Q1: 20 points Q2: 30 points Q3: 30 points Q4: 20+5 points Total: 100+5 points
Question text is in black, solutions in blue.
Here are definitions of sets, predicates, and statements used on this exam.
Remember that the score of any quantifier is always to the end of the statement it is in.
Question 2 deals with the following scenario. Every day, Duncan monitors all visitors to the house and evaluates which ones (in his judgment) pose a threat. One day there were exactly five visitors, arriving at five distinct times. In alphabetical order, they were an Amazon driver (ad), the house cleaner (hc), the mail person (mp). a political canvasser (pc), and a UPS driver (ud).
Let V be the set {ad, hc, mp, pc, ud}. Let PT and DA be two unary relations on V, such that PT(x) means "visitor x posed a threat" and DA(x) means "visitor x was driven away by Duncan". Let AB be a binary relation on V, such that AB(x, y) means "visitor x arrived before visitor y", or equivalently "visitor y arrived after visitor x". We assume that AB is a strict total order, so that it is antireflexive, antisymmetric, transitive, and total.
Let N be the set of natural numbers {0, 1, 2, 3,...}.
If a, b, and m are naturals, with m > 0, the notation "a ≡ b (mod m)" means "a is congruent to b, modulo m".
The operator "%" on naturals, as in Java, refers to integer division, so that "x % y" is the remainder on dividing x by y.
The political canvasser posed a threat if and only if they arrived after the house cleaner, and if the house cleaner posed a threat, then the political canvasser also posed a threat and the house cleaner arrived before the political canvasser. These were mostly right.
(PT(pc) ∧ PT(hc)) → AB(hc, pc)
Either the canvasser or the house cleaner posed a threat, or the house cleaner arrived before the canvasser, and it was not the case that the house cleaner arrived before the canvasser and that the canvasser and the house cleaner both posed threats. These were pretty good -- some people messed up the DeMorgan Law on the second half.
∀x:(¬DA(x)) → (¬AB(hc, x)) Here lots of people translated "not after" as "before", which is wrong in the case that x = hc.
There was a visitor that was driven away by Duncan, and every visitor arrived before the house cleaner if and only if they were that visitor. Alternatively: There was exactly one visitor who arrived before the house cleaner, and they were driven away by Duncan. Here a lot of people misstated the condition on z, saying for example that "all visitors after hc were driven away". It's important that your statement makes clear that there was exactly one visitor before the house cleaner. :
∀x:PT(x) → DA(x)
Using the abbreviations q, r, and s, Statement I becomes "(r ↔ s)
∧ (q → (r ∧ s))". Statement II becomes "(r ∧ q)
→ s", and Statement III becomes "(r ∨ q ∨ s)
∧ ¬(s ∧ q ∧ r)".
If q were true, r and s would also be true by I, and this
contradicts III. So q is false. This makes II vacuously true, so
we can ignore it. I says that r and s are equivalent, and if they
were both false all three would be false, contradicting III. So
r and s are both true, the right half of I is satisfied vacuously,
and III is also satisfied. We got about an equal number of
truth table and deductive sequence proofs, most of them pretty
good. I took off 2 out of 10 if you just gave the value of
the three statements in each case, without showing your reasoning.
I also insisted that your arguement show that your solution was
the only one. A complete truth table does this as a matter of
course.
Let x be an arbitrary visitor. If x = hc, we know that PT(x) is false from part (a), and thus PT(x) → DA(x) is satisfied vacuously. By Instantiation on V, let y be the visitor such that DA(y) and ∀z:AB(z, hc)↔ (y=z). If AB(x, hc), specify this last statement to x to get AB(x, hc) ↔ (x=y), which tells us that x = y and thus also that DA(x) is true. This satisfies PT(x) → DA(x) trivially. By the totality of the relation AB, the only remaining case is AB(hc, x). By Specification of IV to x, we have (¬DA(x)) → (¬ AB(hc, x)), which by Contrapositive gives us AB(hc, x) → DA(x). Now Modus Ponens gives us DA(x), so again PT(x) → DA(x) is satisfied trivially. Since x was arbitrary and we have covered all cases, we have proved ∀x:PT(x) → DA(x) by Generalization, and this is Statement VI.
TRUE. We know DA(mp), but we have information about PT only for hc and pc.
TRUE. Run the Euclidean Algorithm on n and n+1 for one step, and you reach 1.
TRUE, there are exactly four edges. This partial order is a linear order.
TRUE. This is a standard result about surjections and size of finite sets.
TRUE. The new relation is still reflexive as R(x, x) is true for all x. It is still antisymmetric as we've added no pairs and thus created no new obligations. It is still transitive as R(x, y) and R(y, z) can both be true only if all three are in A or all three are in B. (Here is where we use the fact that A and B are disjoint.)
TRUE. A total order is defined to be a partial order that also has the totality property.
FALSE. Any partition is the set of classes for some equivalence relation.
TRUE. A surjection would hit all real numbers, but this hits only the square roots of naturals. It's an injection because every natural has a different square root.
FALSE. It is ∃x:¬DA(x).
FALSE. That function is (g ○ f)(x). The other would be 2(4√x)2 + 1.
TRUE. This is just the rule of Right Joining.
FALSE. Without violating p → q, we could have (p ∨ r) be true because of r, and have q be false.
FALSE. It is not transitive, since R(0, 1) and R(1, 0) are true but R(0, 0) is false. It is symmetric and not reflexive.
77 = 7 × 11, and 100 = 2 × 2 ×. 5 × 5. Since no prime occurs in both factorizations, the two numbers are relatively prime to one another.
Brute force actually works here because 7 × 11 × 13 = 1001, telling us that (77 × 13) - (10 × 100) = 1. But let's do it properly: 100 = 1(100) + 0(77), 77 = 0(100) + 1(77), 23 = 1(100) - 1(77), 8 = -3(100) + 4(77), 7 = 7(100) - 9(77), and 1 = -10(100) + 13(77).
This follows from the result of (d) that g has an inverse. But directly, we know that 13 × 77 is congruent to 1, modulo 100. So any x in X is equal to g(13x), making g onto. And if x and y are two different elements of X, g(x) and g(y) are different because g(y) = g(x) + g(y-x), and g(y-x) = 77(y-x) cannot be 0 modulo 100 if y-x ≠ 0, since 77 and 100 have a least common multiple of 7700 and 77(y-x) < 7700.
The function h(y) = 13y is an inverse of g, because h(g(y)) and g(h(y)) are both congruent to (77 × 13)y modulo 100, and this is congruent to y modulo 100, since 77 × 13 = 1001 is congruent to 1.
We know by the Inverse Theorem that a has a multiplicative inverse modulo 100 if and only if it is relatively prime to 100. If a has no such inverse, then fa,b cannot be a bijection for any b because fa,0 fails to be one-to-one, making fa,b not one-to-one. But if a has an inverse c, then fa,b has the inverse fc, -cb, because a(cx-cb) + b = x - b + b = x and c(ax+b) - cb = x + cb - cb = x, both equations being taken modulo 100.
Last modified 2 March 2020