\text{.}\) In that location are two cases: Showtime, if \(y\) is even, then let \(due north = y+3\text{.}\) Since \(y\) is fifty-fifty, \(northward\) is odd, so \(f(north) = n-3 = y+3-iii = y\) equally desired. 2nd, if \(y\) is odd, then allow \(n = y-1\text{.}\) Since \(y\) is odd, \(north\) is even, so \(f(n) = n+1 = y-one+i = y\) every bit needed. Therefore \(f\) is surjective.

14

At the end of the semester a teacher assigns letter grades to each of her students. Is this a office? If and then, what sets make upwardly the domain and codomain, and is the office injective, surjective, bijective, or neither?

Solution

Yes, this is a function, if you choose the domain and codomain correctly. The domain volition be the set of students, and the codomain will be the set of possible grades. The function is well-nigh certainly not injective, because it is likely that two students will become the same form. The function might be surjective – it will exist if at that place is at least one student who gets each grade.

fifteen

In the game of Hearts, four players are each dealt 13 cards from a deck of 52. Is this a office? If so, what sets make upward the domain and codomain, and is the office injective, surjective, bijective, or neither?

xvi

Suppose vii players are playing 5-card stud. Each histrion initially receives 5 cards from a deck of 52. Is this a function? If and then, what sets make up the domain and codomain, and is the function injective, surjective, bijective, or neither?

Solution

This cannot exist a office. If the domain were the fix of cards, then information technology is not a part considering not every bill of fare gets dealt to a player. If the domain were the set of players, it would non exist a office because a single player would get mapped to multiple cards. Since this is not a role, it doesn't brand sense to say whether it is injective/surjective/bijective.

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Source: http://discrete.openmathbooks.org/dmoi2/sec_intro-functions.html

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