CSCI 301 'Quiz' 2 - Solutions

Complete the missing entries in this partially filled truth table:

Solution:

For the following pair of logical expressions, determine whether they are logically equivalent; justify your answer with a truth table or, if they are equivalent you may justify with a sequence of manipulations using known equivalences. \((\neg P \land Q) \lor (P \land \neg Q)\) and \(\neg (P \Leftrightarrow Q)\)

Solution: Let’s verify using a truth table:

The last two columns are the same, so the two are logically equivalent.

Are the following two statements logically equivalent? Briefly explain.

1. 4 divides \(a\) if \(a\) is even

2. if \(a\) is not even, it is not divisible by 4

Solution: No. Let $Q = $ “4 divides \(a\)” and $P = $ “\(a\) is even”.

• Statement (a) says \(Q\) if \(P\), which is equivalent to “if \(P\), then \(Q\)”, or \(P \Rightarrow Q\).
• Statement (b) says “if \(\neg P\) then \(\neg Q\)”, or \(\neg P \Rightarrow \neg Q\), which is not equivalent to \(P \Rightarrow Q\).

Translate each statement into symbolic form using quantifiers:

1. Some integers are perfect squares Solution: \(\exists x \in \mathbb{Z}, \exists y \in \mathbb{Z}, x = y^2\)

2. For every positive real number, there exists a smaller positive real number Solution: \(\forall x \in \mathbb{R}^+, \exists y \in \mathbb{R}^+, y < x\)

3. Every prime number greater than 2 is odd

Solution: Sometimes \(\mathbb{P}\) is used to represent the set of primes. If we use this notation, then: \(\forall p \in \mathbb{P}, (p > 2) \Rightarrow (p \text{ is odd})\)

Alternative: \(\forall p \in \mathbb{P}, (p > 2) \Rightarrow O(p)\) We could also just say something like \(\forall p\), \(p\) is prime $ ~p > 2 O(p)$.

Translate each symbolic statement into English, and indicate whether they are true or false:

1. \(\forall x \in \mathbb{Z}, \exists y \in \mathbb{Z}, x = 2y \lor x = 2y + 1\)

   Solution: “For every integer x, there exists an integer y such that x is either equal to 2y or equal to 2y+1.”

   This statement is true - integers are either even or odd.

2. \(\exists x \in \mathbb{R}, \forall y \in \mathbb{R}, x \cdot y = y\)

   Solution: “There exists a real number x such that for all real numbers y, x times y equals y.”

   This statement is true. The value x = 1 satisfies this condition, as 1 · y = y for all real numbers y.

Translate each statement to symbols, negate it, simplify the negation, and express the result in English:

1. All natural numbers are positive

   Solution:

   • Original: \(\forall n \in \mathbb{N}, n > 0\)
   • Negation: \(\neg(\forall n \in \mathbb{N}, n > 0) \equiv \exists n \in \mathbb{N}, \neg(n > 0) \equiv \exists n \in \mathbb{N}, n \leq 0\)
   • In English: “There exists a natural number that is not positive (i.e., less than or equal to zero).”

2. There exists a real number that is both positive and negative

   Solution:

   • Original: \(\exists x \in \mathbb{R}, (x > 0) \land (x < 0)\)
   • Negation: \(\neg(\exists x \in \mathbb{R}, (x > 0) \land (x < 0)) \equiv \forall x \in \mathbb{R}, \neg((x > 0) \land (x < 0)) \equiv \forall x \in \mathbb{R}, (x \leq 0) \lor (x \geq 0)\)
   • Simplified: \(\forall x \in \mathbb{R}, \text{true}\) (since any real number satisfies either x ≤ 0 or x ≥ 0)
   • In English: “Every real number is either non-positive or non-negative.” (Which is a tautology, always true)

3. \(a\) is positive and \(b\) is negative.

   Solution:

• Original statement in symbols: \((a > 0) \land (b < 0)\)
• Negation: \(\neg((a > 0) \land (b < 0)) \equiv \neg(a > 0) \lor \neg(b < 0) \equiv (a \leq 0) \lor (b \geq 0)\)
• In English: “Either \(a\) is non-positive or \(b\) is non-negative.”