Exercises – Short Day

Exercises – Short Day#

Exercise 1: Repetition Exercise about 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 one can place a biimplication arrow 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 by the use of 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 that the above is a tautology.

Hint

In words one can pronounce \(\neg P \wedge \neg Q\) as “not \(P\) and also not \(Q\)”. Alternatively, one can say “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#

  1. Solve the equation \(|x|=-x+1\).

  2. Solve the equation \(|x|=2x+1\).

  3. Solve the equation \(3|2x-1|=-4x+3\).

  4. 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

  1. \(x=\frac{1}{2}\)

  2. \(x=-\frac{1}{3}\)

  3. \(x=0 \vee x=\frac{3}{5}\)

  4. \(x=\frac{4}{3} \vee x=\frac{2}{7}\)

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