Download Math 3000 Section 003 Intro to Abstract Math Homework 8

Math 3000 Section 003 Intro to Abstract Math Homework 8
Department of Mathematical and Statistical Sciences
University of Colorado Denver, Spring 2012
Solutions (April 26, 2012)
Please note that these solutions are only suggestions; different answers and proofs are also always possible.
• Section 9.1: The Definition of a Function
1. Exercise 9.6: In each of the following, a function fi : Ai → R (1 ≤ i ≤ 5) is defined, where the
domain Ai consists of all real numbers x for which fi (x) is defined. In each case, determine the
domain Ai and the range of fi .
√
x
(a) f1 (x) = 1 + x2 (b) f2 (x) = 1 − x1 (c) f3 (x) = 3x − 1 (d) f4 (x) = x3 − 8 (e) f5 (x) = x−3
Solution: (a) dom f1 = R, ran f1 = {y ∈ R : y ≥ 1}; (b) dom f2 = R − {0} = {x ∈ R : x 6= 0},
ran f2 = R − {1} = {y ∈ R : y 6= 1}; (c) dom f3 = {x ∈ R : x ≥ 1/3}, ran f3 = R+
0 = {y ∈ R : y ≥ 0};
(d) dom f4 = ran f4 = R; (e) dom f5 = R − {3} = {x ∈ R : x 6= 3}, ran f5 = R − {1} = {y ∈ R : y 6= 1}.
• Section 9.2: The Set of All Functions from A to B
2. Exercise 9.10:
(a) Give an example of two sets A and B such that |B A | = 8.
(b) For the sets A and B given in (a), provide an example of an element in B A .
Solution: For (a), all examples are either of the form A = {a} and B = {b1 , b2 , . . . , b8 }, or of the form
A = {a1 , a2 , a3 } and B = {b1 , b2 }. For (b), we need to give an example for a function from A to B,
and in the first case all possible eight functions are of the form f = {(a, bi )} for some 1 ≤ i ≤ 8, and in
the second case of the form f = {(a1 , bi ), (a2 , bj ), (a3 , bk )} where 1 ≤ i ≤ 2, 1 ≤ j ≤ 2, and 1 ≤ k ≤ 2.
• Section 9.3: One-to-One and Onto Functions
3. Exercise 9.18: Is there a function f : R → R that is onto but not one-to-one? Explain your answer.
Solution: The correct answer is yes. We discussed in class that functions f : R → R can be onto but
not one-to-one (f (x) = x3 + x2 ) and similarly, one-to-one but not onto (f (x) = ex ), neither one-to-one
nor onto (f (x) = x2 or f (x) = sin x), or both one-to-one and onto (bijective and invertible, f (x) = x).
• Section 9.4: Bijective Functions
4. Exercise 9.22(b): Let f : Z5 → Z5 be a function defined by f ([a]) = [2a + 3]. Determine whether f
is bijective. [If you know what “well-defined” means, also do part (a): Show that f is well-defined.]
Solution: If we recall that Z5 stands for the integers modulo 5 whose only elements are the five residue
classes [a] modulo 5, a = 0, 1, 2, 3, 4, we can explicitly list the element pairs of the function f described
by f ([a]) = [2a + 3]: f = {([0], [3]), ([1], [0]), ([2], [2]), ([3], [4]), ([4], [1])}. Because each residue class
in the codomain is the unique image of exactly one residue class in the domain of f , this function is
bijective. [For (a), to show that f is well-defined, we need to show that the image of [a] under f is
independent of the chosen element in the equivalence class [a]. Hence, let b and c be two elements in
[a] so that a ≡ b (mod 5) and a ≡ c (mod 5). Because congruences are an equivalence relation, this
implies that b ≡ c (mod 5) also. By the laws of modular arithmetic, we obtain that 2b ≡ 2c (mod 5)
and 2b+3 ≡ 2c+3 (mod 5) and thus [2b+3] = [2c+3] showing that f ([b]) = f ([c]). Mission complete.]
• Section 9.5: Composition of Functions
5. Exercise 9.28: For nonempty sets A, B, and C, let f : A → B and g : B → C be functions.
(a) Prove:
If g ◦ f is one-to-one, then f is one-to-one.
using as many of the following proof techniques as possible: direct proof, proof by contrapositive, proof by contradiction.
Math 3000-003 Intro to Abstract Math Homework 7, UC Denver, Spring 2012 (Solutions)
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(b) Disprove: If g ◦ f is one-to-one, then g is one-to-one.
Solution: For (a), you should have found that all three techniques are possible, as illustrated below.
Direct Proof. Let g ◦ f be one-to-one so that a = b whenever (g ◦ f )(a) = (g ◦ f )(b). To show that f
is one-to-one, we need to show that a = b also whenever f (a) = f (b). Hence, let f (a) = f (b), then
g(f (a)) = g(f (b)) because g is a function and thus a = b because g ◦ f is one-to-one.
Proof by Contrapositive. For a proof by contrapositive, we need to show that if f is not one-to-one,
then g ◦ f is not one-to-one. Hence, let f be not one-to-one, then there exist two distinct elements
a 6= b ∈ A such that f (a) = f (b) ∈ B and thus g(f (a)) = g(f (b)) because g is a function. It follows
that (g ◦ f )(a) = (g ◦ f )(b) but a 6= b, showing that g ◦ f is not one-to-one.
Proof by Contradiction. Let g ◦ f be one-to-one, so that a = b whenever (g ◦ f )(a) = (g ◦ f )(b), and
assume that f is not one-to-one. Then there exists a 6= b ∈ A such that f (a) = f (b) ∈ B and thus
g(f (a)) = g(f (b)) because g is a function: contradiction E.
For (b), we give a counterexample and let A = C = {1}, B = {1, 2}, f = {(1, 1)} and g = {(1, 1), (2, 1)}.
Then g ◦ f = f = {(1, 1)} is one-to-one but g is not. Those of you who “proved” that g ◦ f is oneto-one have usually done the subtle mistake to ignore the full domain of g; namely, it is true that the
restriction of g onto the range of f is one-to-one: If g ◦f is one-to-one, then ĝ : ran f → C is one-to-one.
• Section 9.6: Inverse Functions
6. Exercise 9.36: Let A = R − {1} and define f : A → A by f (x) =
x
x−1
for all x ∈ A.
(a) Prove that f is bijective.
(b) Determine f −1 .
(c) Determine f ◦ f ◦ f .
Solution: For (a), we need to show that f is one-to-one and onto. For the former, let (x, y) ∈ A × A
so that f (x) = f (y). It follows that x/(x − 1) = y/(y − 1) which can be rewritten as x(y − 1) = y(x − 1)
and readily reduces to x = y, showing that f is one-to-one. For the latter, let y ∈ A and choose
x = y/(y − 1) which is well defined because 1 ∈
/ A (in particular, note that x = f (y)). It follows that
f (x) =
x
y/(y − 1)
y
=
=
=y
x−1
y/(y − 1) − 1
y − (y − 1)
showing that f is onto (this also shows that f (x) = f (f (y)) = (f ◦ f )(y) = y and thus that f ◦ f = idA ,
which we will use later). Together, this shows that f is bijective and completes the proof for (a).
Alternatively, we could have quoted the result proven in class that a function f is bijective if and only if
its inverse f −1 is a function, and concluded the same result from our solution to part (b): To find f −1 ,
it suffices to write y = x/(x − 1) and then solve this equation for x = y/(y − 1), so that f −1 : A → A
is determined by f −1 (y) = f (y) = y/(y − 1) (note that the work here is basically identical to the work
necessary to prove that f is onto in part (a)). Finally, for (c) we already observed in part (a) that
f ◦ f = idA so that f ◦ f ◦ f = f ◦ idA = idA ◦f = f is identical to the original function. Or do the work:
(x/(x−1))
x
(x/(x−1))−1
x−(x−1)
x
= f (x).
(f ◦ f ◦ f )(x) = =
=
(x/(x−1))
x
x−1
−1
−1
(x/(x−1))−1
x−(x−1)
• Section 9.7: Permutations
1
2
Determine α ◦ β and β −1 .
7. Exercise 9.38: Let α =
2
3
3
4
4
5
5
1
and β =
1
3
2
5
3
2
4
4
5
1
be permutations in S5 .
Math 3000-003 Intro to Abstract Math Homework 7, UC Denver, Spring 2012 (Solutions)
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Solution: First, to determine α ◦ β (remember that this means to first permute according to β and
then according to α), we can sort the columns of αso that all numbers
in its first row align with those
3 5 2 4 1
numbers in the second row of β, yielding that α =
. Then we merely need to combine
4 1 3 5 2
1 2 3 4 5
the first row of β with the second row of α to find that α ◦ β =
(of course, you can
4 1 3 5 2
also figure this out without the tedious sorting procedure). Second, to determine β −1
, we can swap the
1 2 3 4 5
−1
two rows of β and then sort its columns so that its first row is again sorted: β =
.
5 3 1 4 2
• Additional Exercises for Chapter 9
8. Exercise 9.60: Let f : A → B be a function. For a subset C of A, the image of C under f is
the set
f (C) = {f (c) : c ∈ C}.
(Thus f (A) is the range of f .) Let A1 , A2 ⊆ A. Prove the following.
(a) f (A1 ∪ A2 ) = f (A1 ) ∪ f (A2 )
(b) f (A1 ∩ A2 ) ⊆ f (A1 ) ∩ f (A2 )
(c) If f is one-to-one, then f (A1 ∩ A2 ) = f (A1 ) ∩ f (A2 ).
Solution: For (a), we need to show that f (A1 ∪ A2 ) ⊆ f (A1 ) ∪ f (A2 ) and f (A1 ) ∪ f (A2 ) ⊆ f (A1 ∪ A2 ).
For the former, let b ∈ f (A1 ∪ A2 ), then there is a ∈ A1 ∪ A2 such that f (a) = b. It follows that a is in
A1 or A2 (or both), and thus f (a) ∈ f (A1 ) or f (a) ∈ f (A2 ) showing that b = f (a) ∈ f (A1 ) ∪ f (A2 ).
For the reverse, let b ∈ f (A1 ) ∪ f (A2 ) so that b is in f (A1 ) or f (A2 ) (or both). It follows that there is
a ∈ A1 or a ∈ A2 such that f (a) = b, and thus a ∈ A1 ∪ A2 showing that b = f (a) ∈ f (A1 ∪ A2 ).
For (b), let b ∈ f (A1 ∩ A2 ), then there is a ∈ A1 ∩ A2 such that f (a) = b. It follows that a is in both
A1 and A2 , and thus f (a) ∈ f (A1 ) and f (a) ∈ f (A2 ) showing that b = f (a) ∈ f (A1 ) ∩ f (A2 ) also.
Finally for part (c), together with part (b) it suffices to show that f (A1 ) ∩ f (A2 ) ⊆ f (A1 ∩ A2 ) if f is
one-to-one. Hence, let f be one-to-one and b ∈ f (A1 ) ∩ f (A2 ), then there exists a1 ∈ A1 and a2 ∈ A2
such that f (a1 ) = f (a2 ) = b, and a1 = a2 because f is one-to-one. Denote a = a1 = a2 . It follows that
a is an element in both A1 and A2 , and thus a ∈ A1 ∩ A2 showing that b = f (a) ∈ f (A1 ∩ A2 ) also.
There will be no homework from Chapter 10 (Cardinalities), but here is a Dino Cartoon:
Please let me know if you have any questions, comments, corrections, or remarks.