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 negations 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 propositions logically equivalent?
Answer
No.
Exercise 3: Implication vs bi-implication#
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 bi-implication arrows and between which only implication arrows? Explain why not every value of \(x\) shown after the last step is necessarily a solution to the original equation.
Hint
According to Equation (1.22) in Theorem 1.3.4 in the textbook, a bi-implication 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 a different (though still rather laborious) way, again 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 bi-implication 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, bi-implication arrows can be placed between all steps.
Exercise 4: Tautology#
Consider the logical proposition \((P \vee Q)\vee (\neg P \wedge \neg Q)\).
Question a#
Use truth tables to show 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 to interpret \((P \vee Q)\vee (\neg P \wedge \neg Q)\) in words in a similar manner.
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 four equations in this exercise by first introducing a tautology.
Question a#
Solve the equation \(|x|=-x+1\).
Hint
See Example 1.4.2 in the textbook for inspiration.
Hint
For the first equation, rewrite it as \(|x|=-x+1 \wedge ( x<0 \vee x \ge 0)\).
Answer
\(x=\frac{1}{2}\)
Question b#
Solve the equation \(|x|=2x+1\).
Answer
\(x=-\frac{1}{3}\)
Question c#
Solve the equation \(3|2x-1|=-4x+3\).
Answer
\(x=0 \vee x=\frac{3}{5}\)
Question d#
Solve the equation \(|2x+1|=|-5x+3|\).
Answer
\(x=\frac{4}{3} \vee x=\frac{2}{7}\)
Have you finished all the exercises?#
If it went too quickly and you are eager for more, then click here for some challenging optional extra exercises.