25 - Injective and surjective functions

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Injective

passes the horizontal line test
don't have the same second component twice

$f : R - {1} \to R - {4}$

Surjective
Every $y$ (second component) is defined

Bijective = BOTH

Prove that $f : R - {1} \to R - {4}$ given by $f (x) = \frac{4 x - 8}{x - 1}$ is bijective.

Scratch work:
injective
Let $a_{1}, a_{2} \in R - {1}$ . Assume $f (a_{1}) = f (a_{2})$ .

\begin{aligned} \frac{4 a_{1} - 8}{a_{1} - 1} & = \frac{4 a_{2} - 8}{a_{2} - 1} \\ \frac{a_{1} - 2}{a_{1} - 1} & = \frac{a_{2} - 2}{a_{2} - 1} \\ (a_{1} - 2) (a_{2} - 1) & = (a_{1} - 2) (a_{2} - 1) \\ a_{1} a_{2} - 2 a_{2} - a_{1} + 2 & = a_{1} a_{2} - 2 a_{2} - a_{2} + 2 \\ a_{1} & = a_{2} \end{aligned}

Surjective
Let $b \in R - {4}$

\begin{aligned} b & = \frac{4 a - 8}{a - 1} \\ b (a - 1) & = 4 a - 8 \\ b a - b & = 4 a - 8 \\ (b - 4) a & = b - 8 \\ a & = \frac{b - 8}{b - 4} \end{aligned}

Proof.
(Injective)
Let $a_{1}, a_{2} \in R - {1}$ . Assume that $f (a_{1}) = f (a_{2})$ .
Then

\begin{aligned} \frac{4 a_{1} - 8}{a_{1} - 1} & = \frac{4 a_{2} - 8}{a_{2} - 1} \\ Thus \frac{a_{1} - 2}{a_{1} - 1} & = \frac{a_{2} - 2}{a_{2} - 1} \\ Hence (a_{1} - 2) (a_{2} - 1) & = (a_{1} - 2) (a_{2} - 1) \\ Thus a_{1} a_{2} - 2 a_{2} - a_{1} + 2 & = a_{1} a_{2} - 2 a_{2} - a_{2} + 2 \\ So a_{1} & = a_{2} \end{aligned}

(Surjective)
Let $b \in R - {4}$

Define $a = \frac{b - 8}{b - 4}$ . Suppose $a = 1$ . Then $b - 4 = b - 8$ , so $- 4 = - 8$ . Hence we see $a \neq 1$ .
Note that

\begin{aligned} f (a) & = \frac{4 a - 8}{a - 1} \\ = \frac{4 (\frac{b - 8}{b - 4}) - 8}{\frac{b - 8}{b - 4} - 1} \\ = \frac{\frac{4 (b - 8) - 8 (b - 4)}{b - 4}}{\frac{b - 8 - b - 4}{b - 4}} \\ = \frac{4 b}{- 4} \\ = b \end{aligned}

HW 25

Ben Finch

E 25.1
A to B
{ $(1, x), (2, x), (3, x)$ }
Neither injective nor surjective

{ $(1, x), (2, x), (3, y)$ }
Surjective but not injective

{ $(1, x), (2, y), (3, y)$ }
Surjective but not injective

{ $(1, y), (2, y), (3, y)$ }
Neither injective nor surjective

{ $(1, y), (2, x), (3, y)$ }
Surjective but not injective

{ $(1, y), (2, y), (3, x)$ }
Surjective but not injective

{ $(1, y), (2, x), (3, x)$ }
Surjective but not injective

{ $(1, x), (2, y), (3, x)$ }
Surjective but not injective

B to A
{ $(x, 1), (y, 1)$ }
Neither injective nor surjective

{ $(x, 1), (y, 2)$ }
Injective but not surjective

{ $(x, 1), (y, 3)$ }
Injective but not surjective

{ $(x, 2), (y, 1)$ }
Injective but not surjective

{ $(x, 2), (y, 2)$ }
Neither injective nor surjective

{ $(x, 2), (y, 3)$ }
Injective but not surjective

{ $(x, 3), (y, 1)$ }
Injective but not surjective

{ $(x, 3), (y, 2)$ }
Injective but not surjective

{ $(x, 3), (y, 3)$ }
Neither injective nor surjective

E 25.2
a)
We aim to prove that the function $f : Z \to Z$ by $f (n) = 2 n + 1$ is bijective.

Proof.
(Injective)
Let $n_{1}, n_{2} \in Z$ . Assume $f (n_{1}) = f (n_{2})$ . Then

\begin{aligned} 2 n_{1} + 1 & = 2 n_{2} + 1 \\ 2 n_{1} & = 2 n_{2} \\ n_{1} & = n_{2} \end{aligned}

So $f$ is injective.
(Surjective)

Let $m \in Z$ . Define $n = \frac{m - 1}{2}$ . Note that

\begin{aligned} f (n) & = 2 n + 1 \\ = 2 (\frac{m - 1}{2}) + 1 \\ = m - 1 + 1 \\ = m \end{aligned}

So $f$ is surjective.

Since $f$ is both injective and surjective, it is bijective. $◻$

b)
We aim to prove that the function $g : R \to R$ by $g (x) = x^{2} + 2 x + 2$ is neither injective nor surjective.

Proof. (Injective)
Fix $z = 0$ and $y = - 2$ . We see that $g (z) = 2$ and $g (y) = 2$ so $g (x) = g (y)$ . But $x \neq y$ so $g$ is not surjective. $◻$

Proof. (Surjective)
Suppose that there exists some some point $(x, 0)$ within the function $g (x)$ . Then we can use the quadratic formula to find out what $x$ it is. We see

\begin{aligned} x & = \frac{- 2 \pm \sqrt{2^{2} - 4 (2)}}{2} \\ = \frac{- 2 \pm \sqrt{- 4}}{2} \\ = \frac{- 2 \pm 2 i}{2} \\ = - 1 \pm i \end{aligned}

We see that there are no real roots, so there is no point $(x, 0)$ within $g (x)$ . Since 0 is within the codomain but not defined within the function, the function is not surjective. $◻$

c)
We aim to prove that the function $h : Z \to Z$ by $h (n) = n + 3$ is bijective.
Proof.
(Injective)
Let $n_{1}, n_{2} \in Z$ . Assume that $h (n_{1}) = h (n_{2})$ . Then

\begin{aligned} n_{1} + 3 & = n_{2} + 3 \\ n_{1} & = n_{3} \end{aligned}

So $h$ is injective.

(Surjective)
Let $m \in Z$ . Define $n = m - 3$ . We see that that

\begin{aligned} h (n) & = n + 3 \\ = (m - 3) + 3 \\ = m \end{aligned}

So $h$ is surjective.

Since we have proven that $h$ is both surjective and injective, $h$ is bijective. $◻$

E 25.3
Define $f : Z_{5} \to Z_{5}$ by $f (\overset{―}{a}) = \overset{―}{2 a + 3}$ .
a)
Proof.
We aim to prove that for each $\overset{―}{a} \in Z_{5}$ , $f (\overset{―}{a})$ is unique.

$f (\overset{―}{0}) = \overset{―}{2 (0) + 3} = \overset{―}{3}$
$f (\overset{―}{1}) = \overset{―}{2 (1) + 3} = \overset{―}{0}$
$f (\overset{―}{2}) = \overset{―}{2 (2) + 3} = \overset{―}{2}$
$f (\overset{―}{3}) = \overset{―}{2 (3) + 3} = \overset{―}{4}$
$f (\overset{―}{4}) = \overset{―}{2 (4) + 3} = \overset{―}{1}$

Then $f$ is well defined.

b)
Based on part a, we see that $f = {(\overset{―}{0}, \overset{―}{3}), (\overset{―}{1}, \overset{―}{0}), (\overset{―}{2}, \overset{―}{2}), (\overset{―}{3}, \overset{―}{4}), (\overset{―}{4}, \overset{―}{1})}$ . We will prove $f$ is both surjective and injective.

Proof.
(Injective)
We see that no two distinct elements in the domain $Z_{5}$ are mapped to the same element in the codomain $Z_{5}$ . Thus, $f$ is injective since $f (a_{1}) = f (a_{2})$ implies $a_{1} = a_{2}$ . $◻$

(Surjective)
The codomain of $f$ is $Z_{5} = {\overset{―}{0}, \overset{―}{1}, \overset{―}{2}, \overset{―}{3}, \overset{―}{4}}$ . Since every element in $Z_{5}$ appears as a second component in the set of ordered pairs, $f$ is surjective.

Therefore, $f$ is bijective. $◻$

E 25.4
a) $f (x) = x^{2}$
Proof.
(Injective)
Fix $a = 1$ and $b = - 1$ . Then $f (a) = 1$ and $f (b) = 1$ but $b \neq a$ . So $f$ is not injective.

(Surjective)
Take $f (x) = - 1$ which is within the codomain. However, solving for $x$ , we see that $x = i$ which is not within the domain. In other words, squaring any real number cannot result in a negative number. So $f$ is not surjective. $◻$

b) $f (x) = \sqrt{x}$
Proof.
(Injective)
Let $x_{1}, x_{2} \in R$ . Assume that $f (x_{1}) = f (x_{2})$ . We see that

\sqrt{x_{1}} = \sqrt{x_{2}}

Squaring both sides we get

x_{1} = x_{2}

So $f$ is injective. $◻$

(Surjective)

Consider any negative number $f (x) = y$ with $y < 0$ . There is no $x \in R$ such that $\sqrt{x} = y$ because the square root of a nonnegative number is always nonnegative.

Therefore, there is no $x$ that maps to any negative $y$ , which means $f$ is not surjective.
$◻$

f (x) = {\begin{cases} x^{2} & if - 3 \leq x \leq 4, \\ x + 12 & otherwise. \end{cases}

Proof.

(Surjective)
There are two cases.
If $- 3 \leq x \leq 4$
Let $z \in R$ . Define $x = \sqrt{z}$ . We see that

\begin{aligned} f (x) & = x^{2} \\ = (\sqrt{z})^{2} \\ = z \end{aligned}

Since $f (x) = x + 12$ crosses the origin at $x = - 12$ , it is fine that $x^{2}$ will not result in any any negative values.

If $x < - 3$ or $x > 4$ .
Let $y \in R$ . Define $x = y - 12$ . We see that

\begin{aligned} f (x) & = x + 12 \\ = (y - 12) + 12 \\ = y \end{aligned}

So $f$ is surjective

(Injective)
Fix $a = - 12$ and $b = 0$ . Then $f (a) = 0$ and $f (b) = 0$ but $b \neq a$ . So $f$ is not injective.

d) $f (x) = x$
Proof.
(Injective)
Let $x_{1}, x_{2} \in R$ . Assume that $f (x_{1}) = f (x_{2})$ . This gives us:

x_{1} = x_{2}

Therefore, $f$ is injective. $◻$

(Surjective)
We aim to show that for every $y \in R$ , there exists an $x \in R$ such that $f (x) = y$ . This is true because for every $y$ , we can take $x = y$ , and we will have $f (x) = y$ . Therefore, $f$ is surjective. $◻$

E 25.5
a)
Let $x \in R - {2}$ . Then $f (x) = \frac{x - 3}{x - 2}$ . We want to show that $f (x) \neq 1$ .

Suppose $f (x) = 1$ . Then we have:

\frac{x - 3}{x - 2} = 1

Multiplying both sides by $x - 2$ (which is not zero since $x \neq 2$ ), we get:

x - 3 = x - 2

This simplifies to $- 3 = - 2$ , which is a contradiction. Therefore, $f (x) \neq 1$ for any $x \in R - {2}$ , and $f$ is indeed a function from $R - {2}$ to $R - {1}$ .

b) Let $x_{1}, x_{2} \in R - {2}$ . Assume $f (x_{1}) = f (x_{2})$ . We find

\begin{aligned} \frac{x_{1} - 3}{x_{1} - 2} & = \frac{x_{2} - 3}{x_{2} - 2} \\ (x_{2} - 2) (x_{1} - 3) & = (x_{1} - 2) (x_{2} - 3) \\ x_{1} x_{2} - 3 x_{2} - 2 x_{1} + 6 & = x_{1} x_{2} - 3 x_{1} - 2 x_{2} + 6 \\ - 3 x_{2} - 2 x_{1} + 3 x_{1} + 2 x_{2} & = - 3 x_{1} - 2 x_{2} + 3 x_{1} + 2 x_{2} \\ x_{1} - x_{2} & = 0 \\ x_{1} & = x_{2} \end{aligned}

So $f$ is injective.

c)
Let $b \in R - {1}$ . Fix

a = \frac{- 2 b + 3}{1 - b}

We now show that $a$ is in the domain of $f$ , which is $R - {2}$ . Since $b$ is not 1, the denominator $1 - b$ is not zero, and so $a$ is a well-defined real number. Furthermore, $a$ cannot be 2 because if $a$ were 2, then we would have:

2 = \frac{- 2 b + 3}{1 - b}

2 (1 - b) = - 2 b + 3

2 - 2 b = - 2 b + 3

2 = 3

which is a contradiction. Therefore, $a \neq 2$ .

Substituting $a$ back into $f (x)$ , we have:

f (\frac{- 2 b + 3}{1 - b}) = \frac{(\frac{- 2 b + 3}{1 - b}) - 3}{(\frac{- 2 b + 3}{1 - b}) - 2}

Multiplying the numerator and the denominator by $1 - b$ we get:

f (\frac{- 2 b + 3}{1 - b}) = \frac{- 2 b + 3 - 3 (1 - b)}{- 2 b + 3 - 2 (1 - b)}

= \frac{- 2 b + 3 - 3 + 3 b}{- 2 b + 3 - 2 + 2 b}

= \frac{b}{1}

= b

Therefore, for every $b \in R - {1}$ , there exists an $a \in R - {2}$ such that $f (a) = b$ . Hence, $f$ is surjective. $◻$

E 25.6
$f$ is surjective and not injective.

Surjective:
We want to show that for every $a \in Z$ , there exists at least one pair $(m, n) \in Z^{2}$ such that $f (m, n) = a$ . This is equivalent to solving the equation $3 m - 2 n = a$ for integers $m$ and $n$ .

Case 1: $a$ is even, $a = 2 k$ for some $k \in Z$ .
We can choose $m = 0$ and $n = - k$ , so $f (0, - k) = 0 - 2 (- k) = 2 k = a$ .

Case 2: $a$ is odd, $a = 2 k + 1$ for some $k \in Z$ .
We can choose $m = 1$ and $n = - k + 1$ , so $f (1, - k + 1) = 3 (1) - 2 (- k + 1) = 3 + 2 k - 2 = 2 k + 1 = a$ .

Since in both cases, we can find integers $m$ and $n$ such that $f (m, n) = a$ , the function is surjective.

Not injective.
Fix $m_{1} = 2, n_{1} = 3$ and $m_{2} = 4, n_{2} = 6$ . These are distinct pairs in $Z^{2}$ . For these values, $f (2, 3) = 3 (2) - 2 (3) = 6 - 6 = 0$ and $f (4, 6) = 3 (4) - 2 (6) = 12 - 12 = 0$ . Since $f (2, 3) = f (4, 6)$ but $(2, 3) \neq (4, 6)$ , the function is not injective.

Therefore, $f$ is surjective but not injective.

E 25.7
a) $[- 1, 1]$
b) $(0, \infty$ )
c) $(- \infty, \infty)$
d) $[0, \infty)$

E 25.8
a) A function $f$ from $R$ to $R$ assigns to each element $x$ in the domain exactly one element $f (x)$ in the codomain. If a vertical line intersects the graph of $f$ more than once, then there would be two different values of $f (x)$ for the same $x$ , which contradicts the definition of a function. Therefore, since $f$ is a function, every vertical line intersects the graph of $f$ at most once.

b) Since $f$ is a function defined on all of $R$ , for every $x$ in $R$ , there exists a unique $f (x)$ in $R$ . This means that for every vertical line (which corresponds to a value of $x$ ), there must be a point on the graph of $f$ where it intersects that line. If there were a vertical line that did not intersect the graph of $f$ , then there would be an $x$ in $R$ for which $f (x)$ is not defined, which would contradict the fact that $f$ is a function from $R$ to $R$ . Therefore, every vertical line intersects the graph of $f$ at least once.

c) To prove that $f$ is injective using the horizontal line test, we assume that for every horizontal line $y = k$ , the line intersects the graph of $f$ at most once. This means for any two points $a$ and $b$ in the domain, if $f (a) = f (b)$ , then it must be the case that $a = b$ . Thus, no two different $x$ -values map to the same $y$ -value, which by definition means $f$ is injective.

d) For $f$ to be surjective, every $y$ in the codomain $R$ must be the image of at least one $x$ in the domain. To prove this, we assume that for every real number $y$ , there exists a real number $x$ such that $f (x) = y$ . Thus, any horizontal line $y = k$ , for any real number $k$ , will intersect the graph of $f$ at least once, which means $f$ is surjective as it covers the entire codomain without any 'holes'.