Suppose you know that the statement “if Peter is not tall, then Quincy is fat and Robert is skinny” is false. What, if anything, can you conclude about Peter and Robert if you know that Quincy is indeed fat? Explain (you may reference problem 2.6.1).
Make a truth table that includes all three statements in the argument:
\(p\)
\(q\)
\(r\)
\(p \to q\)
\(p \to r\)
\(p \to (q \wedge r)\)
T
T
T
T
T
T
T
T
F
T
F
F
T
F
T
F
T
F
T
F
F
F
F
F
F
T
T
T
T
T
F
T
F
T
T
T
F
F
T
T
T
T
F
F
F
T
T
T
Notice that in every row for which both \(p \to q\) and \(p \to r\) is true, so is \(p \to (q \wedge r)\text{.}\) Therefore, whenever the premises of the argument are true, so is the conclusion. In other words, the deduction rule is valid.
5.
Write the negation, converse and contrapositive for each of the statements below.
If the power goes off, then the food will spoil.
If the door is closed, then the light is off.
\(\forall x (x \lt 1 \to x^2 \lt 1)\text{.}\)
For all natural numbers \(n\text{,}\) if \(n\) is prime, then \(n\) is solitary.
For all functions \(f\text{,}\) if \(f\) is differentiable, then \(f\) is continuous.
For all integers \(a\) and \(b\text{,}\) if \(a\cdot b\) is even, then \(a\) and \(b\) are even.
For every integer \(x\) and every integer \(y\) there is an integer \(n\) such that if \(x > 0\) then \(nx > y\text{.}\)
For all real numbers \(x\) and \(y\text{,}\) if \(xy = 0\) then \(x = 0\) or \(y = 0\text{.}\)
For every student in Math 321, if they do not understand implications, then they will fail the exam.
The statement is true. If \(n\) is an even integer less than or equal to 7, then the only way it could not be negative is if \(n\) was equal to 0, 2, 4, or 6.
There is an integer \(n\) such that \(n\) is even and \(n \le 7\) but \(n\) is not negative and \(n \not\in \{0,2,4,6\}\text{.}\) This is false, since the original statement is true.
For all integers \(n\text{,}\) if \(n\) is not negative and \(n \not\in\{0,2,4,6\}\) then \(n\) is odd or \(n > 7\text{.}\) This is true, since the contrapositive is equivalent to the original statement (which is true).
For all integers \(n\text{,}\) if \(n\) is negative or \(n \in \{0,2,4,6\}\) then \(n\) is even and \(n \le 7\text{.}\) This is false. \(n = -3\) is a counterexample.
7.
Consider the statement: \(\forall x (\forall y (x + y = y) \to \forall z (x\cdot z = 0))\text{.}\)
Explain what the statement says in words. Is this statement true? Be sure to state what you are taking the universe of discourse to be.
Write the converse of the statement, both in words and in symbols. Is the converse true?
Write the contrapositive of the statement, both in words and in symbols. Is the contrapositive true?
Write the negation of the statement, both in words and in symbols. Is the negation true?
For any number \(x\text{,}\) if it is the case that adding any number to \(x\) gives that number back, then multiplying any number by \(x\) will give 0. This is true (of the integers or the reals). The “if” part only holds if \(x = 0\text{,}\) and in that case, anything times \(x\) will be 0.
The converse in words is this: for any number \(x\text{,}\) if everything times \(x\) is zero, then everything added to \(x\) gives itself. Or in symbols: \(\forall x (\forall z (x \cdot z = 0) \to \forall y (x + y = y))\text{.}\) The converse is true: the only number which when multiplied by any other number gives 0 is \(x = 0\text{.}\) And if \(x = 0\text{,}\) then \(x + y = y\text{.}\)
The contrapositive in words is: for any number \(x\text{,}\) if there is some number which when multiplied by \(x\) does not give zero, then there is some number which when added to \(x\) does not give that number. In symbols: \(\forall x (\exists z (x\cdot z \ne 0) \to \exists y (x + y \ne y))\text{.}\) We know the contrapositive must be true because the original implication is true.
The negation: there is a number \(x\) such that any number added to \(x\) gives the number back again, but there is a number you can multiply \(x\) by and not get 0. In symbols: \(\exists x (\forall y (x + y = y) \wedge \exists z (x \cdot z \ne 0))\text{.}\) Of course since the original implication is true, the negation is false.
8.
Write each of the following statements in the form, “if …, then ….” Careful, some of the statements might be false (which is alright for the purposes of this question).
To lose weight, you must exercise.
To lose weight, all you need to do is exercise.
Every American is patriotic.
You are patriotic only if you are American.
The set of rational numbers is a subset of the real numbers.
A number is prime if it is not even.
Either the Broncos will win the Super Bowl, or they won’t play in the Super Bowl.
If a number is not even, then it is prime. (Or the contrapositive: if a number is not prime, then it is even.)
If the Broncos don’t win the Super Bowl, then they didn’t play in the Super Bowl. Alternatively, if the Broncos play in the Super Bowl, then they will win the Super Bowl.
Since \(7k + 3\) is an integer, we see that \(7n\) is odd.
The converse is: for all integers \(n\text{,}\) if \(7n\) is odd, then \(n\) is odd. We will prove this by contrapositive.
Proof.
Let \(n\) be an integer. Assume \(n\) is not odd. Then \(n = 2k\) for some integer \(k\text{.}\) So \(7n = 14k = 2(7k)\) which is to say \(7n\) is even. Therefore \(7n\) is not odd.
11.
You come across four trolls playing bridge. They declare:
Troll 1: All trolls here see at least one knave.
Troll 2: I see at least one troll that sees only knaves.
Troll 3: Some trolls are scared of goats.
Troll 4: All trolls are scared of goats.
Are there any trolls that are not scared of goats? Recall, of course, that all trolls are either knights (who always tell the truth) or knaves (who always lie).