25 - Injective and surjective functions

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Injective

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

1. No
2. Yes
3. Yes
4. Yes

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

Surjective
Every $y$ (second component) is defined

Bijective = BOTH

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

Scratch work:
injective
Let $a_1,a_2\in \mathbb{R}-\{1\}$. Assume $f(a_1)=f(a_2)$.

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

Proof.
(Injective)
Let $a_1,a_2\in \mathbb{R}-\{1\}$. Assume that $f(a_1)=f(a_2)$.
Then

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

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

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 $f:\mathbb{Z}\to \mathbb{Z}$ by $f(n)=2n+1$ is bijective.

Proof.
(Injective)
Let $n_1,n_2\in \mathbb{Z}$. Assume $f(n_1)=f(n_2)$. Then

So $f$ is injective.
(Surjective)

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

So $f$ is surjective.

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

b)
We aim to prove that the function $g:\mathbb{R}\to \mathbb{R}$ by $g(x)=x^2+2x+2$ 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 $x\ne y$ 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

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 $h:\mathbb{Z}\to \mathbb{Z}$ by $h(n)=n+3$ is bijective.
Proof.
(Injective)
Let $n_1,n_2\in \mathbb{Z}$. Assume that $h(n_1)=h(n_2)$. Then

So $h$ is injective.

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

So $h$ is surjective.

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

E 25.3
Define $f:{\mathbb{Z}}_5\to {\mathbb{Z}}_5$ by $f(\bar{a})=\overline{2a+3}$.
a)
Proof.
We aim to prove that for each $\bar{a}\in {\mathbb{Z}}_5$, $f(\bar{a})$ is unique.

1. $f(\bar{0})=\overline{2(0)+3}=\bar{3}$
2. $f(\bar{1})=\overline{2(1)+3}=\bar{0}$
3. $f(\bar{2})=\overline{2(2)+3}=\bar{2}$
4. $f(\bar{3})=\overline{2(3)+3}=\bar{4}$
5. $f(\bar{4})=\overline{2(4)+3}=\bar{1}$

Then $f$ is well defined.

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

Proof.
(Injective)
We see that no two distinct elements in the domain ${\mathbb{Z}}_5$ are mapped to the same element in the codomain ${\mathbb{Z}}_5$. Thus, $f$ is injective since $f(a_1)=f(a_2)$ implies $a_1=a_2$. $◻$

(Surjective)
The codomain of $f$ is ${\mathbb{Z}}_5=\{\bar{0},\bar{1},\bar{2},\bar{3},\bar{4}\}$. Since every element in ${\mathbb{Z}}_5$ 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 $b\ne a$. 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 $x_1,x_2\in \mathbb{R}$. Assume that $f(x_1)=f(x_2)$. We see that

Squaring both sides we get

So $f$ is injective. $◻$

(Surjective)

Consider any negative number $f(x)=y$ with $y<0$. There is no $x\in \mathbb{R}$ 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.
$◻$

c)

Proof.

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

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 $y\in \mathbb{R}$. Define $x=y-12$. We see that

So $f$ is surjective

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

d) $f(x)=x$
Proof.
(Injective)
Let $x_1,x_2\in \mathbb{R}$. Assume that $f(x_1)=f(x_2)$. This gives us:

Therefore, $f$ is injective. $◻$

(Surjective)
We aim to show that for every $y\in \mathbb{R}$, there exists an $x\in \mathbb{R}$ 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 $x\in \mathbb{R}-\{2\}$. Then $f(x)=\frac{x-3}{x-2}$. We want to show that $f(x)\ne 1$.

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

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

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

b) Let $x_1,x_2\in \mathbb{R}-\{2\}$. Assume $f(x_1)=f(x_2)$. We find

So $f$ is injective.

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

We now show that $a$ is in the domain of $f$, which is $\mathbb{R}-\{2\}$. 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:

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

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

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

Therefore, for every $b\in \mathbb{R}-\{1\}$, there exists an $a\in \mathbb{R}-\{2\}$ 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 $a\in \mathbb{Z}$, there exists at least one pair $(m,n)\in {\mathbb{Z}}^2$ such that $f(m,n)=a$. This is equivalent to solving the equation $3m-2n=a$ for integers $m$ and $n$.

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

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

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 ${\mathbb{Z}}^2$. 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 $(2,3)\ne (4,6)$, the function is not injective.

Therefore, $f$ is surjective but not injective.

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

E 25.8
a) A function $f$ from $\mathbb{R}$ to $\mathbb{R}$ 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 $\mathbb{R}$, for every $x$ in $\mathbb{R}$, there exists a unique $f(x)$ in $\mathbb{R}$. 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 $\mathbb{R}$ for which $f(x)$ is not defined, which would contradict the fact that $f$ is a function from $\mathbb{R}$ to $\mathbb{R}$. 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 $\mathbb{R}$ 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'.