- Answer the problems on the exam pages.
- There are five problems for 100 total points. Actual scale was A = 85, C = 55.
- If you need extra space use the back of a page.
- No books, notes, calculators, or collaboration.

Q1: 15 points Q2: 15 points Q3: 35 points Q4: 15 points Q5: 20 points Total: 100 points

Question text is in black, solutions in blue.

- (a) an
**equivalence relation**An equivalence relation on a set A is a relation R ⊆ A × A that is reflexive, symmetric, and transitive. Equivalently, it is the "same set" relation for some partition of A, that is, R(x, y) is true if and only if x and y are in the same set of the partition.

- (b) for two sets to be
**disjoint**They have no element in common, i.e., A and B are disjoint if and only if A ∩ B = ∅.

- (c) what it means for a relation R ⊆ A × B to be
a
**function**Such a relation is a function if and only if for every element a of A, there is

*exactly one*element b of B such that R(a, b) is true. - (d) a
**bijection**A bijection is a function that is both one-to-one and onto.

- (e) the
**Product Rule**for counting setsThe Product Rule says that if A and B are finite sets, |A × B| = |A| times |B|.

- (a) If B has more elements than A, then f is not onto.
TRUE. There cannot be an onto function from a smaller finite set to a larger set.

- (b) If f is not onto, then B has more elements than A.
FALSE. Let A = {x, y} and B = {1, 2} and define f(x) = f(y) = 1. This f is not onto, yet A and B have the same number of elements.

- (c) If A has m elements and B has n elements, then the power
set of A × B has 2
^{m}× 2^{n}elements.FALSE. The correct number is 2

^{|A × B|}= 2^{mn}, not 2^{m}2^{n}= 2^{m+n}. - (d) If I distribute seven treats among four dogs, then at least
one dog gets more than two treats.
FALSE. They might be distributed 2-2-2-1, with no dog having more than two. The Pigeonhole Principle here says that at least one dog must get at least 7/4 treats, hence at least two treats.

- (e) Of all the binary strings of length 5, more have an odd
number of ones than have an even number of ones.
FALSE. There are sixteen of each. We can calculate C(5, 0) + C(5, 2) + C(5, 4) = 1 + 10 + 5 = 16, or observe that flipping the first bit gives a bijection between the even-ones and odd-ones strings. Since there is a bijection between those two sets, they must have the same size.

- (a) {F: F is a subset of X containing Baxter but not Duncan}
Such an F is {B} ∪ Z where Z is any subset of {A, C}. There are 2

^{2}= 4 subsets of a two-element set. - (b) {G: G is an ordered list from X of size 5}
This is |X|

^{5}= 4^{5}= 1024. - (c) {H: H is an unordered list from X of size 7}
By the Stars and Bars argument, such a list corresponds to a string containing 7 stars and 4 - 1 = 3 bars. There are C(10, 3) = 120 such strings.

- (d) {I: I is a permutation from X of length 3,
not containing Baxter}
The permutation is from the three-element set of remaining dogs, so there are P(3, 3) = 6 possible permutations.

- (e) {J: J is a subset of X and the size of J is odd}
We could observe as in Question 2(e) that the odd-size subsets form half the possible subsets, so there are (1/2)2

^{4}= 8 of them. Or we could calculate C(4, 1) + C(4, 3) = 4 + 4 = 8. - (f) {K: K is a binary relation on X (a subset of X × X)}
There are |X| times |X| or 4 times 4 = 16 elements of the set X × X. There are thus 2

^{16}= 65536 possible subsets of this set. - (g) {L: L is a permutation form X of size 5}
A permutation of size 5 would be a one-to-one function from {1, 2, 3, 4, 5} to X, which cannot exist by the Pigeonhole Principle. There are thus 0 such permutations. We could also calculate P(4, 5) = 4*3*2*1*0 = 0.

Let x be an arbitrary element of A ∪ C.

Case 1: x is in A. Since A ⊆ B, x is also in B, and thus in B ∪ D.

Case 2: x is in C. Since C ⊆ D, x is also in D, and thus in B ∪ D.

Since every element of A ∪ C is in B ∪ D, the subset relationship (A ∪ C) ⊆ (B ∪ D) follows.

There are several ways to do this.

Proof 1: Since |B| < |C|, g cannot be onto. Thus there exists an element c of C such that c is not equal to g(b) for any possible b. This c thus cannot be g(f(a)) = h(a) for any possible a. So h is not onto and therefore not a bijection.

Proof 2: Since |A| > |B|, f cannot be one-to-one. Thus there exist two elements a and a' of A such that f(a) = f(a'). Therefore g(f(a)) = g(f(a')), which means h(a) = h(a') by definition, and h is not one-to-one and therefore not a bijection.

Last modified 19 November 2015