**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}