CMPS 260 Spring 2022
Homework #1: Sets, Relations, Proofs
Due: 2pm, Wednesday, February 9

1. Using the Laws of Propositional Logic shown on the relevant web page, prove this theorem:

p ⟹ p∧q   ≡   p⟹q

Hint: Start with the left-hand side of the theorem and use the laws, in step-by-step fashion, to transform it into the expression on the other side. (Use as a model the proof found on the relevant web page.)


2. Follow the same approach as in Problem 1 to prove this theorem:

¬p ⟹ p∧q   ≡   p

Hint: One of the Absorption laws should come in handy.


3. Using the provided Laws of Propositional Logic and the Set Operator Axioms (see the relevant web page), prove that intersection distributes over union. That is, prove

R ∩ (S∪T)   =   (R∩S) ∪ (R∩T)

To do so, show that, for arbitrary x,

x ∈ R ∩ (S ∪ T)   ≡   x ∈ (R ∩ S) ∪ (R ∩ T)

4. Recall these definitions pertaining to a binary relation R on a set A:

For each binary relation described below on the left, indicate in the table to the right which among the listed properties it possesses. Each relation is to be understood as being on the set A = { a, b, c }.

R1: { (b,b), (a,c), (c,a) }
R2: { (a,b), (a,c), (b,c), (c,c) }
R3:     (i.e., empty set)
R4: { (a,b), (c,b) }
R5: { (a,a), (b,b), (c,c), (a,b), (a,c) }
      
Relation
Property R1 R2R3 R4R5
reflexive       
symmetric       
transitive       


5. Let A = { a,b,c } and let P = { reflexivity, symmetry, transitivity }. For each of the three properties in P, a binary relation on A either possesses that property or fails to possess it. There are eight possible combinations of possession/non-possession —corresponding to the eight subsets of P— all of which are achievable.

In the previous problem, five relations over A were given and, for each one, you were to have identified which subset of the properties in P it possessed. Hence, at least three of the eight subsets of P were not represented in that problem. Choose two of those three subsets of P, and, for each one, identify it and describe a binary relation over A possessing exactly the properties in that subset.


6. Let f(n) = Σ1≤i≤n(2i−1). In other words, f(n) = 1 + 3 + ... + (2n−1), the sum of the smallest n positive odd integers.

Prove by mathematical induction that, for all n≥1, f(n) = n2.


7. Let m be an integer. Prove that, if m2 − 6m + 5 is even, then m is odd.

Hint 1: First, transform the given statement into its contrapositive (which is equivalent to it). Then prove that using the "Assume the Antecedant" approach.

Hint 2: Having assumed, or established, that an integer, say r, is even, you are justified in introducing a new variable, say k, and using 2k in place of r.

Hint 3: To demonstrate that an integer is odd, it suffices to show that it is equal to 2k+1, for some integer k.