COMP 2711 Ex 3

1

a

Let $p$ be the proposition that $x$ is irrational
Let $q$ be the proposition that $x^{1/5}$ is irrational

To prove $p\to q$, we assume $\neg q$ is true

Since $x^{1/5}$ is rational

$$\begin{array}{r}{x}^{1/5}=\frac{a}{b}\\ x=\frac{a^5}{b^5}\end{array}$$

We can see that $x$ is rational, i.e. $\neg p$ is true, and $\neg q\to \neg p$

Therefore by proof of contraposition, $p\to q$

b

Let $p$ be the proposition that $x$ is irrational and $xy$ is rational
Let $q$ be the proposition that $y$ is irrational

To prove $p\to q$, we assume $\neg (p\to q)$ is true, i.e. $p\wedge \neg q$ is true

Since $y$ is rational

$$\begin{array}{rl}y& =\frac{a}{b}\\ \frac{1}{y}& =\frac{b}{a}\end{array}$$

Since $xy$ is rational

$$\begin{array}{rl}xy& =\frac{c}{d}\\ x& =\frac{c}{d}\frac{1}{y}\\ x& =\frac{cb}{da}\end{array}$$

We can see that $x$ is rational, i.e. $p$ is false, contradicting the assumption $p\wedge \neg q$ is true

Therefore by proof of contradiction, $p\to q$