Exercises – Short Day#
Exercise 1: Revision Exercise on Negation#
Consider again the logical propositions \(u\), \(v\) and \(w\) from Exercise 1 on Long Day.
Question a#
Write out the truth tables for \(\neg u\), \(\neg v\) and \(\neg w\).
Hint
Use the truth tables that you created in Exercise 1 on Long Day rather than starting over from scratch.
Question b#
Now, let \(P\) and \(Q\) denote the negation of the logical propositions “\(6\) is an odd number” and “\(3<7\)”, respectively.
What are the truth values of the three propositions \(u\), \(v\) and \(w\)?
Answer
\(u\) is \(\mathrm T\), \(v\) is \(\mathrm T\), and \(w\) is \(\mathrm T\).
Exercise 2: Implication#
Question a#
Let \(P\) be a logical proposition. Write out the truth tables for \((P \Rightarrow P) \Rightarrow P\) and \(P \Rightarrow (P \Rightarrow P)\).
Question b#
Are the above two logical propositions equivalent?
Answer
No.
Exercise 3: Implication vs. Biimplication#
Question a#
Solve the linear equation \(x-2 = 3\).
Question b#
A student solves the above linear equation in a rather laborious manner with more steps than necessary. They forget to write implication arrows between the steps along the way and arrive at a rather dubious result. The steps they write down are the following:
\(x-2=3\)
\((x-2)^2 =3^2\)
\(x^2 -4x+4=9\)
\(x^2 -4x -5 =0\)
\(x=-1 \vee x=5\)
Between which steps can we place biimplication arrows and between which only implication arrows? Explain why not every shown value of \(x\) after the last step necessarily is a solution to the original equation.
Hint
According to Equation (1.22) in Theorem 1.3.4 in the textbook, a biimplication arrow can be placed between two logical propositions \(P\) and \(Q\) precisely if \(P\) implies \(Q\) and \(Q\) implies \(P\).
Question c#
Another student solves the above linear equation in another (and still rather laborious) way, still without writing implication arrows between the steps:
\(x-2=3\)
\(x-5 =0\)
\((x-5)^2 =0\)
\(x^2 -10x +25 =0\)
\(x=5\) (there is only one solution to the quadratic equation that appeared)
Between which steps can we place biimplication arrows and between which only implication arrows? Explain why this time every value of \(x\) that is shown after the last step is a solution to the original equation.
Answer
Here, biimplication arrows are to be placed between all steps.
Exercise 4: Tautology#
Consider the logical proposition \((P \vee Q)\vee (\neg P \wedge \neg Q)\).
Question a#
Show using truth tables that this is a tautology.
Hint
If you are in doubt about what a tautology is, then see the text in the textbook between Definition 1.3.1 and Example 1.3.1.
Question b#
Explain in words why the above is a tautology.
Hint
In words \(\neg P \wedge \neg Q\) would be read as “not \(P\) and also not \(Q\)”. Alternatively, it could be read as “neither \(P\) nor \(Q\)”. Now try in a similar manner to interpret \((P \vee Q)\vee (\neg P \wedge \neg Q)\) in words.
Question c#
Repeat Question a but with the proposition \((P \Rightarrow Q)\vee (Q \Rightarrow P)\).
Question d#
Repeat Question a but now with the proposition \((P \Rightarrow Q)\vee (\neg P \Rightarrow Q)\).
Exercise 5: Equations#
Solve the following four equations by first introducing a tautology.
Question a#
Solve the equation \(|x|=-x+1\).
Solve the equation \(|x|=2x+1\).
Solve the equation \(3|2x-1|=-4x+3\).
Solve the equation \(|2x+1|=|-5x+3|\).
Hint
See Example 1.4.2 in the textbook for inspiration.
Hint
For the first equation, rewrite to \(|x|=-x+1 \wedge ( x<0 \vee x \ge 0)\).
Answer
\(x=\frac{1}{2}\)
\(x=-\frac{1}{3}\)
\(x=0 \vee x=\frac{3}{5}\)
\(x=\frac{4}{3} \vee x=\frac{2}{7}\)
Are you Done with all Exercises?#
If it went too quick and you are eager for more then click here for some challenging optional extra exercises.