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Enderton (logic). Most of exercises 6.1.

finite-set-exercises
Joshua Potter 2023-08-18 11:22:23 -06:00
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commit 9fe4f2ee78
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327
Bookshelf/Enderton/Logic.tex
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@ -8,6 +8,7 @@
% Truth table start and final color
\definecolor{TTStart}{gray}{0.95}
\definecolor{TTEnd}{rgb}{1,1,0}
\colorlet{TTInvalid}{Salmon}
\newcolumntype{s}{>{\columncolor{TTStart}}c}
\newcolumntype{e}{>{\columncolor{TTEnd}}c}
@ -711,7 +712,7 @@
Suppose that $\alpha$ is a wff not containing the negation symbol $\neg$.
\subsubsection{\verified{Exercise 1.1.5a}}%
\hyperlabel{ssub:exercise-1.1.5.a}
\hyperlabel{ssub:exercise-1.1.5a}
Show that the length of $\alpha$ (i.e., the number of symbols in the string)
is odd.
@ -719,14 +720,14 @@
$4k + 1$.
\code*{Enderton.Logic.Chapter\_1}
{Enderton.Logic.Chapter\_1.exercise\_1\_1\_5\_a}
{Enderton.Logic.Chapter\_1.exercise\_1\_1\_5a}
\begin{proof}
Define $L$ to be the length function mapping arbitrary
\nameref{ref:well-formed-formula} to its length and let
\begin{equation}
\hyperlabel{ssub:exercise-1.1.5.a-eq1}
\hyperlabel{ssub:exercise-1.1.5a-eq1}
S = \{\phi \mid
\phi \text{ is a wff containing } \neg \text{ or }
\exists k \in \mathbb{N}, L(\phi) = 4k + 1\}.
@ -736,14 +737,14 @@
We then conclude with (iii) the proof of the theorem statement.
\paragraph{(i)}%
\hyperlabel{par:exercise-1.1.5.a-i}
\hyperlabel{par:exercise-1.1.5a-i}
Every sentence symbol has length 1 by definition.
That is, every sentence symbol has length $(4)(0) + 1$.
Hence $S$ contains every sentence symbol.
\paragraph{(ii)}%
\hyperlabel{par:exercise-1.1.5.a-ii}
\hyperlabel{par:exercise-1.1.5a-ii}
Let $\alpha, \beta \in S$.
Then there exists some $k_\alpha$ and $k_\beta$ such that
@ -769,7 +770,7 @@
\paragraph{(iii)}%
By \nameref{par:exercise-1.1.5.a-i} and \nameref{par:exercise-1.1.5.a-ii},
By \nameref{par:exercise-1.1.5a-i} and \nameref{par:exercise-1.1.5a-ii},
the \nameref{sub:induction-principle-1} indicates $S$ is the set of all
wffs.
Thus all well-formed formulas not containing symbol $\neg$ has length
@ -779,14 +780,14 @@
\end{proof}
\subsubsection{\verified{Exercise 1.1.5b}}%
\hyperlabel{ssub:exercise-1.1.5-b}
\hyperlabel{ssub:exercise-1.1.5b}
Show that more than a quarter of the symbols are sentence symbols.
\textit{Suggestion}: Apply induction to show that the number of sentence
symbols is of the form $k + 1$.
\code*{Enderton.Logic.Chapter\_1}
{Enderton.Logic.Chapter\_1.exercise\_1\_1\_5\_b}
{Enderton.Logic.Chapter\_1.exercise\_1\_1\_5b}
\begin{proof}
@ -847,16 +848,14 @@
\begin{proof}
Yes.
To prove, consider the following truth table:
$$
\begin{array}{s|c|s|c|s|e|s}
$$\begin{array}{s|c|s|c|s|e|s}
(((P & \Rightarrow & Q) & \Rightarrow & P) & \Rightarrow & P) \\
\hline
T & T & T & T & T & T & T \\
T & F & F & T & T & T & T \\
F & T & T & F & F & T & F \\
F & T & F & F & F & T & F
\end{array}
$$
\end{array}$$
\end{proof}
\subsection{\pending{Exercise 1.2.2b}}%
@ -894,16 +893,14 @@
By definition,
$$\sigma_{k + 2} = ((\sigma_{k} \Rightarrow P) \Rightarrow P).$$
Consider the truth table of the above:
$$
\begin{array}{c|c|s|e|s}
$$\begin{array}{c|c|s|e|s}
((\sigma_k & \Rightarrow & P) & \Rightarrow & P) \\
\hline
T & T & T & T & T \\
T & T & T & T & T \\
T & F & F & T & F \\
T & F & F & T & F
\end{array}
$$
\end{array}$$
This shows $\sigma_{k+2}$ is a tautology.
Hence $P(k + 2)$ is true.
@ -927,16 +924,14 @@
Suppose $k = 1$.
Then $\sigma_k = \sigma_1 = ((P \Rightarrow Q) \Rightarrow P)$.
The following truth table shows $\sigma_1$ is not a tautology:
$$
\begin{array}{s|c|s|e|s}
$$\begin{array}{s|c|s|e|s}
(((P & \Rightarrow & Q) & \Rightarrow & P) \\
\hline
T & T & T & T & T \\
T & F & F & T & T \\
F & T & T & F & F \\
F & T & F & F & F
\end{array}
$$
\end{array}$$
\subparagraph{Case 2}%
@ -945,40 +940,65 @@
By definition, $$\sigma_k = (\sigma_{k-1} \Rightarrow P).$$
By \eqref{par:exercise-1.2.2b-i}, $\sigma_{k-1}$ is a tautology.
The following truth table shows $\sigma_k$ is not:
$$
\begin{array}{c|e|s}
$$\begin{array}{c|e|s}
(\sigma_{k-1} & \Rightarrow & P) \\
\hline
T & T & T \\
T & T & T \\
T & F & F \\
T & F & F
\end{array}
$$
\end{array}$$
\end{proof}
\subsection{\sorry{Exercise 1.2.3a}}%
\subsection{\pending{Exercise 1.2.3a}}%
\hyperlabel{sub:exercise-1.2.3a}
Determine whether or not $((P \Rightarrow Q) \lor (Q \Rightarrow P))$ is a
tautology.
\begin{proof}
TODO
Consider the following truth table:
$$\begin{array}{s|c|s|e|s|c|s}
((P & \Rightarrow & Q) & \lor & (Q & \Rightarrow & P)) \\
\hline
T & T & T & T & T & T & T \\
T & F & F & T & F & T & T \\
F & T & T & T & T & F & F \\
F & T & F & T & F & T & F
\end{array}$$
The above makes it immediately evident that
$((P \Rightarrow Q) \lor (Q \Rightarrow P))$ is a tautology.
\end{proof}
\subsection{\sorry{Exercise 1.2.3b}}%
\subsection{\pending{Exercise 1.2.3b}}%
\hyperlabel{sub:exercise-1.2.3b}
Determine whether or not $((P \land Q) \Rightarrow R)$ tautologically implies
$((P \Rightarrow R) \lor (Q \Rightarrow R))$.
\begin{proof}
TODO
Consider the following truth table:
$$\begin{array}{s|s|s|e|e}
P & Q & R &
((P \land Q) \Rightarrow R) &
((P \Rightarrow R) \lor (Q \Rightarrow R)) \\
\hline
T & T & T & T & T \\
T & T & F & F & F \\
T & F & T & T & T \\
T & F & F & T & T \\
F & T & T & T & T \\
F & T & F & T & T \\
F & F & T & T & T \\
F & F & F & T & T
\end{array}$$
The above makes it immediately evident that
$((P \land Q) \Rightarrow R)$ tautologically implies
$((P \Rightarrow R) \lor (Q \Rightarrow R))$.
\end{proof}
\subsection{\sorry{Exercise 1.2.4}}%
\subsection{\pending{Exercise 1.2.4}}%
\hyperlabel{sub:exercise-1.2.4}
Show that the following hold:
@ -992,10 +1012,98 @@
together with the one possibly new member $\alpha$.)
\begin{proof}
TODO
\paragraph{(a)}%
We prove each direction of the biconditional.
\subparagraph{($\Rightarrow$)}%
Assume $\Sigma; \alpha \vDash \beta$.
Let $v$ be a truth assignment for the sentence symbols in
$\Sigma; \alpha$ and $\beta$.
Then if $v$ satisfies every member of $\Sigma$ and $\alpha$, it must
also satisfy $\beta$.
Denote $\bar{v}(\Sigma)$ as the proposition that $v$ satisfies every
member of $\Sigma$ and consider the following truth table:
$$\begin{array}{s|s|s|e}
\bar{v}(\Sigma) & \bar{v}(\alpha) & \bar{v}(\beta) &
\bar{v}((\alpha \Rightarrow \beta)) \\
\hline
T & T & T & T \\
\rowcolor{TTInvalid}
T & T & F & F \\
T & F & T & T \\
T & F & F & T \\
F & T & T & T \\
F & T & F & F \\
F & F & T & T \\
F & F & F & T
\end{array}$$
The red row denotes a contradiction: it is not possible for
$\bar{v}(\Sigma)$ and $\bar{v}(\alpha)$ to be true but
$\bar{v}(\beta)$ to be false.
All remaining rows show that when $\bar{v}(\Sigma)$ is true, so is
$\bar{v}((\alpha \Rightarrow \beta))$.
Thus $\Sigma \vDash (\alpha \Rightarrow \beta)$.
\subparagraph{($\Leftarrow$)}%
Assume $\Sigma \vDash (\alpha \Rightarrow \beta)$.
Let $v$ be a \nameref{ref:truth-assignment} for the sentence symbols in
$\Sigma$, $\alpha$, and $\beta$.
Then if $v$ satisfies every member of $\Sigma$, it must also satisfy
$(\alpha \Rightarrow \beta)$.
Denote $\bar{v}(\Sigma)$ as the proposition that $v$ satisfies every
member of $\Sigma$.
By definition, $\bar{v}((\alpha \Rightarrow \beta)) = T$ if and only if
$\bar{v}(\alpha) = F$ or $\bar{v}(\alpha)$ and $\bar{v}(\beta)$ are
both true.
Thus the only situation in which both $\bar{v}(\Sigma) = T$ and
$\bar{v}(\alpha) = T$ corresponds to when $\bar{v}(\beta) = T$.
Hence $\Sigma; \alpha \vDash \beta$.
\paragraph{(b)}%
We prove each direction of the biconditional.
\subparagraph{($\Rightarrow$)}%
Suppose $\alpha \vDash \Dashv \beta$.
Let $v$ be a \nameref{ref:truth-assignment} for the sentence symbols in
$\alpha$ and $\beta$.
Consider the following truth table:
$$\begin{array}{s|e|s}
(\alpha & \Leftrightarrow & \beta) \\
\hline
T & T & T \\
\rowcolor{TTInvalid}
T & F & F \\
\rowcolor{TTInvalid}
F & F & T \\
F & T & F
\end{array}$$
The red rows indicate possibilites that cannot occur, for
$\alpha \vDash \beta$ and $\beta \vDash \alpha$ by hypothesis.
Of the remaining rows, $(\alpha \Leftrightarrow \beta)$ is true.
Hence $\vDash (\alpha \Leftrightarrow \beta)$.
\subparagraph{($\Leftarrow$)}%
Assume $\vDash (\alpha \Rightarrow \beta)$.
Let $v$ be a \nameref{ref:truth-assignment} for the sentence symbols in
$\alpha$ and $\beta$.
By definition, $\bar{v}((\alpha \Leftrightarrow \beta)) = T$ if and only
if $\bar{v}(\alpha) = \bar{v}(\beta)$.
Thus if $\bar{v}(\alpha)$ is true, so must $\bar{v}(\beta)$.
That is, $\alpha \vDash \beta$.
Likewise, if $\bar{v}(\beta)$ is true, so must $\bar{v}(\alpha)$.
Therefore $\beta \vDash \alpha$.
Hence $\alpha \vDash \Dashv \beta$.
\end{proof}
\subsection{\sorry{Exercise 1.2.5}}%
\subsection{\pending{Exercise 1.2.5}}%
\hyperlabel{sub:exercise-1.2.5}
Prove or refute each of the following assertions:
@ -1007,10 +1115,62 @@
\end{enumerate}
\begin{proof}
TODO
\paragraph{(a)}%
WLOG, suppose $\Sigma \vDash \alpha$.
That is, every truth assignment for sentence symbols found in $\Sigma$ and
$\alpha$ that satisfies every member of $\Sigma$ also satisfies
$\alpha$.
Let $v$ be one of these truth assignments.
Denote $\bar{v}(\Sigma)$ as the proposition that $v$ satisfies every
member of $\Sigma$ and consider the following truth table:
$$\begin{array}{s|s|s|e}
\bar{v}(\Sigma) & \bar{v}(\alpha) & \bar{v}(\beta) &
\bar{v}((\alpha \lor \beta)) \\
\hline
T & T & T & T \\
T & T & F & T \\
\rowcolor{TTInvalid}
T & F & T & T \\
\rowcolor{TTInvalid}
T & F & F & T \\
F & T & T & T \\
F & T & F & T \\
F & F & T & T \\
F & F & F & F
\end{array}$$
The red rows indicate possiblities that cannot occur since
$\Sigma \vDash \alpha$ by hypothesis.
All remaining rows show that when $\bar{v}(\Sigma)$ is true, so is
$\bar{v}((\alpha \lor \beta))$.
Hence $\Sigma \vDash (\alpha \lor \beta)$.
\paragraph{(b)}%
We proceed by counterexample.
Suppose $\Sigma = \emptyset$.
That is, assume $(\alpha \lor \beta)$ is a tautology, i.e.
$\vDash (\alpha \lor \beta)$.
Consider the following truth table:
$$\begin{array}{s|e|s}
(\alpha & \lor & \beta) \\
\hline
T & T & T \\
T & T & F \\
F & T & T \\
\rowcolor{TTInvalid}
F & F & F
\end{array}$$
The red row indicates an impossibility, since $(\alpha \lor \beta)$ should
always be true by hypothesis.
But this table also clearly demonstrates that $\not\vDash \alpha$ and
$\not\vDash \beta$.
Thus the conditional statement proposed must not be generally true.
\end{proof}
\subsection{\sorry{Exercise 1.2.6a}}%
\subsection{\pending{Exercise 1.2.6a}}%
\hyperlabel{sub:exercise-1.2.6a}
Show that if $v_1$ and $v_2$ are \nameref{ref:truth-assignment}s which agree
@ -1019,7 +1179,104 @@
Use the \nameref{sub:induction-principle-1}.
\begin{proof}
TODO
Let $\sigma$ map a \nameref{ref:well-formed-formula} $\phi$ to the set of
sentence symbols found in $\phi$.
Define
\begin{equation}
\hyperlabel{sub:exercise-1.2.6a-eq1}
S = \{\phi \mid ((\sigma(\phi) = \sigma(\alpha)) \Rightarrow
(\bar{v}_1(\phi) = \bar{v}_2(\phi)))\}.
\end{equation}
We prove that (i) the set of sentence symbols is found in $\phi$ and (ii)
$S$ is closed under the five \nameref{ref:formula-building-operations}.
Afterward we show that (iii) our theorem statement holds.
\paragraph{(i)}%
\hyperlabel{par:exercise-1.2.6a-i}
Let $A_n$ denote an arbitrary sentence symbol.
Suppose $\sigma(A_n) = \{A_n\} = \sigma(\alpha)$.
But then $\bar{v}_1(A_n) = \bar{v}_2(A_n)$ since, by hypothesis, $v_1$
and $v_2$ agree on all the sentence symbols found in $\alpha$.
Hence $S$ contains all the sentence symbols.
\paragraph{(ii)}%
\hyperlabel{par:exercise-1.2.6a-ii}
Let $\beta, \gamma \in S$.
There are three cases to consider:
\subparagraph{Case 1}%
\hyperlabel{spar:exercise-1.2.6a-ii-1}
Suppose $\sigma(\beta) \neq \sigma(\alpha)$.
By definition, $\mathcal{E}_{\neg}(\beta) = (\neg\beta)$.
Then clearly $\sigma(\mathcal{E}_{\neg}(\beta)) \neq \sigma(\alpha)$.
Therefore $\mathcal{E}_{\neg}(\beta) \in S$.
Likewise,
$\mathcal{E}_{\square}(\beta, \gamma) =
(\beta \mathop{\square} \gamma)$
where $\square$ is one of the binary connectives $\land$, $\lor$,
$\Rightarrow$, $\Leftrightarrow$.
Again, it clearly follows that
$\mathcal{E}_{\square}(\beta, \gamma) \neq \sigma(\alpha)$.
Thus $\mathcal{E}_{\square}(\beta, \gamma) \in S$.
\subparagraph{Case 2}%
Suppose $\sigma(\gamma) \neq \sigma(\alpha)$.
This case mirrors \nameref{spar:exercise-1.2.6a-ii-1}.
\subparagraph{Case 3}%
Suppose $\sigma(\beta) = \sigma(\alpha) = \sigma(\alpha)$.
By definition, $\mathcal{E}_{\neg}(\beta) = (\neg\beta)$.
Then clearly $\sigma(\mathcal{E}_{\neg}(\beta)) = \sigma(\alpha)$.
Since
\begin{align*}
\bar{v}_1((\neg\beta))
& = (\neg\bar{v}_1(\beta)) \\
& = (\neg\bar{v}_2(\beta)) & \eqref{sub:exercise-1.2.6a-eq1} \\
& = \bar{v}_2((\neg\beta)),
\end{align*}
it follows that $\mathcal{E}_{\neg}(\beta) \in S$.
Likewise,
$\mathcal{E}_{\square}(\beta, \gamma) =
(\beta \mathop{\square} \gamma)$
where $\square$ is one of the binary connectives $\land$, $\lor$,
$\Rightarrow$, $\Leftrightarrow$.
Again, it clearly follows that
$\mathcal{E}_{\square}(\beta, \gamma) \neq \sigma(\alpha)$.
Since
\begin{align*}
\bar{v}_1((\alpha \mathop{\square} \beta))
& = (\bar{v}_1(\alpha) \mathop{\square} \bar{v}_1(\beta)) \\
& = (\bar{v}_2(\alpha) \mathop{\square} \bar{v}_2(\beta)
& \eqref{sub:exercise-1.2.6a-eq1} \\
& = \bar{v}_2((\alpha \mathop{\square} \beta)),
\end{align*}
it follows that $\mathcal{E}_{\square}(\beta, \gamma) \in S$.
\subparagraph{Subconclusion}%
The above three cases are exhaustive.
Thus it follows $S$ is closed under the five formula-buildiong
operations.
\paragraph{(iii)}%
By \nameref{par:exercise-1.2.6a-i} and \nameref{par:exercise-1.2.6a-ii},
the \nameref{sub:induction-principle-1} indicates $S$ is the set of all
wffs.
Since $\alpha$ is a well-formed formula, it follows $\alpha \in S$.
Therefore
$$((\sigma(\alpha) = \sigma(\alpha)) \Rightarrow
\bar{v}_1(\alpha) = \bar{v}_2(\alpha)).$$
The antecedent clearly holds true.
Hence $\bar{v}_1(\alpha) = \bar{v}_2(\alpha)$ as expected.
\end{proof}
\subsection{\sorry{Exercise 1.2.6b}}%
@ -1030,7 +1287,7 @@
Show that $\Sigma \vDash \tau$ iff every truth assignment for $\mathcal{S}$
which satisfies every member of $\Sigma$ also satisfies $\tau$.
(This is an easy consequence of part (a). The point of part (b) is that we do
not need to worry aboutgetting the domain of a truth assignment
not need to worry about getting the domain of a truth assignment
\textit{exactly} perfect, as long as it is big enough. For example, one
option would be always to use truth assignments on the set of \textit{all}
sentence symbols. The drawback is that these are infinite objects, and there

4
Bookshelf/Enderton/Logic/Chapter_1.lean
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@ -327,7 +327,7 @@ length of `α` (i.e., the number of symbols in the string) is odd.
*Suggestion*: Apply induction to show that the length is of the form `4k + 1`.
-/
theorem exercise_1_1_5_a (α : Wff) (hα : ¬α.hasNotSymbol)
theorem exercise_1_1_5a (α : Wff) (hα : ¬α.hasNotSymbol)
: Odd α.length := by
suffices ∃ k : ℕ, α.length = 4 * k + 1 by
have ⟨k, hk⟩ := this
@ -363,7 +363,7 @@ than a quarter of the symbols are sentence symbols.
*Suggestion*: Apply induction to show that the number of sentence symbols is
`k + 1`.
-/
theorem exercise_1_1_5_b (α : Wff) (hα : ¬α.hasNotSymbol)
theorem exercise_1_1_5b (α : Wff) (hα : ¬α.hasNotSymbol)
: α.sentenceSymbolCount > (Nat.cast α.length : ℝ) / 4 := by
rw [
α.length_eq_sum_symbol_count,

220
Bookshelf/Enderton/Set.tex
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@ -8603,14 +8603,83 @@
\section{Exercises 6}%
\hyperlabel{sec:exercises-6}
\subsection{\sorry{Exercise 6.1}}%
\subsection{\unverified{Exercise 6.1}}%
\hyperlabel{sub:exercise-6-1}
Show that the equation $$f(m, n) = 2^m(2n + 1) - 1$$ defines a one-one-one
correspondence between $\omega \times \omega$ and $\omega$.
\begin{proof}
TODO
We prove that (i) $f$ is one-to-one and (ii) $f$ is onto $\omega$.
\paragraph{(i)}%
Let $y \in \ran{f}$.
Then there exists $(m_1, n_1) \in \omega \times \omega$ such that
$f(m_1, n_1) = y$.
Suppose there exists $(m_2, n_2) \in \omega \times \omega$ such that
$f(m_2, n_2) = y$.
We show that $m_1 = m_2$ and $n_1 = n_2$.
By \nameref{sub:trichotomy-law-natural-numbers}, we know either
$m_1 \leq m_2$ or $m_2 \leq m_1$ and either $n_1 \leq n_2$ or
$n_2 \leq n_1$.
WLOG, assume $m_1 \leq m_2$ and $n_1 \leq n_2$.
Then by \nameref{sub:exercise-4.23}, there exists some $p \in \omega$ such
that $m_1 + p = m_2$.
Likewise, there exists some $q \in \omega$ such that $n_1 + q = n_2$.
Therefore
\begin{align*}
& f(m_1, n_1) = f(m_2, n_2) \\
& \iff 2^{m_1}(2n_1 + 1) - 1 = 2^{m_2}(2n_2 + 1) - 1 \\
& \iff 2^{m_1}(2n_1 + 1) = 2^{m_2}(2n_2 + 1) \\
& \iff 2^{m_1}(2n_1 + 1) = 2^{m_1+p}(2(n_1 + q) + 1) \\
& \iff 2^{m_1}(2n_1 + 1) = 2^{m_1}2^p(2n_1 + 2q + 1) \\
& \iff 2n_1 + 1 = 2^p(2n_1 + 2q + 1) \\
& \iff 2n_1 + 1 = 2^{p+1}n_1 + 2^{p+1}q + 2^p.
\end{align*}
Notice though that the right-hand side of the above equality is smallest
when $p = 0$ and $q = 0$.
Since when evaluated at $p = 0$ and $q = 0$, the equality holds, it
follows this value of $p$ and $q$ are the only possible values that
$p$ and $q$ can take on.
Thus $m_1 + p = m_1 = m_2$ and $n_1 + q = n_1 = n_2$.
\paragraph{(ii)}%
We show $\ran{f} = \omega$ by the
\nameref{sub:strong-induction-principle-natural-numbers}.
It is clear that $\ran{f}$ is a subset of $\omega$.
Let $x \in \omega$ and suppose every number less than $x$ is in $\ran{f}$.
All that remains to be shown is that $x \in \ran{f}$.
By \nameref{sub:exercise-4.14}, there are two cases to consider:
\subparagraph{Case 1}%
Suppose $x$ is an even number.
Then $x = 2 \cdot y$ for some $y \in \omega$.
Then $f(0, y) = 2^0(2y + 1) - 1 = 2y = x$.
Hence $x \in \ran{f}$.
\subparagraph{Case 2}%
Suppose $x$ is an odd number.
Then $x = (2 \cdot y) + 1$ for some $y$.
It immediately follows $x - 1$ is an even number, meaning there exists
some $z$ such that $x - 1 = 2z$.
Since $z \in x$, the induction hypothesis states that there exists
some $m, n \in \omega$ such that $$f(m, n) = z = 2^m(2n + 1) - 1.$$
Therefore
\begin{align*}
f(m + 1, n)
& = 2^{m+1}(2n + 1) - 1 \\
& = (2)(2^m)(2n + 1) - 1 \\
& = 2z + 1 \\
& = x.
\end{align*}
Hence $x \in \ran{f}$.
\end{proof}
\subsection{\unverified{Exercise 6.2}}%
@ -8652,7 +8721,7 @@
Hence $J$ is correctly defined.
\end{proof}
\subsection{\sorry{Exercise 6.3}}%
\subsection{\unverified{Exercise 6.3}}%
\hyperlabel{sub:exercise-6-3}
Find a one-to-one correspondence between the open unit interval $\ioo{0}{1}$
@ -8660,7 +8729,150 @@
irrationals.
\begin{proof}
TODO
\begin{figure}[ht]
\label{sub:exercise-6-3-fig1}
\includegraphics[width=0.6\textwidth]{exercise-6.3.png}
\centering
\end{figure}
Consider function $f \colon (0, 1) \rightarrow \mathbb{R}$ given by
$$f(x) = \begin{cases}
\frac{1}{2x} - 1 & \text{if } x \leq \frac{1}{2} \\
\frac{1}{2(x - 1)} + 1 & \text{otherwise}.
\end{cases}$$
We prove that (i) $f$ is a one-to-one into $\mathbb{R}$, (ii) $f$ is onto
$\mathbb{R}$, and (iii) $f$ takes rationals to rationals and irrationals
to irrationals.
\paragraph{(i)}%
Before proceeding, consider the solutions to the following identity:
\begin{align*}
\frac{1}{2x} - 1 = \frac{1}{2(x - 1)} + 1
& \iff -8x^2 + 8x - 2 = 0 \\
& \iff 4x^2 - 4x + 1 = 0.
\end{align*}
Applying the quadratic equation shows $x = 1 / 2$ is the only solution.
Thus for any $x_1, x_2 \in \ioo{0}{1}$ such that $f(x_1) = f(x_2)$, it
must be that $x_1, x_2 \leq 1 / 2$ or $x_1, x_2 > 1 / 2$.
We now prove $f$ is one-to-one.
Let $y \in \ran{f}$.
Then there exists some $x_1 \in \ioo{0}{1}$ such that $f(x_1) = y$.
Suppose there exists some $x_2 \in \ioo{0}{1}$ such that $f(x_1) = f(x_2)$.
We prove that $x_1 = x_2$ by case analysis:
\subparagraph{Case 1}%
Suppose $x_1, x_2 \leq 1 / 2$.
Then $$\frac{1}{2x_1} - 1 = \frac{1}{2x_2} - 1$$ which straightforwardly
simplifies to $x_1 = x_2$.
\subparagraph{Case 2}%
Suppose $x_1, x_2 > 1 / 2$.
Then $$\frac{1}{2(x_1 - 1)} + 1 = \frac{1}{2(x_2 - 1)} + 1$$ also
straightfowardly simplies to $x_1 = x_2$.
\subparagraph{Subconclusion}%
The above cases are exhaustive.
Therefore $f(x_1) = f(x_2)$ implies $x_1 = x_2$.
Hence $f$ is one-to-one.
\paragraph{(ii)}%
Let $y \in \mathbb{R}$.
We prove that there exists an $x \in (0, 1)$ such that $f(x) = y$.
There are three cases we consider:
\subparagraph{Case 1}%
Suppose $y = 0$.
Then $f(1 / 2) = 0 = y$ is a readily identifiable solution.
\subparagraph{Case 2}%
Suppose $y > 0$.
Consider $x = \frac{1}{2(y + 1)}$.
We note that $x < 1 / 2$ meaning
\begin{align*}
f(x)
& = f\left(\frac{1}{2(y + 1)}\right) \\
& = \frac{1}{2(\frac{1}{2(y + 1)})} - 1 \\
& = (y + 1) - 1 \\
& = y.
\end{align*}
\subparagraph{Case 3}%
Suppose $y < 0$.
Consider $x = \frac{1}{2(y - 1)} + 1$.
We note that $x > 1 / 2$ meaning
\begin{align*}
f(x)
& = f\left(\frac{1}{2(y - 1)} + 1\right) \\
& = \frac{1}{2((\frac{1}{2(y - 1)} + 1) - 1)} + 1 \\
& = (y - 1) + 1 \\
& = y.
\end{align*}
\subparagraph{Subconclusion}%
The above three cases are exhaustive.
Thus $\ran{f} = \mathbb{R}$, i.e. $f$ is onto $\mathbb{R}$.
\paragraph{(iii)}%
Let $x \in (0, 1)$.
There are two cases to consider:
\subparagraph{Case 1}%
Suppose $x$ is a rational number.
Then there exist integers $m$ and $n$ such that $x = m / n$.
If $x \leq 1 / 2$, then
\begin{align*}
f(x)
& = f(m / n) \\
& = \frac{1}{2\left(\frac{m}{n}\right)} - 1 \\
& = \frac{n}{2m} - 1 \\
& = \frac{n - 2m}{2m}.
\end{align*}
Since $n - 2m$ and $2m$ are integers, $f(x)$ is a rational number.
If instead $x > 1 / 2$, then
\begin{align*}
f(x)
& = f(m / n) \\
& = \frac{1}{2\left(\frac{m}{n} - 1\right)} + 1 \\
& = \frac{n}{2(m - n)} + 1 \\
& = \frac{n + 2(m - n)}{2(m - n)}.
\end{align*}
Since $n + 2(m - n)$ and $2(m - n)$ are integers, $f(x)$ is again a
rational number.
Thus $f$ maps every rational number to a rational number.
\subparagraph{Case 2}%
Suppose $x$ is an irrational number.
First, consider the case where $x \leq 1 / 2$ and, for the sake of
contradiction, suppose $f(x)$ was a rational number.
Then there exist integers $m$ and $n$ such that
$$f(x) = \frac{1}{2x} - 1 = \frac{m}{n}.$$
But this would imply $$x = \frac{n}{2(m + n)},$$ a contradiction.
Thus $f(x)$ must be irrational.
Likewise, consider the case where $x > 1 / 2$.
Again, for the sake of contradiction, suppose $f(x)$ was a rational
number.
Then there exist integers $m$ and $n$ such that
$$f(x) = \frac{1}{2(x - 1)} + 1 = \frac{m}{n}.$$
But this would imply $$x = \frac{n}{2(m - n)} + 1,$$ a contradiction.
Thus $f(x)$ must be irrational.
Hence $f$ maps every irrational number to an irrational number.
\end{proof}
\subsection{\sorry{Exercise 6.4}}%

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\newcommand\suitdivider{$$\spadesuit\;\spadesuit\;\spadesuit$$}
% ========================================
% Linking
% Links
% ========================================
\hypersetup{colorlinks=true, linkcolor=blue, urlcolor=blue}