# How to Prove It - Solutions

## Chapter - 5, Functions

### Summary

• Function : Suppose $F$ is a relation from $A$ to $B$. Then $F$ is called a function from $A$ to $B$ if for every $a \in A$ there is exactly one $b \in B$ such that $(a, b) \in F$. In other words, $F$ is a function from $A$ to $B$ means:
$\forall a \in A \exists ! b \in B((a,b) \in F)$.
To indicate that $F$ is a function from $A$ to $B$, we will write $F : A \to B$.
• Suppose $f$ and $g$ are functions from $A$ to $B$. If $\forall a \in A( f (a) = g(a))$, then $f = g$. This theorem can be used in proving two functions are equal.
• As can be seen from the definition of function, funtionsa are special kind of relations. Thus all the theorems of relations applies to functions also. Domain of the function $f$ from $A$ to $B$ must contain each element of the range as first coordinate of an ordered pair in $f$. The elements of the range of $f$ will be the second coordinates of all the ordered pairs in $f$. Thus range need not be all of $B$. Also this second coordinate is also sometimes called as image of the first coordinate.
• As noted earlier, since function is a relation, all definitions that apply to relation also aplly to function. Thus definition of composition of relation also apply here as follows:
• Composition of function: Suppose $f : A \to B$ and $g : B \to C$. Then $g \circ f : A \to C$, and for any $a \in A$, the value of $g \circ f$ at $a$ is given by the formula $(g \circ f )(a) = g( f (a))$.

Soln1

(a) True

(b) False

(c) True

Soln2

(a) False. Because there is no image for point $d$.

(b) $f$ is not a function. Because for eg: “ape” has $3$ images. $g$ is a function.

(c) True.

Soln3

(a) $f(a) = b, f(b) = b, f(c) = a$.

(b) $f(2) = 2^2 - 2 \times 2 = 0$.

(c) $f(\pi) = 3, f(- \pi) = -4$.

Soln4

(a) Capital of Italy = Rome.

(b) $\{1,2,3\} \setminus \{1,3\} = \{2\}$.

(c) $f(2) = (3,1)$.

Soln5

$L \circ H$: It is an identity function on all countries.

$H \circ L$: This function maps a city $c$ to another city $C$ such that $C$ is the capital of the country of of the city $c$.

Soln6

$f \circ g (x) = f(g(x)) = \frac 1 {(2x-1)^2 + 2} = \frac 1 {4x^2 -4x + 3}$.

$g \circ f (x) = g(f(x)) = 2 \times { \frac 1 {x^2 + 2} } - 1$.
$= \frac 2 {x^2 + 2} - 1$.
$= \frac {2 - x^2 - 2} {x^2 + 2}$.
$= \frac {-x^2} {x^2 + 2}$.

Soln7

(a)

To prove $f \upharpoonright C$ is a function,we need to prove that for every $c \in C$ there exists a unique $b \in B$ such that $(c,b) \in f \upharpoonright C$.

Existence:

Suppose $c \in C$. Since $C \subseteq A$, $c \in A$. Since $f : A \to B$, it follows that there exist an element $b \in B$ such that $(c,b) \in f$. Thus $b \in B$. It follows that $(c,b) \in C \times B$. Since $(c,b) \in f$ and $(c,b) \in C \times B$. It follows that $(c,b) \in f \upharpoonright C$.

Uniqueness:

Suppose $(c,b_1) \in f \upharpoonright C$ and $(c,b_2) \in f \upharpoonright C$. Thus $(c,b_1) \in f$ and $(c,b_2) \in f$. Since $f$ is a function, it follows that $b_1 = b_2$. Thus there exist only one element $b$ such that $(c,b) \in f \upharpoonright C$.

Suppose $(c,b) \in f \upharpoonright C$. Thus $b = f(c)$. Since $(c,b) \in f \upharpoonright C$, $b = f \upharpoonright C (c)$. Thus $b = f(c) = f \upharpoonright C (c)$.

(b)

($\to$)Suppose $g = f \upharpoonright C$. Suppose $(c,b) \in g$. Thus $(c,b) \in f \upharpoonright C$. It follows that $(c,b) \in f$. Thus $g \subseteq f$.

($\leftarrow$)Suppose $g \subseteq f$. Suppose $c \in C$. Since $g: C \to B$, there must exist an element $b \in B$ such that $b = g(c)$. Thus $(c,b) \in g$. Since $g \subseteq f$, it follows $(c,b) \in f$. Thus $b = f(c)$. Since $b \in B$ and $c \in C$, it follows $(c,b) \in C \times B$. Since $(c,b) \in C \times B$ and $(c,b) \in f$, if follows that $(c,b) \in f \upharpoonright C$. Thus $b = g(c) = f(c) = f \upharpoonright C(c)$. Since $c$ is arbitrary, it follows that $g = f \upharpoonright C$.

(c)

We have $h : \mathbb R \to \mathbb Z$ and $g = \mathbb Z \to \mathbb Z$. By the definition of restiction, we have $h \upharpoonright \mathbb Z = h \cap (\mathbb Z \times \mathbb Z)$. Thus $h \cap (\mathbb Z \times \mathbb Z) = \{(x,y) \in \mathbb R \times \mathbb Z \, \vert \, y = 2x+3 \} \cap (\mathbb Z \times \mathbb Z)$. Thus we have $h \cap (\mathbb Z \times \mathbb Z) = \{(x,y) \in \mathbb Z \times \mathbb Z \, \vert \, y = 2x+3 \}$. Thus $h \upharpoonright \mathbb Z = h \cap (\mathbb Z \times \mathbb Z) = g$.

Soln8

To prove this we need to prove existence and uniqueness.

Existence:

Since $i_A$ is reflexive, transitive and symmetric, it is a equivalence relation. Also since for each value $x$ in $A$ there is a unique value $x$ \in A such that $(x,x) \in i_A$. Thus $i_A$ is a function and a equivalence relation.

Uniqueness:

Suppose there exists a function say $f$, such that it is a equivalence relation. Suppose $(x,y) \in f$. Thus we have:

$(x,y) \in f$. iff
Since $f$ is symmetric,
$(x,y) \in f \land (y,x) \in f$. iff
Since $f$ is transitive,
$(x,x) \in f$. iff
Since $f$ is a function and $(x,y) \in f$ and $(x,x) \in f$,
$x = y$

Since $(x,y)$ are arbitrary, it follows that $f = i_A$.

Thus $i_A$ is the only unique function.

Soln9

(a)

To prove that $f \cup g$ is a function from $A \cup B$ to $C$, we need to prove that for all $x \in A \cup B$, there exist a image of $x$ in $C$. Also we need to prove that for each $x$, this image is unique. Thus we need to do existence and uniqueness proof.

Existence:

Suppose $x \in A \cup B$. Thus we have following cases:

• Case 1: $x \in A$.
Since $x \in A$ and since $f$ is a function on $A \to C$. Thus there exist a $y \in C$ such that $(x,y) \in f$. Thus $(x,y) \in f \cup g$. It follows that $\exists y \in B((y,x) \in f \cup g)$.

• Case 2: $x \in B$. Since $x \in B$ and since $g$ is a function on $B \to C$. Thus there exist a $y \in C$ such that $(x,y) \in g$. Thus $(x,y) \in f \cup g$. It follows that $\exists y \in B((y,x) \in f \cup g)$.

Thus from both cases $\exists y \in B((y,x) \in f \cup g)$. Since $x$ is arbitrary, it follows that $\forall x \in A \cup B \exists y \in C((y,x) \in f \cup g)$.

Uniqueness:

Suppose $x \in A \cup B$ and $y_1, y_2 \in C$ such that $(x,y_1) \in f \cup g$ and $(x,y_2) \in f \cup g$. Thus we have following possible cases:

• Case 1: $(x,y_1) \in f$ and $(x,y_2) \in f$.
Since $f$ is a function, it follows that $y_1 = y_2$.

• Case 2: $(x,y_1) \in g$ and $(x,y_2) \in g$.
Since $g$ is a function, it follows that $y_1 = y_2$.

• Case 3: $(x,y_1) \in f$ and $(x,y_2) \in g$.
Thus $x \in A$ and $x \in B$. But this is not possible since $A \cap B = \phi$.

• Case 4: $(x,y_1) \in g$ and $(x,y_2) \in f$.
Thus $x \in B$ and $x \in A$. But this is not possible since $A \cap B = \phi$.

Thus from all possible cases, $y_1 = y_2$. Thus there is only one image of $x \in A \cup B$ in the set $C$. Thus $f \cup g : A \cup B \to C$ is a function.

(b)

($\to$)Suppose $f \cup g: A \cup B \to C$. Suppose $(x,y) \in f \upharpoonright (A \cap B)$. It follows that $(x,y) \in f$ and $(x,y) \in (A \cap B) \times C$. Since $(x,y) \in f$, it follows that $(x,y) \in f \cup g$. Also since $x \in B$, and $g: B \to C$, there exist an element $y'$ such that $(x,y') \in g$. Since $(x,y') \in g$, it follows that $(x,y') \in f \cup g$. Since $f \cup g: A \cup B \to C$, it follows that $y = y'$. Thus $(x,y) \in g$. Since $(x,y) \in (A \cap B) \times C$ and $(x,y) \in g$, it follows that $(x,y) \in f \upharpoonright (A \cap B)$. Since $(x,y)$ are arbitrary, it follows $f \upharpoonright (A \cap B) \subseteq g \upharpoonright (A \cap B)$.
Using similar argument it can be proved that $g \upharpoonright (A \cap B) \subseteq f \upharpoonright (A \cap B)$. Thus we can conclude that $f \upharpoonright (A \cap B) = g \upharpoonright (A \cap B)$.

($\leftarrow$)Suppose $f \upharpoonright (A \cap B) = g \upharpoonright (A \cap B)$. To prove $f \cup g: A \cup B \to C$, we need to do existence and uniqueness proofs. Clearly in this case existence proof is exactly same as in part(a). Thus we shall only do the uniqueness proof:

Suppose $(x,y_1) \in f \cup g$ and $(x,y_2) \in f \cup g$. Thus we have following possible cases:

• Case 1: $(x,y_1) \in f$ and $(x,y_2) \in f$.
Since $f$ is a function, it follows that $y_1 = y_2$.

• Case 2: $(x,y_1) \in g$ and $(x,y_2) \in g$.
Since $g$ is a function, it follows that $y_1 = y_2$.

• Case 3: $(x,y_1) \in f$ and $(x,y_2) \in g$.
Thus $x \in A \cap B$. It follows that $(x,y_1) \in (A \cap B) \times C$ and $(x,y_2) \in (A \cap B) \times C$. Since $(x,y_1) \in f$ and
$(x,y_1) \in (A \cap B) \times C$, it follows that $(x,y_1) \in f \upharpoonright (A \cap B)$. Similarly since $(x,y_2) \in g$ and
$(x,y_2) \in (A \cap B) \times C$, it follows that $(x,y_2) \in g \upharpoonright (A \cap B)$. Since $f \upharpoonright (A \cap B) = g \upharpoonright (A \cap B)$, if follows that $(x,y_1)$ and $(x,y_2)$ both exist in $f$ and $g$. Thus $y_1 = y_2$.

• Case 4: $(x,y_1) \in g$ and $(x,y_2) \in f$.
This case is exactly same as Case 3, just need to swap $f$ and $g$ in the case 3.

Thus from all the possible cases $y_1 = y_2$.

Thus $f \cup g: A \cup B \to C$.

Soln10

(a)

Existence:

Suppose $b \in B$. Since $Dom(S) = B$, there must exist $c \in C$ such that $(b,c) \in S$. Since $b$ is arbitrary, it follows that for all $b \in B$ there exist an element $c \in C$ such that $(b,c) \in S$.

Uniqueness:

Suppose $b \in B$ and $c_1, c_2 \in C$ such that $(b,c_1) \in S$ and $(b,c_2) \in S$. Since $Ran(R) = B$, it follows that there must exist an element $a \in A$ such that $(a,b) \in R$. Thus $(a,c_1) \in S \circ R$ and $(a,c_2) \in S \circ R$. But $S \circ R$ is a function, it follows that $c_1 = c_2$.

Thus we can conclude that $S$ is a function.

(b)

$A = \{1,2\}$
$B = \{a,b,c\}$
$C = \{u,v\}$

$R = \{ (1,a), (1,b), (2,c) \}$.
$S = \{ (a,u), (b,u), (c,v) \}$.
$S \circ R = \{ (1,u), (2,v) \}$.

Clearly, $S$ and $S \circ R$ are functions but $R$ is not a function.

Soln11

(a)

Suppose $S$ is reflexive. Suppose $x \in A$ is an arbitrary element. Since $f: A \to B$, thus $f(x) \in B$. Since $S$ is relation on $B$ and $S$ is reflexive, it follows that $(f(x),f(x)) \in S$. Thus $(x,x) \in R$. Thus $R$ is reflexive.

(b)

Suppose $S$ is symmetric. Suppose $(x,y) \in R$. Thus $(f(x), f(y)) \in S$. Since $S$ is symmetric, it follows that $(f(y),f(x)) \in S$. Thus $(y,x) \in R$. Thus $R$ symmetric.

(c)

Suppose $S$ is transitive. Suppose $x,y,z \in A$ such that $(x,y) \in R$ and $(y,z) \in R$. Thus $(f(x),f(y)) \in S$ and $(f(y),(f(z))) \in S$. Since $S$ is transitive, it follows that $(f(x),f(z)) \in S$. Thus $(x,z) \in R$. Thus $R$ is transitive.

Soln12

(a) False.

Counter example:

$A = \{1,2,3\}, B = \{a,b,c,d\}, f = \{(1,a),(2,b),(3,c)\}, R = \{(1,1), (2,2), (3,3), (4,4) \}$.

Clearly $(d,d) \notin S$, because there is no element $u$ in $A$ such that $f(x) = d$.

(b) True.

Suppose $R$ is symmetric. Suppose $(x,y) \in S$. Thus there must exist some $u \in A$ such that $f(u) = x$ and there must exist some $v \in A$ such that $f(v) = x$ and $(u,v) \in R$. Since $R$ is symmetric, it follows that $(v,u) \in R$. Since $y = f(v)$ and $x = f(u)$, it follows that $(y,x) \in S$. Thus $S$ is symmetric.

(c) False.

$A = \{1.2.3.4\}$.
$B = \{u,v,w\}$.
$f = \{(1,u),(2,v),(3,v),(4,w)\}$.
$R = \{(1,2), (3,4) \}$.

Clearly $R$ is transitive. Clearly $(u,v) \in S$, because $(1,2) \in R$. Similarly $(v,w) \in S$, since $(3,4) \in R$. But $(u,w) \notin S$ because $(1,4) \notin R$.

Soln13

(a)

True. Suppose $R$ is reflexive. Suppose $f \in \mathcal F$. Suppose $x \in A$. Since $f$ is a function, $f(x) \in B$ will be a unique value. Since $f(x) \in B$, and $R$ is reflexive, it follows that $(f(x), f(x)) \in R$. Since $x$ is arbitrary, $\forall x \in A((f(x), f(x)) \in R)$. Thus $(f,f) \in S$. Thus $S$ is reflexive.

(b)

True. Suppose $R$ is symmetric. Suppose $(f,g) \in \mathcal F$. Thus $\forall x \in A((f(x), g(x)) \in R)$. Since $R$ is transitive, it follows that $\forall x \in A((g(x), f(x)) \in R)$. Thus $(g,f) \in S$. Thus $R$ is symmetric.

(c)

True. Suppose $R$ is transitive. Suppose $(f,g) \in \mathcal F$ and $(g,h) \in \mathcal F$. Thus $\forall x \in A((f(x), g(x)) \in R)$ and Thus $\forall x \in A((g(x), h(x)) \in R)$. Since $R$ is transitive, it follows that $\forall x \in A((f(x), h(x)) \in R)$. Thus $(f,h) \in S$. Thus $S$ is transitive.

Soln14

(a)

Suppose $x \in A$. Thus $f(x) = a$. Also, $f \circ g (x) = f (g(x)) = a$. Thus $f \circ g (x) = f(x)$. Since $x$ is arbitrary, it follows $f \circ g = f$(using theorem for that talks about two functions are equals if for all values of $x$, corresponding images are also equal).

(b)

Suppose $f \circ g = f$. Since $g$ is arbitrary, $g$ can be any function. Suppose $g$ is a constant function such that $\forall x \in A(g(x) = b)$. Suppose $x \in A$ is an arbitrary element. Thus $f \circ g (x) = f(g(x)) = f(b)$. Since $f \circ g = f$, it follows $f \circ g(x) = f(x)$. It follows that $f(b) = f(x)$. Since $f$ is a function, $f(b)$ is a unique value. Since $x$ is arbitrary, it follows that $\forall x \in A(f(x) = f(b)$. Thus $f(x)$ is a constant function.

Soln15

(a)

Clearly for $a = 0$, we have $\vert x \vert = x$. Thus $(f,g) \in R$.

(b)

Reflexive:

Suppose $f \in \mathcal F$. Suppose $a$ is any real number, clearly $f(x) = f(x)$ for all $x > a$. Thus $(f,f) \in R$. Thus $R$ is reflexive.

Transitive

Suppose $(f,g) \in R$ and $(g,h) \in R$. Since $(f,g) \in R$, it follows that there is an element $a \in R$ such that $\forall x > a(f(x) = g(x))$. Similarly since $(g,h) \in R$, it follows that there is an element $b \in R$ such that $\forall x > b (g(x) = h(x))$.

Now there are two cases, $a \ge b$ or $b > a$. If $a \ge b$, then suppose $x > a$ is any arbitrary number. Then $f(x) = g(x)$ and $g(x) = h(x)$. Thus $f(x) = h(x)$.
Similarly if $b > a$, and suppose $x > b$ is any arbitrary number. Then also we have $f(x) = g(x) = h(x)$.
Thus in either case there exist a number $c$ such that $\forall x > c (f(x) = h(x))$. Thus $R$ is transitive.

Symmetric:

Suppose $(f,g) \in R$. Thus there is an element $a \in R$ such that $\forall x > a(f(x) = g(x))$. Or we can also say that $\forall x > a(g(x) = f(x))$. It follows that $(g,f) \in R$. Thus $R$ is symmetric.

Soln16

(a)

Suppose $a = 3, c = 8$. Then for any $x > a$,

$\vert f(x) \vert = \vert 7x + 3 \vert = 7x + 3 < 7x + x = 8x < 8x^2 = c \vert g(x) \vert$.

Thus $f \in O(g)$.

We will prove it by contradiction. Suppose $g \in O(f)$. Then for some $a \in Z^+$ and $c \in R^+$, we have $\forall x > a (\vert g(x) \vert\le c \vert f(x) \vert)$. Thus $\forall x > a (x^2 \le c (7x + 3)$. Suppose $x > 10c$ is an arbitrary number such that $x > a$. Thus $x^2 > 10cx$. Also, since $g \in O(f)$,we have $x^2 \le c(7x+3) < c(7x + 3x) = 10cx$. Thus $x^2 < 10cx$. But $x^2 > 10cx$. Thus we have a contradiction. Thus $f \notin O(g)$.

(b)

If a relation is preorder, it means it is reflexive and transitive. Thus we will prove these two properties to prove the relation as preorder.

Suppose $f \in \mathcal F$. Thus for $a = 1$ and $c = 1$, we have $\forall x > 1 (f(x) \le f(x)$. Thus $(f,f) \in S$. Thus $S$ is reflexive.

Suppose $(f,g) \in S$ and $(g,h) \in S$. Thus there exist $a_1, c_1, a_2, c_2$ such that $\forall x > a_1 (\vert f(x) \vert \le c_1 \vert g(x) \vert)$ and $\forall x > a_2 (\vert g(x) \vert \le c_2 \vert h(x) \vert)$. Suppose $a$ is the maximum of $a_1, a_2$. Suppose $x > a$, it follows that $\vert f(x) \vert \le c_1 \vert g(x) \vert \le c_1 c_2 \vert h(x) \vert$. Since $x$ is arbitrary, we have $\forall x > a (\vert f(x) \vert \le c \vert h(x) \vert)$, where $c = c_1 c_2$. Thus $(f,h) \in S$. Thus $S$ is transitive.

To prove that $S$ is not partial order, we shall prove that $S$ is not antisymmetric. We will take a counter example to prove $S$ is not symmetric. Suppose $f(x) = x$ and $g(x) = 2x$. Then $\forall x > 0 (\vert f(x) \vert \le \vert g(x) \vert )$ and $\forall x > 0 (\vert g(x) \vert \le 2 \vert f(x) \vert )$. Thus $(f,g) \in S$ and $(g,f) \in S$ but $f \ne g$. Thus $S$ is not antisymmetric.

(c)

Suppose $f_1 \in O(g)$ and $f_2 \in O(g)$. Thus there exist $a_1, c_1, a_2, c_2$ such that $\forall x > a_1 (\vert f_1(x) \vert \le c_1 \vert g(x) \vert)$ and $\forall x > a_2 (\vert f_2(x) \vert \le c_2 \vert g(x) \vert)$. Suppose $a$ is maximum of $a_1, a_2$. Suppose $x > a$ is any arbitrary number. Thus we have $\vert f_1(x) \vert \le c_1 \vert g(x) \vert$ and $\vert f_2(x) \vert \le c_2 \vert g(x) \vert$. Now $f(x) = sf_1(x) + tf_2(x)$. Thus by triangular inequality, we have $\vert sf_1(x) + tf_2(x) \vert \le \vert s \vert \vert f_1(x) \vert + \vert t \vert \vert f_2(x) \vert$.
Since $\vert f_1(x) \vert \le c_1 \vert g(x) \vert$ and $\vert f_2(x) \vert \le c_2 \vert g(x) \vert$, it follows that $\vert sf_1(x) + tf_2(x) \vert \le \vert s \vert \vert f_1(x) \vert + \vert t \vert \vert f_2(x) \vert \le \vert s \vert c_1 \vert g(x) \vert + \vert t \vert c_2 \vert g(x) \vert \le (c_1 \vert s \vert + c_2 \vert t \vert ) \vert g(x) \vert$. Thus $\vert f(x) \vert \le c \vert g(x) \vert$, where $c = (c_1 \vert s \vert + c_2 \vert t \vert )$. Thus $f \in O(g)$.

Soln17

(a)

Suppose $x \in A$. Since $g: A \to B$, $g(x) = g(x)$. Thus $(x,x) \in R$. Thus $R$ is reflexive.

Suppose $(x,y) \in R$. Thus $g(x) = g(y)$. Or we can also say $g(y) = g(x)$. Thus $(y,x) \in R$. Thus $R$ is symmetric.

Suppose $(x,y) \in R$ and $(y,z) \in R$. Thus $g(x) = g(y) = g(z)$. Since $g(x) = g(z)$, it follows $(x,z) \in R$. Thus $R$ is transitive.

Since $R$ is reflexive, symmetric and transitive, it follows that $R$ is an equivalence relation.

(b)

Suppose $R$ is an equivalence relation on $A$. Suppose $g: A \to A/R$ and $g(x) = [x]_R$.

($\to$)Suppose $(x,y) \in R$. Since $R$ is an equivalence relation, it follows that $[x]_R = [y]_R$. Thus $g(x) = g(y)$. Thus $R \subseteq \{(x,y) \in A \times A \, \vert \, g(x) = g(y) \}$.

($\leftarrow$)suppose $(x,y) \in \{(x,y) \in A \times A \, \vert \, g(x) = g(y) \}$. Since $g(x) = [x]_R$, it follows that $[x]_R = [y]_R$. Thus $(x,y) \in R$. Thus $\{(x,y) \in A \times A \, \vert \, g(x) = g(y) \} \subseteq R$.

Thus from both directions, we have $R = \{(x,y) \in A \times A \, \vert \, g(x) = g(y) \}$.

Soln18

(a)

Suppose $R$ is compatible with $f$.

Suppose $h = \{ (X,y) \in A/R \times B \, \vert \, \exists x \in X (f(x) = y) \}$.

Existence:

Suppose $X \in A/R$ and $y \in B$ such that $(X,y) \in h$. Thus there exist an $x$ such that $x \in X$ and $y = f(x)$. Since $x \in X$ and $X \in A/R$, it follows that $X = [x]_R$. Thus $([x]_R,f(x)) \in h$. Or $h([x]_R) = f(x)$. Thus such a relation exists.

Uniqueness:

Suppose there is another relation $k$. Suppose $(X,y) \in k$ such that $X = [x]_R$, $y = f(x)$ and $h(X) = f(x)$. Suppose $p \in [x]_R$. Thus $pRx$. Since $R$ is compatible with $f$, it follows that $f(p) = f(x)$. Thus there exists a $p$ such that $p \in [x]_R$ and $p = f(x)$. It follows that $([x]_p,f(x)) \in h$(where h is the function we defined). Or $(X,y) \in h$. Since $(X,y)$ are arbitrary, it follows that $k = h$.

Note: Not confident about the solution.

(b)

Suppose $x \in A$ and $y \in A$. Suppose $xRy$. Since $x \in A$, $h([x]_R) = f(x)$. Similarly, since $y \in A$, $h([y]_R) = f(y)$. Since $xRy$, it follows that $[x]_R = [y]_R$. Since $f$ is a function, it follows $h([x]_R) = h([y]_R)$. Thus $f(x) = f(y)$. Thus if $xRy$, then $f(x) = f(y)$. Since $x,y$ are arbitrary, it follows that $\forall x \in A \forall y \in A(xRy \to f(x) = f(y))$. Thus $R$ is compatible with $f$.

Soln19

(a)

Suppose $xRy$, then $x - y = 5k$. Thus $x = y + 5k$. Thus $x^2 = y^2 + 10yk + 25k^2$, or $x^2 - y^2 = 5k(2y + 5k)$. Thus $(x^2,y^2) \in R$. Or we can also say that if $xRy$ then $x^2Ry^2$.

Now consider the function $h = \{ (X,Y) \in Z/R \times Z/R \, \vert \, \exists x \in X \exists y \in Y(y = x^2) \}$.

Now we will prove that $h$ is the unique function to satisfy $h([x]_R) = [x^2]_R$.

Existence:

Suppose $(X,Y) \in h$. Thus there exists an element $x \in X$ and $y \in Y$ such that $y = x^2$. Since $R$ is an equivalence relation, it follows that $X = [x]_R$ and $Y = [y]_R = [x^2]_R$. Thus $([x]_R,[x^2]_R) \in h$, or $h([x]_R) = [x^2]_R$.

Uniqueness:

Suppose $k: Z/R \to Z/R$ is a function such that $k([x]_R) = [x^2]_R$. Suppose $p \in [x]_R$. Thus $pRx$. We proved above that if $pRx$ then $p^2Rx^2$. Thus $[x^2]_R = [p^2]_R$. It is also clear that $p^2 \in [p^2]_R$, or $p^2 \in [x^2]_R$. Thus there exist a $p \in [x]_R$ such $p^2 \in [x^2]_R$. Thus $([x]_R,[x^2]_R) \in h$. Since $([x]_R,[x^2]_R)$ is arbitrary, it follows that $k = h$.

(b)

Here it will be enough to show just one counter example. Suppose $x = 0, y = 5$. Clearly $(x,y) \in R$. If $h: Z/R \to Z/R$ is a function, it follows that, $h([o]_R) = h(_R$. Now $h(_R) = [2^0]_R = _R$ and $h(_R) = [2^5]_R = _R$. Thus $_R = _R$. But this is not correct because $32 - 1 \ne 5k$. Thus $h$ is not a function.