6. Let A, B be finite sets. Show that if JA\ B| = |B\ AJ, then |A| = |B|.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
q6
![12:32
ull 5G
Midterm2ReviewSol...
there exists z EX such that f(z) = y. Since fogof = f, we have
S(g(u)) = S(g(S(x)) = S(x) = y.
It follows that fog- idy.
Now, suppose that fog = idy. Let y €Y be given, and set r = g(y). Then
S(z) = f(9(v)) = idy(y) = y.
It follows that f is surjective.
(c) Suppose that f is injective. Let z € X be given, and set y = f(z). By our
assumption that fogof = f, we have
f(z) = f(9(f(2)).
Since f is injective, this implies that g(y) = 9(f(z)) = r. It follows that g is surjective.
5. Let f: X + Y be a function.
(a) Show that A C SU(A)) for any A € P(X).
(b) Show that jU(B)) C B for any BE P(Y).
6. Let A, B be finite sets. Show that if |A \ B| = |B\ A|, then |A| = |B|.-
7. Prove that among a group of n people, there are two people with the same number of
friends in the group (assume that being friends is sy mmet rie, i.e. if person A is friends
with person B, then person B is friends with person A)
8. Let S = [0, 2] x (0, 2] C R. Show if ACS with |A| = 5, then there are a, be A with
a f b such that a and b are at most a distance v2 apart.
Math 109
Midterm 2 Pract ice Problems
9. Let AC {1,2,..., 2n} with |A| = n+1. Prove that there exist distinet a, be A such
that a divides b. (Hint: for a € A, consider the largest odd integer that divides a. You
may use the fact that if n > 1 has no odd divisors greater than 1, then n = 2* for
some k e Z,.)
Solution: Define a function
S:A+ {m € N2 m is odd}
f(a) = max{( e Zje divides a and ( is odd}
Note that f is well-defined since for any a € {1,2,..., 2n}, 1 < max{l € Z]e divides a and é is odd}
since 1|a, and max{{ € Z[e divides a and e is odd} < 2n since a < 2n. Since
|A| = n+1>n = |{m e N2,|m is odd},
we know that there exist a, be A such that a b and f(a) = f(b) by the pigeonhole
principle. Without loss of generality, we may assume that a > b. Since f(a)la and
S(b)|b, we know that
a = f(a)ls
b = f(b)(
for some 1, l2 € Z. Furt hermore, by definition of S(a), f(b), we know that 1, 6z have
no odd divisors greater than 1, so by the hint in the problem we can write = 2
and (2 = 2 for some ki, kz € Z. Since a > b, we have ki > kz. Hence,
a = f(a)t, = f(a)2 = f(b)2 = f (b)2 -kagka- 2k,
so that h divides a.
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Transcribed Image Text:12:32
ull 5G
Midterm2ReviewSol...
there exists z EX such that f(z) = y. Since fogof = f, we have
S(g(u)) = S(g(S(x)) = S(x) = y.
It follows that fog- idy.
Now, suppose that fog = idy. Let y €Y be given, and set r = g(y). Then
S(z) = f(9(v)) = idy(y) = y.
It follows that f is surjective.
(c) Suppose that f is injective. Let z € X be given, and set y = f(z). By our
assumption that fogof = f, we have
f(z) = f(9(f(2)).
Since f is injective, this implies that g(y) = 9(f(z)) = r. It follows that g is surjective.
5. Let f: X + Y be a function.
(a) Show that A C SU(A)) for any A € P(X).
(b) Show that jU(B)) C B for any BE P(Y).
6. Let A, B be finite sets. Show that if |A \ B| = |B\ A|, then |A| = |B|.-
7. Prove that among a group of n people, there are two people with the same number of
friends in the group (assume that being friends is sy mmet rie, i.e. if person A is friends
with person B, then person B is friends with person A)
8. Let S = [0, 2] x (0, 2] C R. Show if ACS with |A| = 5, then there are a, be A with
a f b such that a and b are at most a distance v2 apart.
Math 109
Midterm 2 Pract ice Problems
9. Let AC {1,2,..., 2n} with |A| = n+1. Prove that there exist distinet a, be A such
that a divides b. (Hint: for a € A, consider the largest odd integer that divides a. You
may use the fact that if n > 1 has no odd divisors greater than 1, then n = 2* for
some k e Z,.)
Solution: Define a function
S:A+ {m € N2 m is odd}
f(a) = max{( e Zje divides a and ( is odd}
Note that f is well-defined since for any a € {1,2,..., 2n}, 1 < max{l € Z]e divides a and é is odd}
since 1|a, and max{{ € Z[e divides a and e is odd} < 2n since a < 2n. Since
|A| = n+1>n = |{m e N2,|m is odd},
we know that there exist a, be A such that a b and f(a) = f(b) by the pigeonhole
principle. Without loss of generality, we may assume that a > b. Since f(a)la and
S(b)|b, we know that
a = f(a)ls
b = f(b)(
for some 1, l2 € Z. Furt hermore, by definition of S(a), f(b), we know that 1, 6z have
no odd divisors greater than 1, so by the hint in the problem we can write = 2
and (2 = 2 for some ki, kz € Z. Since a > b, we have ki > kz. Hence,
a = f(a)t, = f(a)2 = f(b)2 = f (b)2 -kagka- 2k,
so that h divides a.
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