Chapter 4 Logical Arguments
Exercises Practice Problems
1. Implications.
Write down the contrapositive, converse and inverse of the implication: "if \(x < 3\) then \(x^2 < 9\text{.}\)" Which of these four statements are true? Which ones are false? Solution.
statement: \(x < 3\) then \(x^2 < 9\text{.}\) This is false, since \(-4 < 3\) and \((-4)^2 = 16 > 9\)
contrapositive: if \(x^2 \geq 9\) then \(x \geq 3\text{.}\) This is false for the same reason.
converse: if \(x^2 < 9\) then \(x < 3\text{.}\) This is true because the first statement forces \(-3 \leq x \leq 3\text{.}\)
inverse: if \(x \geq 3\) then \(x^2 \geq 9\) . This is true.
2. Quantifiers.
Let \(X\) be the set of people. Let \(P(x,y)\) be the statement "\(x\) loves \(y\text{.}\)" Rewrite each of the following statements using formal quantifiers and variables.
Everybody loves somebody.
Somebody loves everybody.
Let \(X\) be the set of all people. Let \(P(x,y)\) be the statement "\(x\) loves \(y\text{.}\)"
\(\displaystyle \forall x \in X, \exists y \in Y, P(x,y)\)
\(\displaystyle \exists x \in X, \forall y \in Y, P(x,y)\)
3. Negating Quantifiers.
Negate the formal statements you made in the previous problem, then translate them into everyday English. Do these negations make sense, when compared to the original statements? Solution.
\(\exists x \in X, \forall y \in Y, \neg P(x,y)\text{,}\) which means "Somebody hates everybody."
\(\forall x \in X, \exists y \in Y, \neg P(x,y)\text{,}\) which means "Everybody hates somebody."
4. Fooling Around.
Negate the statement "You can fool all of the people all of the time" by first creating a formal statement using quantifiers. Hint: you are considering two sets: people and times. Solution.
Let \(X\) be the set of people and let \(T\) be the set of times. Let \(P(x,t)\) be the statement "you can fool person \(x\) at time \(t\text{.}\) The original statement is
The simple negation of the statement leads to the adage "You can't fool all of the people all of the time." What if we try to say this in precise mathematical language? We get
which means there is a person and a time where you fail to fool them.
5. Equivalent Statements.
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Use a truth table to show that \(\neg (P \Rightarrow Q)\) is equivalent to \(P \wedge \neg Q\text{.}\)
\begin{equation*} \begin{array}{c|c||c|c|c|c} P & Q & P \Rightarrow Q & \neg(P \Rightarrow Q) & \neg Q & P \wedge \neg Q \\ \hline T & T & & & & \\ T & F & & & & \\ F & T & & & &\\ F & F & & & &\\ \end{array} \end{equation*}\begin{equation*} \begin{array}{c|c||c|c|c|c} P & Q & P \Rightarrow Q & \neg(P \Rightarrow Q) & \neg Q & P \wedge \neg Q \\ \hline T & T & T & F & F& F\\ T & F & F &T & T& T\\ F & T & T &F & F&F\\ F & F & T &F & T &F\\ \end{array} \end{equation*}The columns for \(\neg(P \Rightarrow Q) \) and \(P \wedge \neg Q\) are identical. So these statements are equivalent.
Negate the statement "For all \(x \in \R\text{,}\) if \(x\) is an integer then \(x\) is positive." Use part (a) to find an equivalent statement, and then explain why providing one counterexample is enough to disprove that \(P \Rightarrow Q\) Solution.
The negation is "There exists an integer that is not positive." Note that "not positive" isn't the same as "negative" because 0 is neither positive nor negative.
This uses the existential quantifiers (at least one), so we just need one example of a non-positive integer.
6. Evaluating an Argument.
Determine the validity of the following arguments. Include a truth table to justify your conclusion.
If I stay up late at night, then I will be tired in the morning. I am tired this morning. Therefore, I stayed up late last night. Solution.
Let \(P = \) "I stay up late" and let \(Q = \) "I am tired in the morning."
\begin{equation*} \begin{array}{c|c|c|c|c} P & Q & P \rightarrow Q & Q \wedge (P \rightarrow Q) & \big( Q \wedge (P \rightarrow Q) \big) \rightarrow P \\ \hline T & T & T & T & T \\ T & F & F & F & T \\ F & T & T & T & F \\ F & F & T & F & T \\ \end{array} \end{equation*}The third line of the final column is false. Therefore the argument is not valid. This is because there are other reasons that I might be tired (perhaps I ran a marathon yesterday).
If I stay up late at night, then I will be tired in the morning. I am not tired this morning. Therefore, I did not stay up late last night. Solution.
Once again, let \(P = \) "I stay up late" and let \(Q = \) "I am tired in the morning."
\begin{equation*} \begin{array}{c|c|c|c|c|c|c} P & Q & P \rightarrow Q & \neg Q & \neg Q \wedge (P \rightarrow Q) & \neg P & \big( \neg Q \wedge (P \rightarrow Q) \big) \rightarrow \neg P \\ \hline T & T & T & F & F & F & T \\ T & F & F & T & F & F & T \\ F & T & T & F & T & T & T \\ F & F & T & T & F & T & T \\ \end{array} \end{equation*}The final column only contains true values. Therefore the argument is valid.
7. Evaluating Another Argument.
Determine the validity of the following argument. Include a truth table to justify your conclusion.
If I work hard, then I earn lots of money. If I earn lots of money, then I pay high taxes. Therefore, if I pay high taxes, then I have worked hard. Solution.
Let \(P = \) "I work hard" and \(Q = \) "I earn lots of money" and \(R = \) "I pay high taxes."
The fifth and seven lines in the final column are false, so the argument is invalid. This line captures the fact that there are other reasons (besides working hard) that I can pay high taxes. For example, I could win the lottery.