Proof Practical Class, Week 24

Question One §

Question §

Let the following statements be denoted as…

$$\begin{array}{rl}R& =\text{it is raining}\\ S& =\text{the sun is shining}\\ B& =\text{there is a rainbow}\end{array}$$

Thus, translate the following statements into logical formulae:

1. There is a rainbow only if it is raining but the sun is shining.
2. There is no rainbow although it is raining and the sun is shining.

And vice versa:

3. $(\neg B\wedge R)⟹\neg S$.
4. $S\vee (R\wedge B)$.

Solution §

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Question Two §

Question §

Use truth tables to prove the logical equivalences…

1. $\neg (P⟹Q)\equiv P\wedge \neg Q$.
2. $(P\wedge Q)\vee R\equiv (P\vee R)\wedge (Q\vee R)$.

Solution §

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Question Three §

Question §

Use the properties of logical operations to simplify the expression $\neg P\wedge \neg (Q\wedge \neg P)$.

Solution §

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Question Four §

Question §

Prove by contradiction:

1. if $k$ is an integer such that $k^2$ is not divisible by $3$, then $k$ is not divisible by $2$.
2. $\sqrt{5}\text{centernot}\in \mathbb{Q}$.

Solution §

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Question Five §

Question §

Use the Cantor-Bernstein-Schroder theorem to show that $|A|=|B|$ by producing injective mappings $A\to B$ and $B\to A$, where $A$ is the square $A=[0,1]\times [0,1]$ and $B$ is the disc $B=\{(x,y):x^2+y^2\le 1\}$.

Hint: use a geometric construction by uniformly “shrinking” one object and then placing within the other.

Solution §

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