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Section1.3Exercises

1

Suppose that

A={x:xN and x is even},B={x:xN and x is prime},C={x:xN and x is a multiple of 5}.\begin{align*} A & = \{ x : x \in \mathbb N \text{ and } x \text{ is even} \},\\ B & = \{x : x \in \mathbb N \text{ and } x \text{ is prime}\},\\ C & = \{ x : x \in \mathbb N \text{ and } x \text{ is a multiple of 5}\}. \end{align*}

Describe each of the following sets.

  1. ABA \cap B

  2. BCB \cap C

  3. ABA \cup B

  4. A(BC)A \cap (B \cup C)

Hint

(a) AB={2};A \cap B = \{ 2 \}\text{;} (b) BC={5}.B \cap C = \{ 5 \}\text{.}

2

If A={a,b,c},A = \{ a, b, c \}\text{,} B={1,2,3},B = \{ 1, 2, 3 \}\text{,} C={x},C = \{ x \}\text{,} and D=,D = \emptyset\text{,} list all of the elements in each of the following sets.

  1. A×BA \times B

  2. B×AB \times A

  3. A×B×CA \times B \times C

  4. A×DA \times D

Hint

(a) A×B={(a,1),(a,2),(a,3),(b,1),(b,2),(b,3),(c,1),(c,2),(c,3)};A \times B = \{ (a,1), (a,2), (a,3), (b,1), (b,2), (b,3), (c,1), (c,2), (c,3) \}\text{;} (d) A×D=.A \times D = \emptyset\text{.}

3

Find an example of two nonempty sets AA and BB for which A×B=B×AA \times B = B \times A is true.

4

Prove A=AA \cup \emptyset = A and A=.A \cap \emptyset = \emptyset\text{.}

5

Prove AB=BAA \cup B = B \cup A and AB=BA.A \cap B = B \cap A\text{.}

6

Prove A(BC)=(AB)(AC).A \cup (B \cap C) = (A \cup B) \cap (A \cup C)\text{.}

Hint

If xA(BC),x \in A \cup (B \cap C)\text{,} then either xAx \in A or xBC.x \in B \cap C\text{.} Thus, xABx \in A \cup B and AC.A \cup C\text{.} Hence, x(AB)(AC).x \in (A \cup B) \cap (A \cup C)\text{.} Therefore, A(BC)(AB)(AC).A \cup (B \cap C) \subset (A \cup B) \cap (A \cup C)\text{.} Conversely, if x(AB)(AC),x \in (A \cup B) \cap (A \cup C)\text{,} then xABx \in A \cup B and AC.A \cup C\text{.} Thus, xAx \in A or xx is in both BB and C.C\text{.} So xA(BC)x \in A \cup (B \cap C) and therefore (AB)(AC)A(BC).(A \cup B) \cap (A \cup C) \subset A \cup (B \cap C)\text{.} Hence, A(BC)=(AB)(AC).A \cup (B \cap C) = (A \cup B) \cap (A \cup C)\text{.}

7

Prove A(BC)=(AB)(AC).A \cap (B \cup C) = (A \cap B) \cup (A \cap C)\text{.}

8

Prove ABA \subset B if and only if AB=A.A \cap B = A\text{.}

9

Prove (AB)=AB.(A \cap B)' = A' \cup B'\text{.}

10

Prove AB=(AB)(AB)(BA).A \cup B = (A \cap B) \cup (A \setminus B) \cup (B \setminus A)\text{.}

Hint

(AB)(AB)(BA)=(AB)(AB)(BA)=[A(BB)](BA)=A(BA)=(AB)(AA)=AB.(A \cap B) \cup (A \setminus B) \cup (B \setminus A) = (A \cap B) \cup (A \cap B') \cup (B \cap A') = [A \cap (B \cup B')] \cup (B \cap A') = A \cup (B \cap A') = (A \cup B) \cap (A \cup A') = A \cup B\text{.}

11

Prove (AB)×C=(A×C)(B×C).(A \cup B) \times C = (A \times C ) \cup (B \times C)\text{.}

12

Prove (AB)B=.(A \cap B) \setminus B = \emptyset\text{.}

13

Prove (AB)B=AB.(A \cup B) \setminus B = A \setminus B\text{.}

14

Prove A(BC)=(AB)(AC).A \setminus (B \cup C) = (A \setminus B) \cap (A \setminus C)\text{.}

Hint

A(BC)=A(BC)=(AA)(BC)=(AB)(AC)=(AB)(AC).A \setminus (B \cup C) = A \cap (B \cup C)' = (A \cap A) \cap (B' \cap C') = (A \cap B') \cap (A \cap C') = (A \setminus B) \cap (A \setminus C)\text{.}

15

Prove A(BC)=(AB)(AC).A \cap (B \setminus C) = (A \cap B) \setminus (A \cap C)\text{.}

16

Prove (AB)(BA)=(AB)(AB).(A \setminus B) \cup (B \setminus A) = (A \cup B) \setminus (A \cap B)\text{.}

17

Which of the following relations f:QQf: {\mathbb Q} \rightarrow {\mathbb Q} define a mapping? In each case, supply a reason why ff is or is not a mapping.

  1. f(p/q)=p+1p2\displaystyle f(p/q) = \frac{p+ 1}{p - 2}

  2. f(p/q)=3p3q\displaystyle f(p/q) = \frac{3p}{3q}

  3. f(p/q)=p+qq2\displaystyle f(p/q) = \frac{p+q}{q^2}

  4. f(p/q)=3p27q2pq\displaystyle f(p/q) = \frac{3 p^2}{7 q^2} - \frac{p}{q}

Hint

(a) Not a map since f(2/3)f(2/3) is undefined; (b) this is a map; (c) not a map, since f(1/2)=3/4f(1/2) = 3/4 but f(2/4)=3/8;f(2/4)=3/8\text{;} (d) this is a map.

18

Determine which of the following functions are one-to-one and which are onto. If the function is not onto, determine its range.

  1. f:RRf: {\mathbb R} \rightarrow {\mathbb R} defined by f(x)=exf(x) = e^x

  2. f:ZZf: {\mathbb Z} \rightarrow {\mathbb Z} defined by f(n)=n2+3f(n) = n^2 + 3

  3. f:RRf: {\mathbb R} \rightarrow {\mathbb R} defined by f(x)=sinxf(x) = \sin x

  4. f:ZZf: {\mathbb Z} \rightarrow {\mathbb Z} defined by f(x)=x2f(x) = x^2

Hint

(a) ff is one-to-one but not onto. f(R)={xR:x>0}.f({\mathbb R} ) = \{ x \in {\mathbb R} : x \gt 0 \}\text{.} (c) ff is neither one-to-one nor onto. f(R)={x:1x1}.f(\mathbb R) = \{ x : -1 \leq x \leq 1 \}\text{.}

19

Let f:ABf :A \rightarrow B and g:BCg : B \rightarrow C be invertible mappings; that is, mappings such that f1f^{-1} and g1g^{-1} exist. Show that (gf)1=f1g1.(g \circ f)^{-1} =f^{-1} \circ g^{-1}\text{.}

20
  1. Define a function f:NNf: {\mathbb N} \rightarrow {\mathbb N} that is one-to-one but not onto.

  2. Define a function f:NNf: {\mathbb N} \rightarrow {\mathbb N} that is onto but not one-to-one.

Hint

(a) f(n)=n+1.f(n) = n + 1\text{.}

21

Prove the relation defined on R2{\mathbb R}^2 by (x1,y1)(x2,y2)(x_1, y_1 ) \sim (x_2, y_2) if x12+y12=x22+y22x_1^2 + y_1^2 = x_2^2 + y_2^2 is an equivalence relation.

22

Let f:ABf : A \rightarrow B and g:BCg : B \rightarrow C be maps.

  1. If ff and gg are both one-to-one functions, show that gfg \circ f is one-to-one.

  2. If gfg \circ f is onto, show that gg is onto.

  3. If gfg \circ f is one-to-one, show that ff is one-to-one.

  4. If gfg \circ f is one-to-one and ff is onto, show that gg is one-to-one.

  5. If gfg \circ f is onto and gg is one-to-one, show that ff is onto.

Hint

(a) Let x,yA.x, y \in A\text{.} Then g(f(x))=(gf)(x)=(gf)(y)=g(f(y)).g(f(x)) = (g \circ f)(x) = (g \circ f)(y) = g(f(y))\text{.} Thus, f(x)=f(y)f(x) = f(y) and x=y,x = y\text{,} so gfg \circ f is one-to-one. (b) Let cC,c \in C\text{,} then c=(gf)(x)=g(f(x))c = (g \circ f)(x) = g(f(x)) for some xA.x \in A\text{.} Since f(x)B,f(x) \in B\text{,} gg is onto.

23

Define a function on the real numbers by

f(x)=x+1x1.\begin{equation*} f(x) = \frac{x + 1}{x - 1}. \end{equation*}

What are the domain and range of f?f\text{?} What is the inverse of f?f\text{?} Compute ff1f \circ f^{-1} and f1f.f^{-1} \circ f\text{.}

Hint

f1(x)=(x+1)/(x1).f^{-1}(x) = (x+1)/(x-1)\text{.}

24

Let f:XYf: X \rightarrow Y be a map with A1,A2XA_1, A_2 \subset X and B1,B2Y.B_1, B_2 \subset Y\text{.}

  1. Prove f(A1A2)=f(A1)f(A2).f( A_1 \cup A_2 ) = f( A_1) \cup f( A_2 )\text{.}

  2. Prove f(A1A2)f(A1)f(A2).f( A_1 \cap A_2 ) \subset f( A_1) \cap f( A_2 )\text{.} Give an example in which equality fails.

  3. Prove f1(B1B2)=f1(B1)f1(B2),f^{-1}( B_1 \cup B_2 ) = f^{-1}( B_1) \cup f^{-1}(B_2 )\text{,} where

    f1(B)={xX:f(x)B}.\begin{equation*} f^{-1}(B) = \{ x \in X : f(x) \in B \}. \end{equation*}
  4. Prove f1(B1B2)=f1(B1)f1(B2).f^{-1}( B_1 \cap B_2 ) = f^{-1}( B_1) \cap f^{-1}( B_2 )\text{.}

  5. Prove f1(YB1)=Xf1(B1).f^{-1}( Y \setminus B_1 ) = X \setminus f^{-1}( B_1)\text{.}

Hint

(a) Let yf(A1A2).y \in f(A_1 \cup A_2)\text{.} Then there exists an xA1A2x \in A_1 \cup A_2 such that f(x)=y.f(x) = y\text{.} Hence, yf(A1)y \in f(A_1) or f(A2).f(A_2) \text{.} Therefore, yf(A1)f(A2).y \in f(A_1) \cup f(A_2)\text{.} Consequently, f(A1A2)f(A1)f(A2).f(A_1 \cup A_2) \subset f(A_1) \cup f(A_2)\text{.} Conversely, if yf(A1)f(A2),y \in f(A_1) \cup f(A_2)\text{,} then yf(A1)y \in f(A_1) or f(A2).f(A_2)\text{.} Hence, there exists an xx in A1A_1 or A2A_2 such that f(x)=y.f(x) = y\text{.} Thus, there exists an xA1A2x \in A_1 \cup A_2 such that f(x)=y.f(x) = y\text{.} Therefore, f(A1)f(A2)f(A1A2),f(A_1) \cup f(A_2) \subset f(A_1 \cup A_2)\text{,} and f(A1A2)=f(A1)f(A2).f(A_1 \cup A_2) = f(A_1) \cup f(A_2)\text{.}

25

Determine whether or not the following relations are equivalence relations on the given set. If the relation is an equivalence relation, describe the partition given by it. If the relation is not an equivalence relation, state why it fails to be one.

  1. xyx \sim y in R{\mathbb R} if xyx \geq y

  2. mnm \sim n in Z{\mathbb Z} if mn>0mn > 0

  3. xyx \sim y in R{\mathbb R} if xy4|x - y| \leq 4

  4. mnm \sim n in Z{\mathbb Z} if mn(mod6)m \equiv n \pmod{6}

Hint

(a) The relation fails to be symmetric. (b) The relation is not reflexive, since 0 is not equivalent to itself. (c) The relation is not transitive.

26

Define a relation \sim on R2{\mathbb R}^2 by stating that (a,b)(c,d)(a, b) \sim (c, d) if and only if a2+b2c2+d2.a^2 + b^2 \leq c^2 + d^2\text{.} Show that \sim is reflexive and transitive but not symmetric.

27

Show that an m×nm \times n matrix gives rise to a well-defined map from Rn{\mathbb R}^n to Rm.{\mathbb R}^m\text{.}

28

Find the error in the following argument by providing a counterexample. “The reflexive property is redundant in the axioms for an equivalence relation. If xy,x \sim y\text{,} then yxy \sim x by the symmetric property. Using the transitive property, we can deduce that xx.x \sim x\text{.}

Hint

Let X=N{2}X = {\mathbb N} \cup \{ \sqrt{2}\, \} and define xyx \sim y if x+yN.x + y \in {\mathbb N}\text{.}

29Projective Real Line

Define a relation on R2{(0,0)}{\mathbb R}^2 \setminus \{ (0,0) \} by letting (x1,y1)(x2,y2)(x_1, y_1) \sim (x_2, y_2) if there exists a nonzero real number λ\lambda such that (x1,y1)=(λx2,λy2).(x_1, y_1) = ( \lambda x_2, \lambda y_2)\text{.} Prove that \sim defines an equivalence relation on R2(0,0).{\mathbb R}^2 \setminus (0,0)\text{.} What are the corresponding equivalence classes? This equivalence relation defines the projective line, denoted by P(R),{\mathbb P}({\mathbb R}) \text{,} which is very important in geometry.