Add proofs and corrections to chapters 1–10

blueprint/src/chapter/chapter01.tex

@@ -1,18 +1,18 @@
 \chapter{Six proofs of the infinity of primes}
 
 \begin{theorem}[Euclid's proof]
-    \label{thm:eulids_proof}
+    \label{thm:euclids_proof}
     \lean{infinity_of_primes₁}
     \leanok
     A finite set $\{p_1, \dots, p_r\}$ cannot be the collection of
     \emph{all} prime numbers.
 \end{theorem}
 \begin{proof}
     \leanok
-    For any finite set $\{p_1, \dots, p_r\}$, consider the number
-    $n = p_1p_2\dots p_r + 1$. This $n$ has a prime divisor $p$.
-    But $p$ is not one of the $p_i$s: otherwise $p$ would be a divisor of $n$ and of the product
-    $p_1p_2\dots p_r$, and thus also of the difference $n - p_1p_2\dots p_r = 1$,
+    For any finite set $\{p_1, \dots, p_r\}$ of primes, consider the number
+    $n = p_1 p_2 \cdots p_r + 1$. This $n$ has a prime divisor $p$.
+    But $p$ is not one of the $p_i$'s: otherwise $p$ would be a divisor of $n$ and of the product
+    $p_1p_2\dots p_r$, and thus also of the difference $n - p_1 p_2 \cdots p_r = 1$,
     which is impossible. So a finite set $\{p_1, \dots, p_r\}$ cannot be the collection of
     \emph{all} prime numbers.
 \end{proof}
@@ -29,7 +29,6 @@ \chapter{Six proofs of the infinity of primes}
     Let us first look at the Fermat numbers \(F_n = 2^{2^n} +1\) for \(n = 0,1,2,\dots\).
     We will show that any two Fermat numbers are relatively prime;
     hence there must be infinitely many primes. To this end, we verify the recursion
-
     \[
     \prod_{k=0}^{n-1} F_k = F_n - 2,
     \]
@@ -75,43 +74,34 @@ \chapter{Six proofs of the infinity of primes}
     Let \(\pi(x) := \#\{p \leq x : p \in \mathbb{P}\}\) be the number of primes that are less than or equal to the real number \(x\).
     We number the primes \(\mathbb{P} = \{p_1, p_2, p_3, \dots \}\) in increasing order.
     Consider the natural logarithm \(\log x\), defined as
-
     \[
     \log x = \int_1^x \frac{1}{t} dt.
     \]
 
     Now we compare the area below the graph of \(f(t) = \frac{1}{t}\) with an upper step function.
     (See also the appendix for this method.) Thus for \(n \leq x < n+1\) we have
-
     \[
     \log x \leq 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n-1} + \frac{1}{n} \leq \sum \frac{1}{m},
     \]
     where the sum extends over all \(m \in \mathbb{N}\) which have only prime divisors \(p \leq x\).
 
     Since every such \(m\) can be written in a unique way as a product of the form \(\prod_{p \leq x} p^{k_p}\),
     we see that the last sum is equal to
-
     \[
     \prod_{p \in \mathbb{P}, p \leq x} \left( \sum_{k \geq 0} \frac{1}{p^k} \right).
     \]
-
     The inner sum is a geometric series with ratio \(\frac{1}{p}\), hence
-
     \[
     \log x \leq \prod_{p \leq x} \frac{1}{1 - \frac{1}{p}} = \prod_{p \leq x} \frac{p}{p - 1} = \prod_{k=1}^{\pi(x)} \frac{p_k}{p_k - 1}.
     \]
-
     Now clearly \(p_k \geq k+1\), and thus
-
     \[
     \frac{p_k}{p_k - 1} = 1 + \frac{1}{p_k - 1} \leq 1 + \frac{1}{k} = \frac{k+1}{k},
     \]
     and therefore
-
     \[
     \log x \leq \prod_{k=1}^{\pi(x)} \frac{k+1}{k} = \pi(x) + 1.
     \]
-
     Everybody knows that \(\log x\) is not bounded, so we conclude that \(\pi(x)\) is unbounded as well, and so there are infinitely many primes.
 \end{proof}
 
@@ -123,35 +113,29 @@ \chapter{Six proofs of the infinity of primes}
 \end{theorem}
 \begin{proof}
     Consider the following curious topology on the set \(\mathbb{Z}\) of integers. For \(a, b \in \mathbb{Z}, b > 0\), we set
-
     \[
     N_{a,b} = \{a + nb : n \in \mathbb{Z}\}.
     \]
-
     Each set \(N_{a,b}\) is a two-way infinite arithmetic progression.
     Now call a set \(O \subseteq \mathbb{Z}\) open if either \(O\) is empty, or if to every \(a \in O\) there exists some \(b > 0\) with \(N_{a,b} \subseteq O\).
     Clearly, the union of open sets is open again. If \(O_1, O_2\) are open, and \(a \in O_1 \cap O_2\) with \(N_{a,b_1} \subseteq O_1\) and \(N_{a,b_2} \subseteq O_2\),
     then \(a \in N_{a, b_1 b_2} \subseteq O_1 \cap O_2\). So we conclude that any finite intersection of open sets is again open.
     Therefore, this family of open sets induces a bona fide topology on \(\mathbb{Z}\).
 
     Let us note two facts:
-
     \begin{itemize}
         \item[(A)] Any nonempty open set is infinite.
         \item[(B)] Any set \(N_{a,b}\) is closed as well.
     \end{itemize}
 
     Indeed, the first fact follows from the definition. For the second, we observe
-
     \[
     N_{a,b} = \mathbb{Z} \setminus \bigcup_{i=1}^{b-1} N_{a+i,b},
     \]
-
     which proves that \(N_{a,b}\) is the complement of an open set and hence closed.
 
     So far, the primes have not yet entered the picture — but here they come.
     Since any number \(n \neq 1, -1\) has a prime divisor \(p\), and hence is contained in \(N_{0,p}\), we conclude
-
     \[
     \mathbb{Z} \setminus \{1, -1\} = \bigcup_{p \in \mathbb{P}} N_{0,p}.
     \]
@@ -174,32 +158,27 @@ \chapter{Six proofs of the infinity of primes}
     Let \(p_1, p_2, p_3, \dots\) be the sequence of primes in increasing order, and assume that \(\sum_{p \in \mathbb{P}} \frac{1}{p}\) converges.
     Then there must be a natural number \(k\) such that \(\sum_{i \geq k+1} \frac{1}{p_i} < \frac{1}{2}\).
     Let us call \(p_1, \dots, p_k\) the small primes, and \(p_{k+1}, p_{k+2}, \dots\) the big primes. For an arbitrary natural number \(N\), we therefore find
-
     \[
     \sum_{i \geq k+1} \frac{N}{p_i} < \frac{N}{2}. \tag{1}
     \]
 
     Let \(N_b\) be the number of positive integers \(n \leq N\) which are divisible by at least one big prime,
     and \(N_s\) the number of positive integers \(n \leq N\) which have only small prime divisors. We are going to show that for a suitable \(N\)
-
     \[
     N_b + N_s < N,
     \]
-
     which will be our desired contradiction, since by definition \(N_b + N_s\) would have to be equal to \(N\).
 
     To estimate \(N_b\), note that \(\left\lfloor \frac{N}{p_i} \right\rfloor\) counts the positive
     integers \(n \leq N\) which are multiples of \(p_i\). Hence by (1) we obtain
-
     \[
     N_b \leq \sum_{i \geq k+1} \left\lfloor \frac{N}{p_i} \right\rfloor < \frac{N}{2}. \tag{2}
     \]
-
+    
     Let us now look at \(N_s\). We write every \(n \leq N\) which has only small prime
     divisors in the form \(n = a_n b_n^2\), where \(a_n\) is the square-free part. Every \(a_n\)
-    is thus a product of different small primes, and we conclude that there are precisely \(2^k\)
+    is thus a product of \emph{different} small primes, and we conclude that there are precisely \(2^k\)
     different square-free parts. Furthermore, as \(b_n^2 \leq n \leq N\), we find that there are at most \(\sqrt{N}\) different square parts, and so
-
     \[
     N_s \leq 2^k \sqrt{N}.
     \]
@@ -218,7 +197,41 @@ \section{Appendix: Infinitely many more proofs} \label{appendix:more_primes}
     then the set \(\mathbb{P}_S\) of primes that divide some member of $S$ is infinite.
 \end{theorem}
 \begin{proof}
-    %TODO: add proof here
+	We may assume that \(f(n)\) is monotonely increasing. Otherwise, replace \(f(n)\) by \(F(n) = \max_{i \leq n} f(i)\); you can easily check that with this \(F(n)\) the sequence \(S\) again satisfies the subexponential growth condition.
+
+	Let us suppose for a contradiction that \(\mathbb{P}_S = \{p_1, \dots, p_k\}\) is finite. For \(n \in \mathbb{N}\), let
+	\[
+	s_n = \varepsilon_n p_1^{\alpha_1} \cdots p_k^{\alpha_k}, \quad \text{with } \varepsilon_n \in \{1, 0, -1\}, \alpha_i \ge 0,
+	\]
+	where the \(\alpha_i = \alpha_i(n)\) depend on \(n\). (For \(s_n = 0\) we can put \(\alpha_i = 0\) for all \(i\).) Then
+	\[
+	2^{\alpha_1 + \dots + \alpha_k} \leq |s_n| \leq 2^{2^{f(n)}} \quad \text{for } s_n \ne 0,
+	\]
+	and thus by taking the binary logarithm
+	\[
+	0 \leq \alpha_i \leq \alpha_1 + \dots + \alpha_k \leq 2^{f(n)} \quad \text{for } 1 \leq i \leq k.
+	\]
+
+	Hence there are not more than \(2^{f(n)} + 1\) different possible values for each \(\alpha_i = \alpha_i(n)\). Since \(f\) is monotone, this gives a first estimate
+	\[
+	\#\{\text{distinct } |s_n| \ne 0 \text{ for } n \leq N\} \leq (2^{f(N)} + 1)^k \leq 2^{(f(N)+1)k}.
+	\]
+
+	On the other hand, since \(S\) is almost injective only \(c\) terms in the sequence can be equal to 0, and each nonzero absolute value can occur at most \(2c\) times, so we get the lower estimate
+	\[
+	\#\{\text{distinct } |s_n| \ne 0 \text{ for } n \leq N\} \ge \frac{N-c}{2c}.
+	\]
+
+	Altogether, this gives
+	\[
+	\frac{N-c}{2c} \leq 2^{k(f(N)+1)}.
+	\]
+	Taking again the logarithm with base 2 on both sides, we obtain
+	\[
+	\log_2(N-c) - \log_2(2c) \leq k(f(N)+1) \quad \text{for all } N.
+	\]
+
+	This, however, is plainly false for large \(N\), as \(k\) and \(c\) are constants, so \(\frac{\log_2(N-c)}{\log_2 N}\) goes to 1 for \(N \to \infty\), while \(\frac{f(N)}{\log_2 N}\) goes to 0.
 \end{proof}
 
 % This is just to group things together more nicely in the dependency graph.
@@ -229,6 +242,6 @@ \section{Appendix: Infinitely many more proofs} \label{appendix:more_primes}
 \end{theorem}
 \begin{proof}
     See theorems in this chapter.
-    \uses{thm:eulids_proof, thm:second_proof, thm:third_proof, thm:fourth_proof,
+    \uses{thm:euclids_proof, thm:second_proof, thm:third_proof, thm:fourth_proof,
         thm:fifth_proof, thm:sixth_proof, thm:infty_proof}
 \end{proof}