Q1: 20 points Q2: 20 points Q3: 30 points Q4: 30 points Total: 100 points
Question text is in black, solutions in blue.
A one-to-one function is one where each element of the domain maps to at most one element of the co-domain.
An antisymmetric relation on an set A is a binary relation R ⊆ A × A such that for any elements x and y, R(x, y) and R(y, x) are both true only if x = y. That is, there are not two distinct elements each of which is related to the other by R.
A partial order on a set A is a binary relation R ⊆ A × A that is reflexive, antisymmetric, and transitive.
The Cartesian product A × B is the set of all ordered pairs (x, y) such that x ∈ A and y ∈ B.
C(n, r) is the number of size-r subsets of an n-element set.
There are two ways to draw this. Best is to draw B and C intersecting, then draw A inside of the region common to B and C. You could also draw the usual three interlocking circles for A, B, and C, and indicate that the three regions within the A circle, that are not in both the B and C circle, are empty.
This statement cannot be proved true or false from the information given. It could be true -- it is certainly true if all three sets are empty. But it also could be false, for example if A = ∅ and B = C = {x}.
This statement must be true. A counterexample would be an element x that was in A but not in B ∩ C. But if x were such an element, x ∈ A would imply both x ∈ B and x ∈ C, and thus imply x ∈ (B ∩ C). The existence of such an element leads to a contradiction (x is both in and not in (B ∩ C)) so there is no such element.
Here the double-headed arrows represent an arrow in each direction.
There should also be a loop on each of the four vertices, which I didn't
draw because of the mess.
[b]<----->[c]
^ ^^
| / |
| / |
| / |
| / |
| / |
| / |
| / |
VV V
[d]<----->[n]
R is reflexive, since every dog's name has a letter in common with itself and thus R(x, x) is true for every dog x. (If we had a dog whose name was the empty string, the relation would not be reflexive, but we don't.)
R is symmetric, since "dog x has a letter in common with dog y" and "dog y has a letter in common with dog x" are exactly the same statement, so that "R(x, y) <--> R(y, x)" is always true.
This R is not transitive, since R(b, c) and R(c, n) are true but R(b, n) is false. (For a different set of dogs, the relation might be transitive.)
Two ways to count: (1) the integer range {x,..., y} always has size y - x + 1 (not y - x, as you have to count both x and y) so here it is 999 - 100 + 1 = 900. (2) There are 9 choices for the first digit, 10 for the second digit, and 10 for the third digit, so by the Product Rule there are 9 × 10 × 10 = 900 in all.
You could say that exactly one-fifth of the numbers, or 180, are divisible by 5 and you would be right, but this is a risky method in general because it does not work for all ranges. Better is to note that we still have 9 choices for the first digit and 10 for the second digit, but only 2 choices for the third digit, 0 and 5. So the total is 9 × 10 × 2 = 180.
Two arguments: (1) There are 9 ways to choose the first digit, 9 ways to choose the second digit (the ten digits, but without the one chosen for the first digit) and 2 choices for the third, for 9 × 9 × 2 = 162 total. (2) We can take the answer for (b) and subtract the size of the set of numbers that are divisble by 5 and have their first two digits the same. These are the numbers xx0 and xx5, where x ∈ {1, 2, 3,..., 9} -- there are 18 of these so we have 180 - 18 = 162 total. Note that an arithmetic expression like 9 × 10 × 2 - 9 × 1 × 2 would be a full-credit answer -- you don't have to complete the arithmetic.
Last modified 6 November 2013