Exercise 1: If $r$ is rational ($r \neq 0$) and $x$ is irrational, prove that $r + x$ and $rx$ are irrational.

Suppose $r \in \mathbb{Q}$ and $x \in \mathbb{I}$ but $r+x,\ rx \in \mathbb{Q}$. Then we have $r +x = \frac{a}{b}$ for some $a,b \in \mathbb{Z}$. This implies that $x = \frac{a}{b} - r$. Since $r \in \mathbb{Q}$, $r = \frac{c}{d}$ for some $c, d \in \mathbb{Z}$. Therefore $x = \frac{a}{b} - \frac{c}{d} = \frac{ad-bc}{bd}$ which implies $x \in \mathbb{Q}$: a contradiction. Therefore, $r + x$ is irrational. Now, consider $rx$. If $rx \in \mathbb{Q}$, then

\begin{align} rx &= \frac{a}{b}\ \text{for}\ a,b \in \mathbb{Z}
\implies x &= \frac{a}{b} \cdot \frac{d}{c}
\implies x &\in \mathbb{Q} \end{align}

which is a contradiction. Therefore, $rx$ is irrational.

Exercise 2: Prove that there is no rational whose square is $12$.

Suppose there exists a rational $p$ such that $p^2 = 12$. Since $p \in \mathbb{Q}$, we can write $p = \frac{m}{n}$ where the fraction $\frac{m}{n}$ is in lowest terms, i.e. $m$ and $n$ are not both even. Then we have $\frac{m^2}{n^2} = 12 \implies m^2=4\cdot 3n^2 \implies \frac{m^2}{4} = 3n^2$ Since $3n^2$ is an integer, $\frac{m^2}{4}$ must be divisible by $4$ and hence even. Since only one of $m$ and $n$ can be even, this implies that $n$ is odd, from which it follows that $3n^2$ is odd. But if $3n^2$ is odd and $\frac{m^2}{4}$ is even then we must have $\frac{m^2}{4} \neq 3n^2$ Therefore, there is no rational whose square is $12$.

Prove Proposition 1.15. (a) We have \begin{align} y = 1\cdot y = \left(\frac{1}{x} \cdot x\right) \cdot y = \frac{1}{x} \cdot (xy) = \frac{1}{x} \cdot(xz) = \left(\frac{1}{x} \cdot x \right) \cdot z = 1 \cdot z = z \end{align} (b) If $xy = x$ then $xy = x \cdot 1$, so by (a) we have $y=1$.
(c) If $xy = 1$ then \begin{align} y = 1 \cdot y = \left(\frac{1}{x} \cdot x \right) \cdot y = \frac{1}{x} \cdot (xy) = \frac{1}{x} \cdot 1 = \frac{1}{x} \end{align} (d) We have \begin{align} \dfrac{1}{(1/x)} = 1 \cdot \dfrac{1}{(1/x)} = \left( \frac{1}{x} \cdot x \right) \cdot \dfrac{1}{(1/x)} = \left(\frac{1}{x} \cdot \dfrac{1}{(1/x)} \right) \cdot x = 1 \cdot x = x \end{align} Let $E$ be a nonempty subset of an ordered set; suppose $\alpha$ is a lower bound of $E$ and $\beta$ is an upper bound of $E$. Prove that $\alpha \leq \beta$. \ Because $E$ is nonempty, there exists $x \in E$. Since $\alpha$ is a lower bound of $E$, we have $\alpha \leq x$. Since $\beta$ is an upper bound of $E$, we have $x \leq \beta$. Therefore $\alpha \leq x \leq \beta \implies \alpha \leq \beta$ Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A = - \sup(-A)$ Let $\alpha = \sup(-A)$. Then for each $x \in -A$ we have $x \leq \alpha$. Therefore, we have $-x \geq -\alpha$ for each $-x \in A$, so $-\alpha$ is a lower bound for $A$. Now, suppose there exists a lower bound $\beta$ of $A$ such that $\beta > -\alpha$. Then \begin{align} -x \geq \beta > \alpha \quad \forall -x \in A \implies x \leq -\beta < \alpha \quad \forall x \in -A \implies \alpha \neq \sup(-A) \end{align} which is a contradiction. Therefore, $-\alpha = \inf A$ which implies that $\inf A = -\sup (-A)$ (a) We have \begin{align} (b^m)^{1/n} = (b^m)^{r/m} =b^r = b^{p/q} = (b^p)^{1/q} \end{align} (b) We have \begin{align} b^{r+s} = b^{\frac{m}{n} + \frac{p}{q}} = b^{\frac{mq + np}{nq}} = (b^{mq}b^{np})^{\frac{1}{nq}} = b^{\frac{m}{n}}b^{\frac{p}{q}} = b^rb^s \end{align} (c) Let $b^t \in B(r)$. We have \begin{align} b^t \cdot b^{r-t} = b^r \end{align} Since $r \geq t$ for every $t$, we know that $b^{r-t} \geq 1 \implies b^r \geq b^t$. Therefore $b^r$ is an upper bound for $B(r)$.

Now, suppose $c$ is an upper bound for $B(r)$ and $c < b^r$. Then $c = b^{r - \varepsilon}$ for some $\varepsilon \in \mathbb{Q}$. Since $\mathbb{Q}$ is dense in $\mathbb{R}$, there exists $t$ such that $r - \varepsilon < t < r$. Therefore $b^t > c$ and $b^t \in B(r)$ so $c$ is not an upper bound for $B(r)$, a contradiction. Therefore, $b^r = \sup B(r)$.
(d) We have \begin{align} b^{x+y} = \sup B(x+y) \end{align} where $x \in B(x)$ and

Exercise Fix $b > 1$, $y > 0$, and prove that there is a unique real $x$ such that $b^x = y$, by completing the following outline. (This $x$ is called the \textit{logarithm of y to the base b}.)
(a) For any positive integer $n$, $b^n - 1 \geq n(b-1)$.
(b) Hence $b - 1 \geq n(b^{1/n}-1)$.
(c) If $t > 1$ and $n > (b-1)/(t-1)$, then $b^{1/n} < t$.
(d) If $w$ is such that $b^w < y$, then $b^{w + (1/n)}y < y$ for sufficiently large $n$.
(e) If $b^2 > y$, then $b^{w - (1/n)} > y$ for sufficiently large $n$.
(f) Let $A$ be the set of all $w$ such that $b^w < y$, and show that $x = \sup A$ satisfies $b^x = y$.
(g) Prove that this $x$ is unique.

(a) Note that

\begin{align} b^n-1 &= b^{n-1}\cdot b-1
&= (b-1)b^{n-1} + b^{n-1} - 1
&= (b-1)b^{n-1} + (b-1)b^{n-2} + b^{n-2} - 1
&\ \ \vdots
& = (b-1)(b^{n-1} + b^{n-2} + \cdots + b + 1) \end{align}

Since each of $b^{n-1}, b^{n-2}, \dots, b, 1 \geq 1$ we have

\begin{align} (b-1)(b^{n-1} + b^{n-2} + \cdots + b + 1) \geq (b-1)\underbrace{(1 + 1 + \cdots + 1)}_{n\text{-times}} = n(b-1) \end{align}

(b) Fix $n$ and set $b = a^{1/n}$. Then by part (a) we have

\begin{align} b^n -1 \geq n(b-1) \implies (a^{1/n})^n - 1 \geq n(a^{1/n} - 1) \implies a-1 \geq n(a^{1/n} -1) \end{align}

(c) Since $\dfrac{b-1}{t-1} \geq n$ we have

\begin{align} \frac{b-1}{t-1}(b^{1/n}-1) < n(b^{1/n}-1) \leq b-1
\implies \frac{1}{t-1}(b^{1/n}-1) < 1
\implies t - 1 > b^{1/n} - 1
\implies t > b^{1/n} \end{align}