diff --git a/blueprint/lean_decls b/blueprint/lean_decls deleted file mode 100644 index 99b1eed..0000000 --- a/blueprint/lean_decls +++ /dev/null @@ -1,52 +0,0 @@ -infinity_of_primes₁ -infinity_of_primes₂ -infinity_of_primes₃ -infinity_of_primes₄ -infinity_of_primes₅ -infinity_of_primes₆ -Asymptotics.infinitely_many_more_proofs -chapter2.exists_prime_lt_and_le_two_mul -chapter2.harmonic_number_bounds -chapter2.bound_factorial -chapter2.bound_binomial_coeff -chapter3.sylvester -chapter3.binomials_coefficients_never_powers -ch04.lemma₁ -ch04.lemma₂ -ch04.theorem₁ -ch04.theorem₂ -book.quadratic_reciprocity.fermat_little -book.quadratic_reciprocity.euler_criterion -book.quadratic_reciprocity.product_rule -book.quadratic_reciprocity.lemma_of_Gauss -book.quadratic_reciprocity.quadratic_reciprocity_1 -book.quadratic_reciprocity.mult_cyclic -book.quadratic_reciprocity.fact_A -book.quadratic_reciprocity.fact_B -book.quadratic_reciprocity.quadratic_reciprocity_2 -wedderburn -chapter7.Theorem₁ -chapter7.Theorem₂ -book.irrational.e_irrational -book.irrational.e_pow_2_irrational -book.irrational.little_lemma -book.irrational.e_pow_4_irrational -book.irrational.lem_aux_i -book.irrational.lem_aux_ii -book.irrational.lem_aux_iii -euler_series -euler_series' -euler_series_3 -euler_series_4 -cauchy_schwarz_inequality -harmonic_geometric_arithmetic₁ -harmonic_geometric_arithmetic₂ -harmonic_geometric_arithmetic₃ -mantel -fundamental_theorem_of_algebra -Matrix.permanent_conjecture -chapter28.pigeon_hole_principle -chapter28.handshaking -chapter44.friendship_theorem -chapter45.theorem_1 -chapter45.ramsey_exists \ No newline at end of file diff --git a/blueprint/src/chapter/chapter01.tex b/blueprint/src/chapter/chapter01.tex index e44764c..f988786 100644 --- a/blueprint/src/chapter/chapter01.tex +++ b/blueprint/src/chapter/chapter01.tex @@ -1,7 +1,7 @@ \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 @@ -9,10 +9,10 @@ \chapter{Six proofs of the infinity of primes} \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 - \[ \prod_{k=0}^{n-1} F_k = F_n - 2, \] @@ -75,14 +74,12 @@ \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 - \[ \log x \leq 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n-1} + \frac{1}{n} \leq \sum \frac{1}{m}, \] @@ -90,28 +87,21 @@ \chapter{Six proofs of the infinity of primes} 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} @@ -123,11 +113,9 @@ \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 \(N_{a,b_2} \subseteq O_2\), @@ -135,23 +123,19 @@ \chapter{Six proofs of the infinity of primes} 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} diff --git a/blueprint/src/chapter/chapter02.tex b/blueprint/src/chapter/chapter02.tex index cccf91a..4feb4b2 100644 --- a/blueprint/src/chapter/chapter02.tex +++ b/blueprint/src/chapter/chapter02.tex @@ -1,6 +1,6 @@ \chapter{Bertrand's postulate} -\begin{theorem} +\begin{theorem}[Bertrand's postulate] \label{thm:bertrands_postulate} \lean{chapter2.exists_prime_lt_and_le_two_mul} \leanok @@ -9,7 +9,91 @@ \chapter{Bertrand's postulate} \end{theorem} \begin{proof} \leanok - TODO: make this follow the book proof more closely! + We will estimate the size of the binomial coefficient $\binom{2n}{n}$ carefully enough to see that if it didn't have any prime factors in the range \(n < p \le 2n\), then it would be ``too small.'' Our argument is in five steps. + + \begin{enumerate} + \item We first prove Bertrand's postulate for $n \le 511$. For this one does not need to check 511 cases: it suffices (this is ``Landau's trick'') to check that + \[ + 2,3,5,7,13,23,43,83,163,317,521 + \] + is a sequence of prime numbers, where each is smaller than twice the previous one. Hence every interval \(\{y : n < y \le 2n\}\), with $n \le 511$, contains one of these 11 primes. + + \item Next we prove that + \begin{equation} \label{eq:prod_prime_le} + \prod_{p \le x} p \le 4^{x-1} \quad \text{for all real } x \ge 2, + \end{equation} + where our notation --- here and in the following --- is meant to imply that the product is taken over all prime numbers $p \le x$. The proof that we present for this fact uses induction on the number of these primes. It is not from Erdős' original paper, but it is also due to Erdős, and it is a true Book Proof. First we note that if \(q\) is the largest prime with \(q \le x\), then + \[ + \prod_{p \le x} p = \prod_{p \le q} p \quad \text{and} \quad 4^{q-1} \le 4^{x-1}. + \] + + Thus it suffices to check \eqref{eq:prod_prime_le} for the case where $x=q$ is a prime number. For $q=2$ we get ``$2 \le 4$,'' so we proceed to consider odd primes $q = 2m + 1$. (Here we may assume, by induction, that \eqref{eq:prod_prime_le} is valid for all integers $x$ in the set $\{2, 3, \dots, 2m\}$.) For $q = 2m + 1$ we split the product and compute + \[ + \prod_{p \le 2m + 1} p = \prod_{p \le m + 1} p \cdot \prod_{m + 1 < p \le 2m + 1} p \le 4^m \binom{2m + 1}{m} \le 4^m 2^{2m} = 4^{2m}. + \] + + All the pieces of this ``one-line computation'' are easy to see. In fact, + \[ + \prod_{p \le m + 1} p \le 4^m + \] + holds by induction. The inequality + \[ + \prod_{m + 1 < p \le 2m + 1} p \le \binom{2m + 1}{m} + \] + follows from the observation that $\binom{2m + 1}{m} = \frac{(2m + 1)!}{m! (m + 1)!}$ is an integer, where the primes that we consider all are factors of the numerator $(2m + 1)!$, but not of the denominator $m! (m + 1)!$. Finally + \[ + \binom{2m + 1}{m} \le 2^{2m} + \] + holds since + \[ + \binom{2m + 1}{m} \quad \text{and} \quad \binom{2m + 1}{m + 1} + \] + are two (equal!) summands that appear in + \[ + \sum_{k = 0}^{2m + 1} \binom{2m + 1}{k} = 2^{2m + 1}. + \] + + \item From Legendre's theorem we get that $\binom{2n}{n} = \frac{(2n)!}{n! n!}$ contains the prime factor $p$ exactly + \[ + \sum_{k \geq 1} \left( \left\lfloor \frac{2n}{p^k} \right\rfloor - 2 \left\lfloor \frac{n}{p^k} \right\rfloor \right) + \] + times. Here each summand is at most 1, since it satisfies + \[ + \left\lfloor \frac{2n}{p^k} \right\rfloor - 2 \left\lfloor \frac{n}{p^k} \right\rfloor < \frac{2n}{p^k} - 2 \left( \frac{n}{p^k} - 1 \right) = 2, + \] + and it is an integer. Furthermore the summands vanish whenever $p^k > 2n$. Thus $\binom{2n}{n}$ contains $p$ exactly + \[ + \sum_{k \geq 1} \left( \left\lfloor \frac{2n}{p^k} \right\rfloor - 2 \left\lfloor \frac{n}{p^k} \right\rfloor \right) \le \max\{r : p^r \le 2n\} + \] + times. Hence the largest power of $p$ that divides $\binom{2n}{n}$ is not larger than $2n$. In particular, primes $p > \sqrt{2n}$ appear at most once in $\binom{2n}{n}$. + + Furthermore --- and this, according to Erdős, is the key fact for his proof --- primes $p$ that satisfy $\frac{2}{3} n < p < n$ do not divide $\binom{2n}{n}$ at all! Indeed, $3p > 2n$ implies (for $n \geq 3$, and hence $p \geq 3$) that $p$ and $2p$ are the only multiples of $p$ that appear as factors in the numerator of $\frac{(2n)!}{n! n!}$, while we get two $p$-factors in the denominator. + + \item Now we are ready to estimate \(\binom{2n}{n}\), benefitting from a suggestion by Raimund Seidel, which nicely improves Erdős' original argument. For \(n \ge 3\), using an estimate for the lower bound, we get + \[ + \frac{4^n}{2n} \le \binom{2n}{n} \le \prod_{p \le \sqrt{2n}} 2n \cdot \prod_{\sqrt{2n} < p \le \frac{2}{3}n} p \cdot \prod_{n < p \le 2n} p. + \] + + Now, there are no more than \(\sqrt{2n}\) primes in the first factor; hence using (1) for the second factor and letting \(P(n)\) denote the number of primes between \(n\) and \(2n\) we get + \[ + \frac{4^n}{2n} < ((2n)^{\sqrt{2n}}) \cdot (4^{\frac{2}{3}n}) \cdot (2n)^{P(n)}, + \] + that is, + \[ + 4^{\frac{n}{3}} < (2n)^{\sqrt{2n}+1+P(n)}. \quad \text{(2)} + \] + + \item Taking the logarithm to base 2, the last inequality is turned into + \[ + P(n) > \frac{2n}{3 \log_2(2n)} - (\sqrt{2n} + 1). \quad \text{(3)} + \] + + It remains to verify that the right-hand side of (3) is positive for \(n\) large enough. We show that this is the case for \(n = 2^9 = 512\) (actually, it holds for \(n = 468\) onward). By writing \(2n - 1 = (\sqrt{2n} - 1)(\sqrt{2n} + 1)\) and cancelling the \((\sqrt{2n} + 1)\)-factor it suffices to show + \[ + \sqrt{2n}-1 > 3 \log_2(2n) \quad \text{for } n \ge 2^9. \quad \text{(4)} + \] + For \(n = 2^9\), (4) becomes \(31 > 30\), and comparing the derivatives \((\sqrt{x}-1)' = \frac{1}{2} \frac{1}{\sqrt{x}}\) and \((3 \log_2 x)' = \frac{3}{\log 2} \frac{1}{x}\) we see that \(\sqrt{x} - 1\) grows faster than \(3 \log_2 x\) for \(x > (\frac{6}{\log 2})^2 \approx 75\) and thus certainly for \(x \ge 2^{10} = 1024.\) \qedhere + \end{enumerate} \end{proof} \section{Appendix: Some estimates} @@ -24,7 +108,22 @@ \section{Appendix: Some estimates} \] \end{theorem} \begin{proof} - TODO + There is a very simple-but-effective method of estimating sums by integrals. For estimating the harmonic numbers + \[ + H_n = \sum_{k=1}^{n} \frac{1}{k} + \] + we draw the figure and derive from it + \[ + H_n - 1 = \sum_{k=2}^{n} \frac{1}{k} < \int_{1}^{n} \frac{1}{t} \, dt = \log n + \] + by comparing the area below the graph of $f(t) = \frac{1}{t} (1 \le t \le n)$ with the area of the dark shaded rectangles, and + \[ + H_n - \frac{1}{n} = \sum_{k=1}^{n-1} \frac{1}{k} > \int_{1}^{n} \frac{1}{t} \, dt = \log n + \] + by comparing with the area of the large rectangles (including the lightly shaded parts). Taken together, this yields + \[ + \log n + \frac{1}{n} < H_n < \log n + 1. \qedhere + \] \end{proof} @@ -34,11 +133,31 @@ \section{Appendix: Some estimates} \leanok For all \(n \in \mathbb{N}\) \[ - n! = n(n -1)! < ne^{n \log n - n + 1}= e\left(\frac n e\right)^n. + e\left(\frac{n}{e}\right)^n < n! < en\left(\frac{n}{e}\right)^n. \] \end{theorem} \begin{proof} - TODO + The same method applied to + \[ + \log(n!) = \log 2 + \log 3 + \dots + \log n = \sum_{k=2}^{n} \log k + \] + yields + \[ + \log((n-1)!) < \int_{1}^{n} \log t \, dt < \log(n!), + \] + where the integral is easily computed: + \[ + \int_{1}^{n} \log t \, dt = \left[t\log t - t\right]_{1}^{n} = n\log n - n + 1. + \] + + Thus we get a lower estimate on $n!$ + \[ + n! > e^{n\log n - n + 1} = e\left(\frac{n}{e}\right)^n + \] + and at the same time an upper estimate + \[ + n! = n(n-1)! < n e^{n\log n - n + 1} = en\left(\frac{n}{e}\right)^n. \qedhere + \] \end{proof} \begin{theorem} @@ -48,5 +167,7 @@ \section{Appendix: Some estimates} \[\binom{n}{k} \le \frac{n^k}{k!} \le \frac{n^k}{2^{k - 1}}\] \end{theorem} \begin{proof} - TODO + \[ + \binom{n}{k} = \frac{n (n-1) \cdots (n-k+1)}{k!} \le \frac{n^k}{k!} \le \frac{n^k}{2^{k-1}}. \qedhere + \] \end{proof} diff --git a/blueprint/src/chapter/chapter03.tex b/blueprint/src/chapter/chapter03.tex index 9bb0957..1e6765c 100644 --- a/blueprint/src/chapter/chapter03.tex +++ b/blueprint/src/chapter/chapter03.tex @@ -24,7 +24,7 @@ \chapter{Binomial coefficients are (almost) never powers} \end{theorem} \begin{proof} \uses{sylvester} - Note first that we may assume \(n \geq 2k\) because of \(\binom{n}{k} = \binom{n}{n-k}\). + Note first that we may assume \(n \ge 2k\) because of \(\binom{n}{k} = \binom{n}{n-k}\). Suppose the theorem is false, and that \(\binom{n}{k} = m^\ell\). The proof, by contradiction, proceeds in the following four steps. \begin{enumerate} @@ -32,45 +32,45 @@ \chapter{Binomial coefficients are (almost) never powers} hence \(p^\ell\) divides \(n(n - 1) \dots (n - k + 1)\). Clearly, only one of the factors \(n - i\) can be a multiple of \(p\) (because \(p > k\)), and we conclude \(p^\ell \mid n - i\), and therefore \[ - n \geq p^\ell > k^\ell \geq k^2. + n \ge p^\ell > k^\ell \ge k^2. \] \item Consider any factor \(n - j\) of the numerator and write it in the form \(n - j = a_j m_j^\ell\), where \(a_j\) is not divisible by any nontrivial \(\ell\)-th power. We note by (1) that \(a_j\) has only prime divisors less than or equal to \(k\). - We want to show next that \(a_i \neq a_j\) for \(i \neq j\). - Assume to the contrary that \(a_i = a_j\) for some \(i < j\). Then \(m_i > m_j + 1\) and - \begin{align} - k & > (n - i) - (n - j) = a_j(m_i^\ell - m_j^\ell) \geq a_j((m_j + 1)^\ell - m_j^\ell)\\ - & > a_j \ell m_j^{\ell-1} \geq \ell (a_j m_j^\ell)^{1/2} \geq \ell (n - k + 1)^{1/2}\\ - & \geq \ell \left(\frac{n}{2}\right)^{1/2} > n^{1/2}, - \end{align} + We want to show next that \(a_i \ne a_j\) for \(i \ne j\). + Assume to the contrary that \(a_i = a_j\) for some \(i < j\). Then \(m_i \ge m_j + 1\) and + \begin{align*} + k &> (n - i) - (n - j) = a_j(m_i^\ell - m_j^\ell) \ge a_j((m_j + 1)^\ell - m_j^\ell)\\ + &> a_j \ell m_j^{\ell-1} \ge \ell (a_j m_j^\ell)^{1/2} \ge \ell (n - k + 1)^{1/2}\\ + &\ge \ell \left(\frac{n}{2} + 1\right)^{1/2} > n^{1/2}, + \end{align*} which contradicts \(n > k^2\) from above. \item Next we prove that the \(a_i\)'s are the integers \(1, 2, \dots, k\) in some order. (According to Erdős, this is the crux of the proof.) Since we already know that they are all distinct, it suffices to prove that \[ - a_0 a_1 \dots a_{k-1} \text{ divides } k!. + a_0 a_1 \cdots a_{k-1} \text{ divides } k!. \] Substituting \(n - j = a_j m_j^\ell\) into the equation \(\binom{n}{k} = m^\ell\), we obtain \[ - a_0 a_1 \dots a_{k-1} (m_0 m_1 \dots m_{k-1})^\ell = k! m^\ell. + a_0 a_1 \cdots a_{k-1} (m_0 m_1 \cdots m_{k-1})^\ell = k! m^\ell. \] - Canceling the common factors of \(m_0 \dots m_{k-1}\) and \(m\) yields + Canceling the common factors of \(m_0 \cdots m_{k-1}\) and \(m\) yields \[ a_0 a_1 \dots a_{k-1} u^\ell = k! v^\ell \] with \(\gcd(u, v) = 1\). It remains to show that \(v = 1\). If not, then \(v\) contains a prime divisor \(p\). Since \(\gcd(u, v) = 1\), \(p\) must be a prime divisor of \(a_0 a_1 \dots a_{k-1}\) and hence is less than or equal to \(k\). - By the theorem of Legendre (see page 8), we know that \(k!\) contains \(p\) to the power \(\sum_{i \geq 1} \left\lfloor \frac{k}{p^i} \right\rfloor\). + By the theorem of Legendre, we know that \(k!\) contains \(p\) to the power \(\sum_{i \ge 1} \left\lfloor \frac{k}{p^i} \right\rfloor\). We now estimate the exponent of \(p\) in \(n(n - 1) \dots (n - k + 1)\). Let \(i\) be a positive integer, and let \(b_1 < b_2 < \dots < b_s\) be the multiples of \(p^i\) among \(n, n - 1, \dots, n - k + 1\). Then \(b_s = b_1 + (s - 1)p^i\) and hence \[ - (s - 1)p^i = b_s - b_1 \leq n - (n - k + 1) = k - 1, + (s - 1)p^i = b_s - b_1 \le n - (n - k + 1) = k - 1, \] which implies \[ - s \leq \left\lfloor \frac{k - 1}{p^i} \right\rfloor + 1 \leq \left\lfloor \frac{k}{p^i} \right\rfloor + 1. + s \le \left\lfloor \frac{k - 1}{p^i} \right\rfloor + 1 \le \left\lfloor \frac{k}{p^i} \right\rfloor + 1. \] So for each \(i\), the number of multiples of \(p^i\) among \(n, \dots, n-k+1\), and hence among the \(a_j\)'s, is bounded by \(\left\lfloor \frac{k}{p^i} \right\rfloor + 1\). This implies that the exponent of \(p\) in \(a_0 a_1 \dots a_{k-1}\) is at most @@ -85,51 +85,49 @@ \chapter{Binomial coefficients are (almost) never powers} Taking both counts together, we find that the exponent of \(p\) in \(v^\ell\) is at most \[ - \sum_{i=1}^{\ell-1} \left( \left\lfloor \frac{k}{p^i} \right\rfloor + 1 \right) - \sum_{i \geq 1} \left\lfloor \frac{k}{p^i} \right\rfloor \leq \ell - 1, + \sum_{i=1}^{\ell-1} \left( \left\lfloor \frac{k}{p^i} \right\rfloor + 1 \right) - \sum_{i \ge 1} \left\lfloor \frac{k}{p^i} \right\rfloor \le \ell - 1, \] and we have our desired contradiction, since \(v^\ell\) is an \(\ell\)-th power. This suffices already to settle the case \(\ell = 2\). - Indeed, since \(k \geq 4\), one of the \(a_i\)'s must be equal to 4, - but the \(a_i\)'s contain no squares. So let us now assume that \(\ell \geq 3\). + Indeed, since \(k \ge 4\), one of the \(a_i\)'s must be equal to 4, + but the \(a_i\)'s contain no squares. So let us now assume that \(\ell \ge 3\). - \item Since \(k \geq 4\), we must have \(a_{i_1} = 1\), \(a_{i_2} = 2\), \(a_{i_3} = 4\) for some \(i_1, i_2, i_3\), that is, + \item Since \(k \ge 4\), we must have \(a_{i_1} = 1\), \(a_{i_2} = 2\), \(a_{i_3} = 4\) for some \(i_1, i_2, i_3\), that is, \[ n - i_1 = m_1^\ell, \quad n - i_2 = 2m_2^\ell, \quad n - i_3 = 4m_3^\ell. \] - We claim that \((n - i_2)^2 \neq (n - i_1)(n - i_3)\). If not, put \(b = n - i_2\) + We claim that \((n - i_2)^2 \ne (n - i_1)(n - i_3)\). If not, put \(b = n - i_2\) and \(n - i_1 = b - x, n - i_3 = b + y\), where \(0 < |x|, |y| < k\). Hence \[ b^2 = (b - x)(b + y) \quad \text{or} \quad (y - x)b = xy, \] where \(x = y\) is plainly impossible. Now we have by part (1) \[ - |xy| = b |y - x| \geq b > n - k > (k - 1)^2 \geq |xy|, + |xy| = b |y - x| \ge b > n - k > (k - 1)^2 \ge |xy|, \] which is absurd. - So we have \(m_2^2 \neq m_1 m_3\), where we assume \(m_2 > m_1 m_3\) (the other case being analogous), + So we have \(m_2^2 \ne m_1 m_3\), where we assume \(m_2^2 > m_1 m_3\) (the other case being analogous), and proceed to our last chain of inequalities. We obtain - \begin{align} + \begin{align*} 2(k - 1)n & > n^2 - (n - k + 1)^2 > (n - i_2)^2 - (n - i_1)(n - i_3)\\ - & = 4[m_2^\ell - (m_1 m_3)^\ell] \geq 4[(m_1 m_3 + 1)^\ell - (m_1 m_3)^\ell] \\ - & \geq 4 \ell m_1^{\ell-1} m_3^{\ell-1}. - \end{align} + & = 4[m_2^{2\ell} - (m_1 m_3)^\ell] \ge 4[(m_1 m_3 + 1)^\ell - (m_1 m_3)^\ell] \\ + & \ge 4 \ell m_1^{\ell-1} m_3^{\ell-1}. + \end{align*} - Since \(\ell \geq 3\) and \(n \geq k^\ell \geq k^3 > 6k\), this yields - - \begin{align} + Since \(\ell \ge 3\) and \(n > k^\ell \ge k^3 > 6k\), this yields + \begin{align*} 2(k - 1)n m_1 m_3 & > 4 \ell m_1^\ell m_3^\ell = \ell (n - i_1)(n - i_3) \\ - & > \ell (n - k + 1)^2 > 3(n - \frac{n}{6})^2 > 2n^2. - \end{align} + & > \ell (n - k + 1)^2 > 3 \left(n - \frac{n}{6}\right)^2 > 2n^2. + \end{align*} Now since \(m_i \le n^{1/\ell} \le n ^{1/3}\) we finally obtain \[ kn^{2/3} \ge km_1m_3 > (k - 1)m_1m_3 > n, \] - or \(k^3>n\). With this contradiction, the proof is complete. - + or \(k^3 > n\). With this contradiction, the proof is complete. \qedhere \end{enumerate} \end{proof} diff --git a/blueprint/src/chapter/chapter04.tex b/blueprint/src/chapter/chapter04.tex index 34a5866..ece39a5 100644 --- a/blueprint/src/chapter/chapter04.tex +++ b/blueprint/src/chapter/chapter04.tex @@ -2,7 +2,8 @@ \chapter{Representing numbers as sums of two squares} \begin{lemma}[Lemma 1] \lean{ch04.lemma₁} - \label{ch04.lemma1} + \label{ch04.lemma₁} + \leanok For primes \(p = 4m + 1\) the equation \(s^2 \equiv -1 (\mod p)\) has two solutions \(s \in \{1, 2, \dots, p - 1\}\), for \(p = 2\) there is one such solution, while for primes of the form \(p = 4m + 3\) there is no solution. @@ -40,6 +41,7 @@ \chapter{Representing numbers as sums of two squares} \begin{lemma}[Lemma 2] \label{ch04.lemma2} \lean{ch04.lemma₂} + \leanok No number \(n = 4m + 3\) is a sum of two squares. \end{lemma} \begin{proof} @@ -52,11 +54,12 @@ \chapter{Representing numbers as sums of two squares} \begin{proposition}[First proof] \label{ch04.proposition1} \lean{ch04.theorem₁} + \leanok Every prime of the form \(p = 4m + 1\) is a sum of two squares, that is, it can be written as \(p = x^2 + y^2\) for some natural numbers \(x,y \in \mathbb{N}\). \end{proposition} \begin{proof} - \uses{ch04.lemma1} + \uses{ch04.lemma₁} Consider the pairs $(x', y')$ of integers with $0 \leq x', y' \leq \sqrt{p}$, that is, $x', y' \in \{0, 1, \dots, \lfloor \sqrt{p} \rfloor\}$. There are $(\lfloor \sqrt{p} \rfloor + 1)^2$ such pairs. @@ -91,11 +94,11 @@ \chapter{Representing numbers as sums of two squares} \begin{proposition}[Second proof] \label{ch04.proposition2} \lean{ch04.theorem₂} + \leanok Every prime of the form \(p = 4m + 1\) is a sum of two squares, that is, it can be written as \(p = x^2 + y^2\) for some natural numbers \(x,y \in \mathbb{N}\). \end{proposition} \begin{proof} - \leanok We study the set \[ S := \{(x, y, z) \in \mathbb{Z}^3 : 4xy + z^2 = p, \ x > 0, \ y > 0\}. @@ -127,7 +130,7 @@ \chapter{Representing numbers as sums of two squares} What do we get from the study of $f$? The main observation is that since $f$ maps the sets $T$ and $U$ to their complements, it also interchanges the elements in $T \setminus U$ with these in $U \setminus T$. That is, there is the same number of solutions in $U$ that are not in $T$ as - there are solutions in $T$ that are not in $U$ — so $T$ and $U$ have the same cardinality. + there are solutions in $T$ that are not in $U$ --- \emph{so $T$ and $U$ have the same cardinality}. \item The second involution that we study is an involution on the set $U$: \[ @@ -158,7 +161,7 @@ \chapter{Representing numbers as sums of two squares} This map is clearly well-defined, and an involution. We combine now our knowledge derived from the other two involutions: The cardinality of $T$ is equal to the cardinality of $U$, which is odd. But if $h$ is an involution on a finite set of odd cardinality, then - \emph{it has a fixed point}: There is a point $(x, y, z) \in T$ with $x = y$, that is, + it \emph{has a fixed point}: There is a point $(x, y, z) \in T$ with $x = y$, that is, a solution of \[ p = 4x^2 + z^2 = (2x)^2 + z^2. @@ -223,5 +226,15 @@ \chapter{Representing numbers as sums of two squares} \end{theorem} \begin{proof} \uses{ch04.lemma2, ch04.proposition1, ch04.proposition2, ch04.proposition3} - TODO + Call a number $n$ \textit{representable} if it is a sum of two squares, that is, if $n = x^2 + y^2$ for some $x, y \in \mathbb{N}_0$. The theorem is a consequence of the following five facts. + \begin{enumerate}[label=(\arabic*)] + \item $1 = 1^2 + 0^2$ and $2 = 1^2 + 1^2$ are representable. Every prime of the form $p = 4m + 1$ is representable. + \item The product of any two representable numbers $n_1 = x_1^2 + y_1^2$ and $n_2 = x_2^2 + y_2^2$ is representable: $n_1 n_2 = (x_1 x_2 + y_1 y_2)^2 + (x_1 y_2 - x_2 y_1)^2$. + \item If $n$ is representable, $n = x^2 + y^2$, then also $nz^2$ is representable, by $nz^2 = (xz)^2 + (yz)^2$. + \end{enumerate} + Facts (1), (2) and (3) together yield the ``if'' part of the theorem. + \begin{enumerate}[start=4,label=(\arabic*)] + \item If $p = 4m + 3$ is a prime that divides a representable number $n = x^2 + y^2$, then $p$ divides both $x$ and $y$, and thus $p^2$ divides $n$. In fact, if we had $x \not\equiv 0 \pmod{p}$, then we could find $\overline{x}$ such that $x\overline{x} \equiv 1 \pmod{p}$, multiply the equation $x^2 + y^2 \equiv 0$ by $\overline{x}^2$, and thus we would obtain that $1 + y^2 \overline{x}^2 = 1 + \left(\overline{x} y\right)^2 \equiv 0 \pmod{p}$, which is impossible for $p = 4m + 3$ by Lemma 1. + \item If $n$ is representable, and $p = 4m + 3$ divides $n$, then $p^2$ divides $n$, and $n/p^2$ is representable. This follows from (4), and completes the proof. \qedhere + \end{enumerate} \end{proof} diff --git a/blueprint/src/chapter/chapter05.tex b/blueprint/src/chapter/chapter05.tex index 01adb6b..f5baff6 100644 --- a/blueprint/src/chapter/chapter05.tex +++ b/blueprint/src/chapter/chapter05.tex @@ -5,13 +5,17 @@ \chapter{The law of quadratic reciprocity} \label{fermats_little} \lean{book.quadratic_reciprocity.fermat_little} \leanok - For \(a \not\equiv 0 \mod p\), + For \(a \not\equiv 0 \pmod{p}\), \[ - a^{p - 1} \equiv 1 \mod p + a^{p - 1} \equiv 1 \pmod{p} \] \end{theorem} \begin{proof} - TODO + Since $\mathbb{Z}_p^* = \mathbb{Z}_p \setminus \{0\}$ is a group with multiplication, the set $\{1a, 2a, 3a, \dots, (p-1)a\}$ runs again through all nonzero residues, + \[ + (1a)(2a)\dots((p-1)a) \equiv 1\cdot 2 \cdots (p-1) \pmod{p} + \] + and hence by dividing by $(p - 1)!$, we get $a^{p - 1} \equiv 1 \pmod{p}$. \end{proof} @@ -19,49 +23,99 @@ \chapter{The law of quadratic reciprocity} \label{euler_criterion} \lean{book.quadratic_reciprocity.euler_criterion} \leanok - For \(a \not\equiv 0 (\mod p)\), + For \(a \not\equiv 0 \pmod{p}\), \[ - (\frac{a}{p}) \equiv a ^{\frac{p-1}{2}} \mod p + (\frac{a}{p}) \equiv a ^{\frac{p-1}{2}} \pmod{p} \] \end{theorem} \begin{proof} \uses{fermats_little} - TODO + From Fermat's little theorem, the polynomial $x^{p-1} - 1 \in \mathbb{Z}_p[x]$ has as roots all nonzero residues. Next we note that + \[ + x^{p-1} - 1 = \left(x^{\frac{p-1}{2}} - 1\right)\left(x^{\frac{p-1}{2}} + 1\right). + \] + + Suppose $a \equiv b^2 \pmod{p}$ is a quadratic residue. Then by Fermat's little theorem $a^{\frac{p-1}{2}} \equiv b^{p-1} \equiv 1 \pmod{p}$. Hence the quadratic residues are precisely the roots of the first factor $x^{\frac{p-1}{2}} - 1$, and the $\frac{p-1}{2}$ nonresidues must thus be the roots of the second factor $x^{\frac{p-1}{2}} + 1$. Comparing this to the definition of the Legendre symbol, we obtain + \[ + (\frac{a}{p}) \equiv a^{\frac{p-1}{2}} \pmod{p}. \qedhere + \] \end{proof} \begin{theorem}[Product Rule] \label{product_rule} \leanok \lean{book.quadratic_reciprocity.product_rule} - \[ + \begin{equation} \label{eq:product_rule} (\frac{ab}{p}) = (\frac{a}{p}) \cdot (\frac{b}{p}) - \] + \end{equation} \end{theorem} \begin{proof} \uses{euler_criterion} - TODO + This obviously holds for the right-hand side of Euler's criterion. \end{proof} \begin{theorem}[Lemma of Gauss] \label{gauss_lemma} \lean{book.quadratic_reciprocity.lemma_of_Gauss} - \leanok - TODO + Suppose $a \not\equiv 0 \pmod{p}$. Take the numbers $1a, 2a, \dots, \frac{p-1}{2}a$ and reduce them modulo $p$ to the residue system smallest in absolute value, $ia \equiv r_i \pmod{p}$ with $-\frac{p-1}{2} \le r_i \le \frac{p-1}{2}$ for all $i$. Then + \[ + (\frac{a}{p}) = (-1)^s, \quad \text{where } s = \#\{i : r_i < 0\}. + \] \end{theorem} \begin{proof} \uses{euler_criterion} - TODO + Suppose $u_1, \dots, u_s$ are the residues smaller than $0$, and that $v_1, \dots, v_{\frac{p-1}{2} - s}$ are those greater than $0$. If $-u_i = v_j$, then $u_i + v_j \equiv 0 \pmod{p}$. Now $u_i \equiv ka, v_j \equiv \ell a \pmod{p}$ implies $p \mid (k + \ell) a$. As $p$ and $a$ are relatively prime, $p$ must divide $k + \ell$ which is impossible, since $k + \ell \le p-1$. Thus the numbers $-u_1, \dots, -u_s$ are between $1$ and $\frac{p-1}{2}$, and are all different from the $v_j$'s; hence $\{-u_1, \dots, -u_s, v_1, \dots, v_{\frac{p-1}{2} - s}\} = \{1, 2, \dots, \frac{p-1}{2}\}$. Therefore + \[ + \prod_{i} (-u_i) \prod_{j} v_j = \left(\frac{p-1}{2}\right)!, + \] + which implies + \[ + (-1)^s \prod_{i} u_i \prod_{j} v_j \equiv \left(\frac{p-1}{2}\right)! \pmod{p}. + \] + + Now remember how we obtained the numbers $u_i$ and $v_j$; they are the residues of $1a, \dots, \frac{p-1}{2}a$. Hence + \[ + \left(\frac{p-1}{2}\right)! \equiv (-1)^s \prod_{i} u_i \prod_{j} v_j \equiv (-1)^s \left(\frac{p-1}{2}\right)! a^{\frac{p-1}{2}} \pmod{p}. + \] + Cancelling $\left(\frac{p-1}{2}\right)!$ together with Euler's criterion gives + \[ + (\frac{a}{p}) \equiv a^{\frac{p-1}{2}} \equiv (-1)^s \pmod{p}, + \] + and therefore $(\frac{a}{p}) = (-1)^s$, since $p$ is odd. \end{proof} \begin{theorem}[Quadratic reciprocity I] \label{quadratic_reciprocity1} \lean{book.quadratic_reciprocity.quadratic_reciprocity_1} \leanok - TODO + Let $p$ and $q$ be different odd primes. Then + \[ + (\frac{q}{p})(\frac{p}{q}) = (-1)^{\frac{p-1}{2} \frac{q-1}{2}}. + \] \end{theorem} \begin{proof} \uses{gauss_lemma} - TODO + The key to our first proof is a counting formula given by Lemma of Gauss. Let $p$ and $q$ be odd primes, and consider $(\frac{q}{p})$. Suppose $iq$ is a multiple of $q$ that reduces to negative residue $r_i < 0$ in the Lemma of Gauss. This means that there is a unique integer $j$ such that $-\frac{p}{2} < iq-jp < 0$. Note that $0 < j < \frac{q}{2}$ since $0 < i < \frac{p}{2}$. In other words, $(\frac{q}{p}) = (-1)^s$, where $s$ is the number of lattice points $(x, y)$, that is, pairs of integers $x$, $y$ satisfying + \begin{equation} \label{eq:s} + 0 < py - qx < \frac{p}{2}, \quad 0 < x < \frac{p}{2}, \quad 0 < y < \frac{q}{2}. + \end{equation} + Similarly, $(\frac{p}{q}) = (-1)^t$ where $t$ is the number of lattice points $(x, y)$ with + \begin{equation} \label{eq:t} + 0 < qx - py < \frac{q}{2}, \quad 0 < x < \frac{p}{2}, \quad 0 < y < \frac{q}{2}. + \end{equation} + Now look at the rectangle with side lengths $\frac{p}{2}$, $\frac{q}{2}$, and draw the two lines parallel to the diagonal $py = qx$, $y = \frac{q}{p}x + \frac{1}{2}$ or $py - qx = \frac{p}{2}$, respectively, $y = \frac{q}{p} \left(x - \frac{1}{2}\right)$ or $qx - py = \frac{q}{2}$. + + The proof is now quickly completed by the following three observations: + \begin{enumerate} + \item There are no lattice points on the diagonal and the two parallels. This is so because $py = qx$ would imply $p \mid x$, which cannot be. For the parallels observe that $py - qx$ is an integer while $\frac{p}{2}$ and $\frac{q}{2}$ are not. + \item The lattice points observing \eqref{eq:s} are precisely the points in the upper strip $0 < py - qx < \frac{p}{2}$, and those of \eqref{eq:t} the points in the lower strip $0 < qx - py < \frac{q}{2}$. Hence the number of lattice points in the two strips is $s + t$. + \item The outer regions $R : py - qx > \frac{p}{2}$ and $S: qx - py > \frac{q}{2}$ contain the \textit{same} number of points. To see this consider the map $\varphi: R \to S$ which maps $(x, y)$ to $\left(\frac{p+1}{2} - x, \frac{q+1}{2} - y\right)$ and check that $\varphi$ is an involution. + \end{enumerate} + + Since the total number of lattice points in the rectangle is $\frac{p-1}{2} \cdot \frac{q-1}{2}$, we infer that $s + t$ and $\frac{p-1}{2} \cdot \frac{q-1}{2}$ have the same parity, and so + \[ + (\frac{q}{p}) (\frac{p}{q}) = (-1)^{s+t} = (-1)^{\frac{p-1}{2} \frac{q-1}{2}}. \qedhere + \] \end{proof} @@ -69,41 +123,112 @@ \chapter{The law of quadratic reciprocity} \label{mult_cyclic} \lean{book.quadratic_reciprocity.mult_cyclic} \leanok - The multiplicative group of a finite field is cyclic + The multiplicative group of a finite field is cyclic. \end{theorem} \begin{proof} - TODO + Let $F^*$ be the multiplicative group of the field $F$, with $|F^*| = n$. Writing $\ord(a)$ for the order of an element, that is, the smallest positive integer $k$ such that $a^k = 1$, we want to find an element $a \in F^*$ with $\ord(a) = n$. If $\ord(b) = d$, then by Lagrange's theorem, $d$ divides $n$. Classifying the elements according to their order, we have + \begin{equation} \label{eq:sum_psi} + n = \sum_{d \mid n} \psi(d), \quad \text{where } \psi(d) = \#\{b\in F^*: \ord(b) = d\}. + \end{equation} + If $\ord(b) = d$, then every element $b^i$ ($i = 1, \dots, d$) satisfies $(b^i)^d = 1$ and is therefore a root of the polynomial $x^d - 1$. But, since $F$ is a field, $x^d - 1$ has at most $d$ roots, and so the elements $b, b^2, \dots, b^d = 1$ are precisely these roots. In particular, every element of order $d$ is of the form $b^i$. + + On the other hand, it is easily checked that $\ord(b^i) = \frac{d}{(i, d)}$, where $(i, d)$ denotes the greatest common divisor of $i$ and $d$. Hence $\ord(b^i) = d$ if and only if $(i, d) = 1$, that is, if $i$ and $d$ are relatively prime. Denoting \textit{Euler's function} by $\varphi(d) = \#\{i : 1 \le i \le d, (i, d) = 1\}$, we thus have $\psi(d) = \varphi(d)$ whenever $\psi(d) > 0$. Looking at \eqref{eq:sum_psi} we find + \[ + n = \sum_{d \mid n} \psi(d) \le \sum_{d \mid n} \varphi(d). + \] + But as we are going to show that + \begin{equation} \label{eq:sum_phi} + \sum_{d \mid n} \varphi(d) = n, + \end{equation} + we must have $\psi(d) = \varphi(d)$ for all $d$. In particular, $\psi(n) = \varphi(n) \ge 1$, and so there is an element of order $n$. + + The following (folklore) proof of \eqref{eq:sum_phi} belongs in the Book as well. Consider the $n$ fractions + \[ + \frac{1}{n}, \frac{2}{n}, \dots, \frac{k}{n}, \dots, \frac{n}{n}, + \] + reduce them to the lowest term $\frac{k}{n} = \frac{i}{d}$ with $1 \le i \le d$, $(i, d) = 1$, $d \mid n$, and check that the denominator $d$ appears precisely $\varphi(d)$ times. \end{proof} \begin{theorem}[A] \label{fact_A} \lean{book.quadratic_reciprocity.fact_A} \leanok - TODO + Let $p$ and $q$ be distinct odd primes, and consider the finite field $F$ with $q^{p-1}$ elements. Then for any $a, b \in F$, $(a + b)^q = a^q + b^q$. \end{theorem} \begin{proof} - \uses{euler_criterion} - TODO + The prime field of $F$ is $\mathbb{Z}_q$, whence $qa = 0$ for any $a \in F$. This implies that $(a + b)^q = a^q + b^q$, since any binomial coefficient $\binom{q}{i}$ is a multiple of $q$ for $0 < i < q$, and thus 0 in $F$. \end{proof} \begin{theorem}[B] \label{fact_B} \lean{book.quadratic_reciprocity.fact_B} \leanok - TODO + For the field $F$ defined in (A), there exists an element $\zeta \in F$ of multiplicative order $p$, that is, $\zeta^p = 1$. Moreover, we have a polynomial decomposition + \[ + x^p - 1 = (x - \zeta) (x - \zeta^2) \cdots (x - \zeta^p). + \] \end{theorem} \begin{proof} - \uses{fermats_little} - TODO + \uses{fermats_little,mult_cyclic} + The multiplicative group $F^* = F \setminus \{0\}$ is cyclic of size $q^{p-1} - 1$. Since by Fermat's little theorem $p$ is a divisor of $q^{p-1} - 1$, there exists an element $\zeta \in F$ of order $p$, that is, $\zeta^p = 1$, and $\zeta$ generates the subgroup $\{\zeta, \zeta^2, \dots, \zeta^p = 1\}$ of $F^*$. Note that any $\zeta^i$ ($i \ne p$) is again a generator. Hence we obtain the polynomial decomposition $x^p - 1 = (x - \zeta) (x - \zeta^2) \cdots (x - \zeta^p)$. \end{proof} \begin{theorem}[Quadratic reciprocity II] \label{quadratic_reciprocity2} \lean{book.quadratic_reciprocity.quadratic_reciprocity_2} \leanok - TODO + Let $p$ and $q$ be different odd primes. Then + \[ + (\frac{q}{p})(\frac{p}{q}) = (-1)^{\frac{p-1}{2} \frac{q-1}{2}}. + \] \end{theorem} \begin{proof} \uses{fact_A, fact_B} - TODO + The second proof does not use Gauss' lemma, instead it employs so-called ``Gauss sums'' in finite fields. Gauss invented them in his study of the equation $x^p - 1 = 0$ and the arithmetical properties of the field $\mathbb{Q}(\zeta)$ (called cyclotomic field), where $\zeta$ is a $p$-th root of unity. They have been the starting point for the search for higher reciprocity laws in general number fields. + + Consider the \textit{Gauss sum} + \[ + G := \sum_{i=1}^{p-1} (\frac{i}{p}) \zeta^i \in F, + \] + where $(\frac{i}{p})$ is the Legendre symbol. For the proof we derive two different expressions for $G^q$ and then set them equal. + + \textbf{First expression.} We have + \begin{equation} \label{eq:first-expression} + G^q = \sum_{i=1}^{p-1} (\frac{i}{p})^q \zeta^{iq} = \sum_{i=1}^{p-1} (\frac{i}{p}) \zeta^{iq} = (\frac{q}{p}) \sum_{i=1}^{p-1} (\frac{iq}{p}) \zeta^{iq} = (\frac{q}{p}) G, + \end{equation} + where the first equality follows from $(a + b)^q = a^q + b^q$, the second uses that $(\frac{i}{p})^q = (\frac{i}{p})$ since $q$ is odd, the third one is derived from \eqref{eq:product_rule}, which yields $(\frac{i}{p}) = (\frac{q}{p}) (\frac{iq}{p})$, and the last one holds since $iq$ runs with $i$ through all nonzero residues modulo $p$. + + \textbf{Second expression.} Suppose we can prove + \begin{equation} \label{eq:G^2} + G^2 = (-1)^{\frac{p-1}{2}} p, + \end{equation} + then we are quickly done. Indeed, + \begin{equation} \label{eq:second-expression} + G^q = G (G^2)^{\frac{q-1}{2}} = G (-1)^{\frac{p-1}{2} \frac{q-1}{2}} p^{\frac{q-1}{2}} = G (\frac{p}{q}) (-1)^{\frac{p-1}{2} \frac{q-1}{2}}. + \end{equation} + Equating the expressions in \eqref{eq:first-expression} and \eqref{eq:second-expression} and cancelling $G$, which is nonzero by \eqref{eq:G^2}, we find $(\frac{q}{p}) = (\frac{p}{q}) (-1)^{\frac{p-1}{2} \frac{q-1}{2}}$, and thus + \[ + (\frac{q}{p}) (\frac{p}{q}) = (-1)^{\frac{p-1}{2} \frac{q-1}{2}}. + \] + + It remains to verify \eqref{eq:G^2}, and for this we first make two simple observations: + \begin{itemize} + \item $\sum_{i=1}^{p} \zeta^i = 0$ and thus $\sum_{i=1}^{p-1} \zeta^i = -1$. Just note that $-\sum_{i=1}^{p} \zeta^i$ is the coefficient of $x^{p-1}$ in $x^p - 1 = \prod_{i=1}^{p} (x - \zeta^i)$, and thus 0. + \item $\sum_{k=1}^{p-1} (\frac{k}{p}) = 0$ and thus $\sum_{k=1}^{p-2} (\frac{k}{p}) = - (\frac{-1}{p})$, since there are equally many quadratic residues and nonresidues. + \end{itemize} + + We have + \[ + G^2 = \left(\sum_{i=1}^{p-1} (\frac{i}{p}) \zeta^i\right) \left(\sum_{j=1}^{p-1} (\frac{j}{p}) \zeta^j\right) = \sum_{i,j} (\frac{ij}{p}) \zeta^{i+j}. + \] + Setting $j \equiv ik \pmod{p}$ we find + \[ + G^2 = \sum_{i,k} (\frac{k}{p}) \zeta^{i(1+k)} = \sum_{k=1}^{p-1} (\frac{k}{p}) \sum_{i=1}^{p-1} \zeta^{(1+k)i}. + \] + + For $k = p - 1 \equiv -1 \pmod{p}$ this gives $(\frac{-1}{p}) (p - 1)$, since $\zeta^{1+k} = 1$. Move $k = p - 1$ in front and write + \[ + G^2 = (\frac{-1}{p}) (p - 1) + \sum_{k=1}^{p-2} (\frac{k}{p}) \sum_{i=1}^{p-1} \zeta^{(1+k)i}. + \] + Since $\zeta^{1+k}$ is a generator of the group for $k \ne p - 1$, the inner sum equals $\sum_{i=1}^{p-1} \zeta^i = -1$ for all $k \ne p - 1$ by our first observation. Hence the second summand is $-\sum_{k=1}^{p-2} (\frac{k}{p}) = (\frac{-1}{p})$ by our second observation. It follows that $G^2 = (\frac{-1}{p}) p$ and thus with Euler's criterion $G^2 = (-1)^{\frac{p-1}{2}} p$, which completes the proof. \end{proof} diff --git a/blueprint/src/chapter/chapter06.tex b/blueprint/src/chapter/chapter06.tex index ded16f1..af72881 100644 --- a/blueprint/src/chapter/chapter06.tex +++ b/blueprint/src/chapter/chapter06.tex @@ -1,11 +1,166 @@ \chapter{Every finite division ring is a field} +\begin{theorem}[Roots of unity] + \label{root_of_unity} + The $n$-th roots of unity are + \[ + \lambda_k = e^{\frac{2k\pi i}{n}} = \cos(2k\pi/n) + i \sin(2k\pi/n), \quad 0 \le k \le n - 1. + \] +\end{theorem} +\begin{proof} + Any complex number $z = x + iy$ may be written in the ``polar'' form + \[ + z = r e^{i\varphi} = r(\cos \varphi + i \sin \varphi), + \] + where $r = |z| = \sqrt{x^2 + y^2}$ is the distance of $z$ to the origin, + and $\varphi$ is the angle measured from the positive $x$-axis. + The $n$-th roots of unity are therefore of the form + \[ + \lambda_k = e^{\frac{2k\pi i}{n}} = \cos(2k\pi/n) + i \sin(2k\pi/n), \quad 0 \le k \le n - 1, + \] + since for all $k$ + \[ + \lambda_k^n = e^{2k\pi i} = \cos(2k\pi) + i \sin(2k\pi) = 1. + \] + We obtain these roots geometrically by inscribing a regular $n$-gon into the unit circle. + Note that $\lambda_k = \zeta^k$ for all $k$, where $\zeta = e^{\frac{2\pi i}{n}}$. + Thus the $n$-th roots of unity form a cyclic group $\{\zeta, \zeta^2, \dots, \zeta^{n-1}, \zeta^n = 1\}$ of order $n$. +\end{proof} + \begin{theorem}[Wedderburn's theorem] \label{wedderburn} \lean{wedderburn} - \leanok - Every finite division ring is commutative + Every finite division ring is commutative. \end{theorem} \begin{proof} - TODO + Our first ingredient comes from a blend of linear algebra and basic group theory. + For an arbitrary element $s \in R$, let $C_s$ be the set $\{x \in R : xs = sx\}$ of elements which commute with $s$; + $C_s$ is called the \emph{centralizer} of $s$. + Clearly, $C_s$ contains $0$ and $1$ and is a sub-division ring of $R$. + The \emph{center} $Z$ is the set of elements which commute with all elements of $R$, thus $Z = \bigcap_{s \in R} C_s$. + In particular, all elements of $Z$ commute, $0$ and $1$ are in $Z$, and so $Z$ is a \emph{finite field}. + Let us set $|Z| = q$. + + We can regard $R$ and $C_s$ as vector spaces over the field $Z$ and deduce that $|R| = q^n$, where $n$ is the dimension of the vector space $R$ over $Z$, + and similarly $|C_s| = q^{n_s}$ for suitable integers $n_s \ge 1$. + + Now let us assume that $R$ is not a field. + This means that for \emph{some} $s \in R$ the centralizer $C_s$ is not all of $R$, or, what is the same, $n_s < n$. + + On the set $R^* := R \setminus \{0\}$ we consider the relation + \[ + r' \sim r :\iff r' = x^{-1}rx \quad \text{for some } x \in R^*. + \] + It is easy to check that $\sim$ is an equivalence relation. Let + \[ + A_s := \{x^{-1}sx : x \in R^*\} + \] + be the equivalence class containing $s$. + We note that $|A_s| = 1$ precisely when $s$ is in the center $Z$. + So by our assumption, there are classes $A_s$ with $|A_s| \ge 2$. + Consider now for $s \in R^*$ the map $f_s : x \mapsto x^{-1}sx$ from $R^*$ onto $A_s$. + For $x, y \in R^*$ we find + \[ + x^{-1}sx = y^{-1}sy \iff (yx^{-1})s = s(yx^{-1}) \iff yx^{-1} \in C_s^* \iff y \in C_s^* x, + \] + for $C_s^* := C_s \setminus \{0\}$, where $C_s^* x = \{zx : z \in C_s^*\}$ has size $|C_s^*|$. + Hence any element $x^{-1}sx$ is the image of precisely $|C_s^*| = q^{n_s} - 1$ elements in $R^*$ under the map $f_s$, + and we deduce $|R^*| = |A_s| |C_s^*|$. In particular, we note that + \[ + \frac{|R^*|}{|C_s^*|} = \frac{q^n - 1}{q^{n_s} - 1} = |A_s| \quad \text{is an \emph{integer} for all } s. + \] + We know that the equivalence classes partition $R^*$. + We now group the central elements $Z^*$ together and denote by $A_1, \dots, A_t$ the equivalence classes containing more than one element. + By our assumption we know $t \ge 1$. + Since $|R^*| = |Z^*| + \sum_{k=1}^t |A_k|$, we have proved the so-called \emph{class formula} + \begin{equation} \label{eq:class_formula} + q^n - 1 = q - 1 + \sum_{k=1}^t \frac{q^n - 1}{q^{n_k} - 1}, + \end{equation} + where we have $1 < \frac{q^n - 1}{q^{n_k} - 1} \in \mathbb{N}$ for all $k$. + + With \eqref{eq:class_formula} we have left abstract algebra and are back to the natural numbers. + Next we claim that $q^{n_k} - 1 \mid q^n - 1$ implies $n_k \mid n$. + Indeed, write $n = a n_k + r$ with $0 \le r < n_k$, then $q^{n_k} - 1 \mid q^{a n_k + r} - 1$ implies + \[ + q^{n_k} - 1 \mid (q^{a n_k + r} - 1) - (q^{n_k} - 1) = q^{n_k} (q^{(a-1)n_k + r} - 1), + \] + and thus $q^{n_k} - 1 \mid q^{(a-1)n_k + r} - 1$, since $q^{n_k}$ and $q^{n_k} - 1$ are relatively prime. + Continuing in this way we find $q^{n_k} - 1 \mid q^r - 1$ with $0 \le r < n_k$, which is only possible for $r = 0$, that is, $n_k \mid n$. + In summary, we note + \begin{equation} \label{eq:nk_divides_n} + n_k \mid n \quad \text{for all } k. + \end{equation} + Now comes the second ingredient: the complex numbers $\mathbb{C}$. + Consider the polynomial $x^n - 1$. + Its roots in $\mathbb{C}$ are called the $n$-th roots of unity. + Since $\lambda^n = 1$, all these roots $\lambda$ have $|\lambda| = 1$ and lie therefore on the unit circle of the complex plane. + In fact, they are precisely the numbers $\lambda_k = e^{\frac{2k\pi i}{n}} = \cos(2k\pi/n) + i \sin(2k\pi/n)$, $0 \le k \le n - 1$. + Some of the roots $\lambda$ satisfy $\lambda^d = 1$ for $d < n$; + for example, the root $\lambda = -1$ satisfies $\lambda^2 = 1$. + For a root $\lambda$, let $d$ be the smallest positive exponent with $\lambda^d = 1$, + that is, $d$ is the order of $\lambda$ in the group of the roots of unity. + Then $d \mid n$, by Lagrange's theorem (``the order of every element of a group divides the order of the group''). + Note that there are roots of order $n$, such as $\lambda_1 = e^{\frac{2\pi i}{n}}$. + + Now we group all roots of order $d$ together and set + \[ + \phi_d(x) := \prod_{\lambda \text{ of order } d} (x - \lambda). + \] + Note that the definition of $\phi_d(x)$ is independent of $n$. + Since every root has some order $d$, we conclude that + \begin{equation} \label{eq:prod_phi} + x^n - 1 = \prod_{d \mid n} \phi_d(x). + \end{equation} + Here is the crucial observation: + The coefficients of the polynomials $\phi_n(x)$ are integers (that is, $\phi_n(x) \in \mathbb{Z}[x]$ for all $n$), + where in addition the constant coefficient is either $1$ or $-1$. + Let us carefully verify this claim. + For $n = 1$ we have $1$ as the only root, and so $\phi_1(x) = x - 1$. + Now we proceed by induction, where we assume $\phi_d(x) \in \mathbb{Z}[x]$ for all $d < n$, + and that the constant coefficient of $\phi_d(x)$ is $1$ or $-1$. + By \eqref{eq:prod_phi}, + \begin{equation} \label{eq:poly_div} + x^n - 1 = p(x) \phi_n(x) + \end{equation} + where $p(x) = \prod_{d \mid n, d < n} \phi_d(x) = \sum_{j=0}^{n-\ell} p_j x^j$, $\phi_n(x) = \sum_{k=0}^{\ell} a_k x^k$, with $p_0 = 1$ or $p_0 = -1$. + Since $-1 = p_0 a_0$, we see $a_0 \in \{1, -1\}$. + Suppose we already know that $a_0, a_1, \dots, a_{k-1} \in \mathbb{Z}$. + Computing the coefficient of $x^k$ on both sides of \eqref{eq:poly_div} we find + \[ + \sum_{j=0}^k p_j a_{k-j} = \sum_{j=1}^k p_j a_{k-j} + p_0 a_k \in \mathbb{Z}. + \] + By assumption, all $a_0, \dots, a_{k-1}$ (and all $p_j$) are in $\mathbb{Z}$. + Thus $p_0 a_k$ and hence $a_k$ must also be integers, since $p_0$ is $1$ or $-1$. + + We are ready for the \emph{coup de gr\^ace}. + Let $n_k \mid n$ be one of the numbers appearing in \eqref{eq:class_formula}. + Then + \[ + x^n - 1 = \prod_{d \mid n} \phi_d(x) = (x^{n_k} - 1) \phi_n(x) \prod_{d \mid n, d \nmid n_k, d \ne n} \phi_d(x). + \] + We conclude that in $\mathbb{Z}$ we have the divisibility relations + \begin{equation} \label{eq:div_relations} + \phi_n(q) \mid q^n - 1 \quad \text{and} \quad \phi_n(q) \mid \frac{q^n - 1}{q^{n_k} - 1}. + \end{equation} + Since \eqref{eq:div_relations} holds for all $k$, we deduce from the class formula \eqref{eq:class_formula} + \[ + \phi_n(q) \mid q - 1, + \] + but this cannot be. + Why? We know $\phi_n(x) = \prod (x - \lambda)$ where $\lambda$ runs through all roots of $x^n - 1$ of order $n$. + Let $\lambda = a + ib$ be one of those roots. + By $n > 1$ (because of $R \ne Z$) we have $\lambda \ne 1$, which implies that the real part $a$ is smaller than $1$. + Now $|\lambda|^2 = a^2 + b^2 = 1$, and hence + \begin{align*} + |q - \lambda|^2 &= |q - a - ib|^2 = (q - a)^2 + b^2 \\ + &= q^2 - 2aq + a^2 + b^2 = q^2 - 2aq + 1 \\ + &> q^2 - 2q + 1 \quad (\text{because of } a < 1) \\ + &= (q - 1)^2, + \end{align*} + and so $|q - \lambda| > q - 1$ holds for all roots of order $n$. + This implies + \[ + |\phi_n(q)| = \prod_{\lambda} |q - \lambda| > q - 1, + \] + which means that $\phi_n(q)$ cannot be a divisor of $q - 1$, contradiction and end of proof. \end{proof} diff --git a/blueprint/src/chapter/chapter07.tex b/blueprint/src/chapter/chapter07.tex index 41b4014..1101b1f 100644 --- a/blueprint/src/chapter/chapter07.tex +++ b/blueprint/src/chapter/chapter07.tex @@ -1,35 +1,289 @@ \chapter{The spectral theorem and Hadamard's determinant problem} +We start with some preliminary facts. Let $O(n) \subseteq \mathbb{R}^{n \times n}$ be the set of real orthogonal matrices of order $n$. Since +\[ +(PQ)^{-1} = Q^{-1} P^{-1} = Q^T P^T = (PQ)^T +\] +for $P, Q \in O(n)$, we see that the set $O(n)$ is a group. Regarding any matrix in $\mathbb{R}^{n \times n}$ as a vector in $\mathbb{R}^{n^2}$, we find that $O(n)$ is a compact set. Indeed, as the columns of an orthogonal matrix $Q = (q_{ij})$ are unit vectors, we have $|q_{ij}| \le 1$ for all $i$ and $j$, thus $O(n)$ is bounded. Furthermore, the set $O(n)$ is defined as a subset of $\mathbb{R}^{n^2}$ by the equations +\[ +x_{i1} x_{j1} + x_{i2} x_{j2} + \dots + x_{in} x_{jn} = \delta_{ij} \quad \text{for } 1 \le i, j \le n, +\] +hence it is closed, and thus compact. + +For any real square matrix $A$ let $\Od(A) = \sum_{i \ne j} a_{ij}^2$ be the sum of the squares of the off-diagonal entries. + \begin{lemma} \label{ch7.lemma} - If $A$ is a real symmetric \(n \times n\) matrix that is not diagonal, that is - \(\Od(A) > 0\), then there exists \(U \in O(n)\) such that \(\Od(U^TAU)<\Od(A)\). + If $A$ is a real symmetric $n \times n$ matrix that is not diagonal, that is, + $\Od(A) > 0$, then there exists $U \in O(n)$ such that $\Od(U^T A U) < \Od(A)$. \end{lemma} \begin{proof} - TODO + We use a very clever method attributed to Carl Gustav Jacob Jacobi. + Suppose that $a_{rs} \ne 0$ for some $r \ne s$. + Then we claim that the matrix $U$ that agrees with the identity matrix except that $u_{rr} = u_{ss} = \cos \vartheta$, + $u_{rs} = \sin \vartheta$, $u_{sr} = -\sin \vartheta$ does the job, for some choice of the (real) angle $\vartheta$:\vspace{1em} + \[ + U = \left(\begin{array}{@{\,}ccccc@{\hspace{12pt}}l} + & \smash{\vbox to 0pt{\vss\hbox{$r$}\vskip 1em}} & & \smash{\vbox to 0pt{\vss\hbox{$s$}\vskip 1em}} & & \\[-1em] + \begin{array}{@{}c@{\,}c@{\,}c@{}} 1&&\\[-1ex]&\ddots &\\[-1ex]&&1\end{array} & & & & & \\ + & \cos \vartheta & & \sin \vartheta & & r \\ + & & \begin{array}{@{}c@{\,}c@{\,}c@{}} 1&&\\[-1ex]&\ddots &\\[-1ex]&&1\end{array} & & & \\ + & -\sin \vartheta & & \cos \vartheta & & s\\ + & & & & \begin{array}{@{}c@{\,}c@{\,}c@{}} 1&&\\[-1ex]&\ddots &\\[-1ex]&&1\end{array} & \\ + \end{array}\kern-20pt\right) + \] + Clearly, $U$ is orthogonal for any $\vartheta$. + + Now let us compute the $(k, \ell)$-entry $b_{k\ell}$ of $U^T A U$. We have + \begin{equation} \label{eq:b_kl} + b_{k\ell} = \sum_{i,j} u_{ik} a_{ij} u_{j\ell}. + \end{equation} + For $k, \ell \notin \{r, s\}$ we get $b_{k\ell} = a_{k\ell}$. Furthermore, we have + \begin{align*} + b_{kr} &= \sum_{i=1}^n u_{ik} \sum_{j=1}^n a_{ij} u_{jr} \\ + &= \sum_{i=1}^n u_{ik} (a_{ir} \cos \vartheta - a_{is} \sin \vartheta) \\ + &= a_{kr} \cos \vartheta - a_{ks} \sin \vartheta \quad (\text{for } k \ne r, s). + \end{align*} + Similarly, one computes + \[ + b_{ks} = a_{kr} \sin \vartheta + a_{ks} \cos \vartheta \quad (\text{for } k \ne r, s). + \] + It follows that + \begin{align*} + b_{kr}^2 + b_{ks}^2 &= a_{kr}^2 \cos^2 \vartheta - 2 a_{kr} a_{ks} \cos \vartheta \sin \vartheta + a_{ks}^2 \sin^2 \vartheta \\ + &\quad + a_{kr}^2 \sin^2 \vartheta + 2 a_{kr} a_{ks} \sin \vartheta \cos \vartheta + a_{ks}^2 \cos^2 \vartheta \\ + &= a_{kr}^2 + a_{ks}^2, + \end{align*} + and by symmetry + \[ + b_{r\ell}^2 + b_{s\ell}^2 = a_{r\ell}^2 + a_{s\ell}^2 \quad (\text{for } \ell \ne r, s). + \] + We conclude that the function $\Od$, which sums the squares of the off-diagonal values, + agrees for $A$ and $U^T A U$ except for the entries at $(r, s)$ and $(s, r)$, for \emph{any} $\vartheta$. + To conclude the proof we now show that $\vartheta_0$ can be chosen suitably as to make $b_{rs} = 0$, which will result in + \[ + \Od(U^T A U) = \Od(A) - 2 a_{rs}^2 < \Od(A) + \] + as required. + + Using \eqref{eq:b_kl} we find + \[ + b_{rs} = (a_{rr} - a_{ss}) \sin \vartheta \cos \vartheta + a_{rs} (\cos^2 \vartheta - \sin^2 \vartheta). + \] + For $\vartheta = 0$ this becomes $a_{rs}$, while for $\vartheta = \pi/2$ it is $-a_{rs}$. + Hence by the intermediate value theorem there is some $\vartheta_0$ between $0$ and $\pi/2$ such that $b_{rs} = 0$, + and we are through. \end{proof} \begin{theorem} \label{diagonalize_real_symmetric} \lean{chapter7.Theorem₁} \leanok - For every real symmetric matrix \(A\) there is a real - orthogonal matrix \(Q\) such that \(Q^{T}AQ\) is diagonal. + For every real symmetric matrix $A$ there is a real + orthogonal matrix $Q$ such that $Q^{T}AQ$ is diagonal. \end{theorem} \begin{proof} \uses{ch7.lemma} - TODO + The theorem follows in three quick steps. Let $A$ be a real symmetric $n \times n$ matrix. + \begin{enumerate} + \item[(A)] Consider the map $f_A : O(n) \to \mathbb{R}^{n \times n}$ with $f_A(P) := P^T A P$. The map $f_A$ is continuous on the compact set $O(n)$, and so the image $f_A(O(n))$ is compact. + \item[(B)] The function $\Od : f_A(O(n)) \to \mathbb{R}$ is continuous, hence it assumes a minimum, say at $D = Q^T A Q \in f_A(O(n))$. + \item[(C)] The value $\Od(D)$ must be zero, and hence $D$ is a diagonal matrix as required. + \end{enumerate} + Indeed, if $\Od(D) > 0$, then applying the Lemma we find $U \in O(n)$ with $\Od(U^T D U) < \Od(D)$. But + \[ + U^T D U = U^T Q^T A Q U = (QU)^T A (QU) + \] + is in $f_A(O(n))$ (remember $O(n)$ is a group!) with $\Od$-value smaller than that of $D$ --- contradiction, and end of proof. +\end{proof} + +\begin{theorem}[Hadamard's inequality] + \label{thm:hadamard_inequality} + \lean{chapter7.Theorem₂} + For any real $n \times n$ matrix $A = (a_{ij})$ with $|a_{ij}| \le 1$, + \[ + |\det A| \le n^{n/2}. + \] +\end{theorem} + +\begin{proof} + The problem to find the maximum value of $\det A$ on the set of all real $n \times n$ matrices $A = (a_{ij})$ with $|a_{ij}| \le 1$ is unsolved. + Since the determinant is a continuous function in the $a_{ij}$ (considered as variables) + and the matrices form a compact set in $\mathbb{R}^{n^2}$, this maximum must exist. + Furthermore, the maximum is attained for some matrix all of whose entries are $+1$ or $-1$, + because the function $\det A$ is linear in each single entry $a_{ij}$ (if we keep all other entries fixed). + Thus we can start with any matrix $A$ and move one entry after the other to $+1$ or to $-1$, + in every single step not decreasing the determinant, until we arrive at a $\pm 1$-matrix. + In the search for the largest determinant we may thus assume that all entries of $A$ are $\pm 1$. + + Here is the trick: + Instead of $A$ we consider the matrix $B = A^T A = (b_{ij})$. + That is, if $c_j = (a_{1j}, a_{2j}, \dots, a_{nj})^T$ denotes the $j$-th column vector of $A$, + then $b_{ij} = \langle c_i, c_j \rangle$, the inner product of $c_i$ and $c_j$. + In particular, + \[ + b_{ii} = \langle c_i, c_i \rangle = n \quad \text{for all } i, + \] + and + \begin{equation} \label{eq:trace_B} + \trace B = \sum_{i=1}^n b_{ii} = n^2, + \end{equation} + which will come in handy in a moment. + + Now we can go to work. + First of all, from $B = A^T A$ we get $|\det A| = \sqrt{\det B}$. + Since multiplication of a column of $A$ by $-1$ turns $\det A$ into $-\det A$, + we see that the maximum problem for $\det A$ is the same as for $\det B$. + Furthermore, we may assume that $A$ is nonsingular, and hence that $B$ is nonsingular as well. + + Since $B = A^T A$ is a symmetric matrix the spectral theorem tells us that for some $Q \in O(n)$, + \begin{equation} \label{eq:QBQ} + \setlength{\arraycolsep}{2pt} + Q^T B Q = Q^T A^T A Q = (AQ)^T (AQ) = + \begin{pmatrix} + \lambda_1 & & & & \\ + & \ddots & & O & \\ + & & \ddots & & \\ + & O & & \ddots & \\ + & & & & \lambda_n + \end{pmatrix}, + \end{equation} + where the $\lambda_i$ are the eigenvalues of $B$. Now, if $d_j$ denotes the $j$-th column vector of $AQ$ (which is nonzero since $A$ is nonsingular), then + \[ + \lambda_j = \langle d_j, d_j \rangle = \sum_{i=1}^n d_{ij}^2 > 0. + \] + Thus $\lambda_1, \dots, \lambda_n$ are positive real numbers and + \[ + \det B = \lambda_1 \cdots \lambda_n, \quad \trace B = \sum_{i=1}^n \lambda_i. + \] + Whenever such a product and sum of positive numbers turn up, it is always a good idea to try the arithmetic-geometric mean inequality. In our case this gives with \eqref{eq:trace_B} + \begin{equation} \label{eq:AM_GM} + \det B = \lambda_1 \cdots \lambda_n \le \left( \frac{\sum_{i=1}^n \lambda_i}{n} \right)^n = \left( \frac{\trace B}{n} \right)^n = n^n, + \end{equation} + and out comes Hadamard's upper bound + \begin{equation} \label{eq:hadamard_bound} + |\det A| \le n^{n/2}. + \end{equation} + When do we have equality in \eqref{eq:hadamard_bound} or, what is the same, in \eqref{eq:AM_GM}? Easy enough: if and only if the geometric mean of the $\lambda_i$'s equals the arithmetic mean, or equivalently, if and only if $\lambda_1 = \dots = \lambda_n = \lambda$. But then $\trace B = n\lambda = n^2$, and so $\lambda_1 = \dots = \lambda_n = n$. Looking at \eqref{eq:QBQ} this means $Q^T B Q = n I_n$, where $I_n$ is the $n \times n$ identity matrix. Now recall $Q^T = Q^{-1}$, multiply by $Q$ on the left, by $Q^{-1}$ on the right, to obtain + \[ + B = n I_n. + \] + Going back to $A$ this means that + \[ + |\det A| = n^{n/2} \iff \langle c_i, c_j \rangle = 0 \quad \text{for } i \ne j. + \] + Matrices $A$ with $\pm 1$-entries that achieve equality in \eqref{eq:hadamard_bound} are aptly called \emph{Hadamard matrices}. So an $n \times n$ matrix $A$ with $\pm 1$-entries is a Hadamard matrix if and only if + \[ + A^T A = A A^T = n I_n. + \] \end{proof} \begin{theorem} - \label{exists_matrix} - \lean{chapter7.Theorem₂} + \label{thm:hadamard_order} + If a Hadamard matrix of size $n \times n$ exists for $n > 2$, then $n$ must be a multiple of $4$. +\end{theorem} + +\begin{proof} + A short argument shows that if $n$ is greater than $2$, then it must be a multiple of $4$. + Indeed, suppose that $A$ is an $n \times n$ Hadamard matrix, $n \ge 2$, whose rows are the vectors $r_1, \dots, r_n$. + Clearly, multiplication of any row or column by $-1$ gives another Hadamard matrix. + So we may assume that the first row consists of $1$'s only. + Since $\langle r_1, r_i \rangle = 0$ for $i \ne 1$, every other row must contain $n/2$ $1$'s and $n/2$ $-1$'s; + in particular, $n$ must be even. + Assume now that $n > 2$ and consider rows $r_2$ and $r_3$, and denote by $a, b, c, d$ the numbers of columns that have + $\begin{pmatrix} +1 \\ +1 \end{pmatrix}$, $\begin{pmatrix} +1 \\ -1 \end{pmatrix}$, $\begin{pmatrix} -1 \\ +1 \end{pmatrix}$, and $\begin{pmatrix} -1 \\ -1 \end{pmatrix}$ + in rows $2$ and $3$, respectively. + Then from $\langle r_1, r_2 \rangle = 0$ and $\langle r_1, r_3 \rangle = 0$ we get + \[ + a+b = c+d = a+c = b+d = n/2, + \] + which gives $b = c, a = d$. + But from $\langle r_2, r_3 \rangle = 0$ we also have $a+d = b+c$, resulting in $2a = 2b$. + We conclude that $a = b = c = d = n/4$. + Thus the order of the Hadamard matrix is either $n = 1$ or $n = 2$, or $n = a+b+c+d = 4a$, a multiple of $4$. +\end{proof} + + +\begin{theorem} + \label{thm:hadamard_existence} + \leanok + Hadamard matrices exist for all $n = 2^m$. +\end{theorem} + +\begin{proof} \leanok - There exists an \(n \times n\) matrix with entries \(\pm 1\) whose - determinant is greater than \(\sqrt{n!}\). + Consider an $m$-set $X$ and index the $2^m$ subsets $C \subseteq X$ in any way $C_1, \dots, C_{2^m}$. The matrix $A = (a_{ij})$ is defined as + \[ + a_{ij} = (-1)^{|C_i \cap C_j|}. + \] + We want to verify $\langle r_i, r_j \rangle = 0$ for $i \ne j$. From the definition, + \[ + \langle r_i, r_j \rangle = \sum_k (-1)^{|C_i \cap C_k| + |C_j \cap C_k|}. + \] + Now, as $C_i \ne C_j$ there exists an element $a \in X$ with $a \in C_i \setminus C_j$ or $a \in C_j \setminus C_i$; + suppose $a \in C_i \setminus C_j$. + Half the subsets of $X$ contain $a$, and half do not. + Let $C$ run through all subsets that contain $a$, then the pairs $\{C, C \setminus \{a\}\}$ will comprise all subsets of $X$. + But for each such pair $\{C, C \setminus \{a\}\}$, $|C_i \cap C| + |C_j \cap C|$ and $|C_i \cap (C \setminus \{a\})| + |C_j \cap (C \setminus \{a\})|$ have different parity, + and so the corresponding terms in the sum will sum to $0$. + But then the whole sum is $0$, as required. +\end{proof} + +\begin{theorem} + \label{thm:det_greater_n_fact} + \lean{chapter7.Theorem₂} + There exists an $n \times n$ matrix with entries $\pm 1$ whose determinant is greater than $\sqrt{n!}$. \end{theorem} + \begin{proof} - \uses{diagonalize_real_symmetric} - TODO + Let us look at all $2^{n^2}$ matrices with $\pm 1$-entries and consider some averages of the determinant. + The arithmetic mean $\frac{1}{2^{n^2}} \sum_A \det A$ is $0$ (clear?), so this is no big help. + But if we consider the \emph{mean square average} instead, + \[ + D_n := \sqrt{\frac{\sum_A (\det A)^2}{2^{n^2}}}, + \] + then things brighten up. + Clearly, + \[ + \max_A \det A \ge D_n, + \] + so this will give us a lower bound for the maximum. + + The following stunningly simple calculation of $D_n^2$ probably appeared first in an article by George Szekeres and Paul Tur\'an. + We learnt it from a beautiful paper of Herb Wilf who heard it from Mark Kac. + In the words of Mark Kac: + ``\emph{Just write $(\det A)^2$ out twice, interchange summation, and everything simplifies.}'' + So we want to do just that. + + From the definition of the determinant we get + \begin{align*} + D_n^2 &= \frac{1}{2^{n^2}} \sum_A \left( \sum_{\pi} (\sign \pi) a_{1\pi(1)} a_{2\pi(2)} \cdots a_{n\pi(n)} \right)^2 \\ + &= \frac{1}{2^{n^2}} \sum_A \sum_{\sigma} \sum_{\tau} (\sign \sigma) (\sign \tau) a_{1\sigma(1)} a_{1\tau(1)} \cdots a_{n\sigma(n)} a_{n\tau(n)}, + \end{align*} + where $\sigma$ and $\tau$ run independently through all permutations of $\{1, \dots, n\}$. + Interchange of summation yields + \[ + D_n^2 = \frac{1}{2^{n^2}} \sum_{\sigma, \tau} (\sign \sigma) (\sign \tau) \left(\sum_A a_{1\sigma(1)} a_{1\tau(1)} \cdots a_{n\sigma(n)} a_{n\tau(n)}\right). + \] + This doesn't look too promising, but wait. + Look at a fixed pair $(\sigma, \tau)$. + The inner sum $\sum_A$ is really a summation over $n^2$ variables, one for each $a_{ij}$: + \begin{equation} \label{eq:sum_A} + \sum_{a_{11} = \pm 1} \sum_{a_{12} = \pm 1} \cdots \sum_{a_{nn} = \pm 1} a_{1\sigma(1)} a_{1\tau(1)} \cdots a_{n\sigma(n)} a_{n\tau(n)}. + \end{equation} + Suppose $\sigma(i) = k \ne \tau(i)$. + Then every summand contains $a_{ik}$, and therefore the whole sum has the \emph{factor} $\sum_{a_{ik} = \pm 1} a_{ik} = 0$, + and hence is $0$ as well. + The only way that the sum fails to be $0$ is when $\sigma = \tau$, and everything simplifies indeed: + For $\sigma = \tau$, the inner product is $1$ as is the term $(\sign \sigma)^2$. + The sum in \eqref{eq:sum_A} is therefore + \[ + \sum_{a_{11} = \pm 1} \cdots \sum_{a_{nn} = \pm 1} 1 = 2^{n^2}, + \] + and wrapping things up we obtain + \[ + D_n^2 = \frac{1}{2^{n^2}} \sum_{\sigma} 2^{n^2} = n!, + \] + and thus the result. \end{proof} diff --git a/blueprint/src/chapter/chapter08.tex b/blueprint/src/chapter/chapter08.tex index 8c4de9f..2b3e552 100644 --- a/blueprint/src/chapter/chapter08.tex +++ b/blueprint/src/chapter/chapter08.tex @@ -5,21 +5,89 @@ \chapter{Some irrational numbers} \label{e_irrational} \lean{book.irrational.e_irrational} \leanok - \(e\) is irrational -\end{theorem} + $e$ is irrational. +\end{theorem} \begin{proof} - TODO + To start with, it is rather easy to see (as did Fourier in 1815) that $e = \sum_{k \ge 0} \frac{1}{k!}$ is irrational. + Indeed, if we had $e = \frac{a}{b}$ for integers $a$ and $b > 0$, then we would get + \[ + n! b e = n! a + \] + for \emph{every} $n \ge 0$. + But this cannot be true, because on the right-hand side we have an integer, while the left-hand side with + \[ + e = 1 + \frac{1}{1!} + \frac{1}{2!} + \dots + \frac{1}{n!} + \frac{1}{(n+1)!} + \frac{1}{(n+2)!} + \frac{1}{(n+3)!} + \dots + \] + decomposes into an integral part + \[ + b n! \left( 1 + \frac{1}{1!} + \frac{1}{2!} + \dots + \frac{1}{n!} \right) + \] + and a second part + \[ + b \left( \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} + \dots \right) + \] + which is \emph{approximately} $\frac{b}{n}$, so that for large $n$ it certainly cannot be integral: + It is larger than $\frac{b}{n+1}$ and smaller than $\frac{b}{n}$, as one can see from a comparison with a geometric series: + \begin{align*} + \frac{1}{n+1} &< \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} + \dots \\ + &< \frac{1}{n+1} + \frac{1}{(n+1)^2} + \frac{1}{(n+1)^3} + \dots = \frac{1}{n}. + \end{align*} \end{proof} \begin{theorem} \label{e_pow_2_irrational} \lean{book.irrational.e_pow_2_irrational} \leanok - \(e ^ 2\) is irrational + $e^2$ is irrational. \end{theorem} \begin{proof} \uses{e_irrational} - TODO + Now one might be led to think that this simple multiply-by-$n!$ trick is not sufficient to show that $e^2$ is irrational. + This is a stronger statement: + $\sqrt{2}$ is an example of a number which is irrational, but whose square is not. + From John Cosgrave we have learned that with two nice ideas/observations (let's call them ``tricks'') one can get two steps further nevertheless: + Each of the tricks is sufficient to show that $e^2$ is irrational, the combination of both of them even yields the same for $e^4$. + The first trick may be found in a one page paper by J. Liouville from 1840 --- and the second one in a two page ``addendum'' + which Liouville published on the next two journal pages. + + Why is $e^2$ irrational? + What can we derive from $e^2 = \frac{a}{b}$? + According to Liouville we should write this as + \[ + b e = a e^{-1}, + \] + substitute the series + \[ + e = 1 + \frac{1}{1} + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \frac{1}{120} + \cdots + \] + and + \[ + e^{-1} = 1 - \frac{1}{1} + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120} \pm \cdots, + \] + and then multiply by $n!$, for a sufficiently large even $n$. + Then we see that $n! b e$ is nearly integral: + \[ + n! b \left( 1 + \frac{1}{1} + \frac{1}{2} + \frac{1}{6} + \dots + \frac{1}{n!} \right) + \] + is an integer, and the rest + \[ + n! b \left( \frac{1}{(n+1)!} + \frac{1}{(n+2)!} + \dots \right) + \] + is approximately $\frac{b}{n}$: + It is larger than $\frac{b}{n+1}$ but smaller than $\frac{b}{n}$, as we have seen above. + + At the same time $n! a e^{-1}$ is nearly integral as well: + Again we get a large integral part, and then a rest + \[ + (-1)^{n+1} n! a \left( \frac{1}{(n+1)!} - \frac{1}{(n+2)!} + \frac{1}{(n+3)!} \mp \cdots \right), + \] + and this is approximately $(-1)^{n+1} \frac{a}{n}$. + More precisely: + for even $n$ the rest is larger than $-\frac{a}{n}$, but smaller than + \[ + -a \left( \frac{1}{n+1} - \frac{1}{(n+1)^2} - \frac{1}{(n+1)^3} - \cdots \right) = - \frac{a}{n+1} \left( 1 - \frac{1}{n} \right) < 0. + \] + But this cannot be true, since for large even $n$ it would imply that $n! a e^{-1}$ is just a bit smaller than an integer, while $n! b e$ is a bit larger than an integer, so $n! a e^{-1} = n! b e$ cannot hold. \end{proof} @@ -27,41 +95,98 @@ \chapter{Some irrational numbers} \label{little_lemma} \lean{book.irrational.little_lemma} \leanok - TODO + For any $n \ge 1$ the integer $n!$ contains the prime factor $2$ at most $n-1$ times --- with equality if (and only if) $n$ is a power of two, $n = 2^m$. \end{theorem} \begin{proof} - TODO + This lemma is not hard to show: + $\lfloor \frac{n}{2} \rfloor$ of the factors of $n!$ are even, $\lfloor \frac{n}{4} \rfloor$ of them are divisible by $4$, and so on. + So if $2^k$ is the largest power of two which satisfies $2^k \le n$, then $n!$ contains the prime factor $2$ exactly + \[ + \left\lfloor \frac{n}{2} \right\rfloor + \left\lfloor \frac{n}{4} \right\rfloor + \dots + \left\lfloor \frac{n}{2^k} \right\rfloor \le \frac{n}{2} + \frac{n}{4} + \dots + \frac{n}{2^k} = n \left( 1 - \frac{1}{2^k} \right) \le n-1 + \] + times, with equality in both inequalities exactly if $n = 2^k$. \end{proof} \begin{theorem} \label{e_pow_4_irrational} \lean{book.irrational.e_pow_4_irrational} \leanok - \(e ^ 4\) is irrational + $e^4$ is irrational. \end{theorem} \begin{proof} \uses{little_lemma, e_pow_2_irrational} - TODO + In order to show that $e^4$ is irrational, we now courageously assume that $e^4 = \frac{a}{b}$ were rational, and write this as + \[ + b e^2 = a e^{-2}. + \] + We could now try to multiply this by $n!$ for some large $n$, and collect the non-integral summands, but this leads to nothing useful: + The sum of the remaining terms on the left-hand side will be approximately $b \frac{2^{n+1}}{n}$, + on the right side $(-1)^{n+1} a \frac{2^{n+1}}{n}$, and both will be very large if $n$ gets large. + + So one has to examine the situation a bit more carefully, and make two little adjustments to the strategy: + First we will not take an \emph{arbitrary} large $n$, but a large power of two, $n = 2^m$; + and secondly we will not multiply by $n!$, but by $\frac{n!}{2^{n-1}}$. + Then we need the little lemma \ref{little_lemma}, a special case of Legendre's theorem (see page 10). + + Let's get back to $b e^2 = a e^{-2}$. + We are looking at + \begin{equation} \label{eq:e4_irrational} + b \frac{n!}{2^{n-1}} e^2 = a \frac{n!}{2^{n-1}} e^{-2} + \end{equation} + and substitute the series + \[ + e^2 = 1 + \frac{2}{1} + \frac{4}{2} + \frac{8}{6} + \dots + \frac{2^r}{r!} + \dots + \] + and + \[ + e^{-2} = 1 - \frac{2}{1} + \frac{4}{2} - \frac{8}{6} \pm \dots + (-1)^r \frac{2^r}{r!} + \dots + \] + For $r \le n$ we get integral summands on both sides, namely + \[ + b \frac{n!}{2^{n-1}} \frac{2^r}{r!} \quad \text{resp.} \quad (-1)^r a \frac{n!}{2^{n-1}} \frac{2^r}{r!}, + \] + where for $r > 0$ the denominator $r!$ contains the prime factor $2$ at most $r-1$ times, while $n!$ contains it \emph{exactly} $n-1$ times. + (So for $r > 0$ the summands are even.) + + And since $n$ is even (we assume that $n = 2^m$), the series that we get for $r \ge n+1$ are + \[ + 2b \left( \frac{2}{n+1} + \frac{4}{(n+1)(n+2)} + \frac{8}{(n+1)(n+2)(n+3)} + \dots \right) + \] + resp. + \[ + 2a \left( - \frac{2}{n+1} + \frac{4}{(n+1)(n+2)} - \frac{8}{(n+1)(n+2)(n+3)} \pm \dots \right). + \] + These series will for large $n$ be roughly $\frac{4b}{n}$ resp. $-\frac{4a}{n}$, as one sees again by comparison with geometric series. + For large $n = 2^m$ this means that the left-hand side of \eqref{eq:e4_irrational} is a bit larger than an integer, + while the right-hand side is a bit smaller --- contradiction! \end{proof} \begin{lemma} \label{lem_aux_i} \lean{book.irrational.lem_aux_i} \leanok - TODO + \upshape For some fixed $n \ge 1$, let + \[ + f(x) = \frac{x^n (1-x)^n}{n!}. + \] + \begin{enumerate} + \item[(i)] \itshape The function $f(x)$ is a polynomial of the form $f(x) = \frac{1}{n!} \sum_{i=n}^{2n} c_i x^i$, where the coefficients $c_i$ are integers. + \end{enumerate} \end{lemma} \begin{proof} - TODO + Part (i) is clear. \end{proof} \begin{lemma} \label{lem_aux_ii} \lean{book.irrational.lem_aux_ii} \leanok - TODO + \begin{enumerate} + \item[(ii)] For $0 < x < 1$ we have $0 < f(x) < \frac{1}{n!}$. + \end{enumerate} \end{lemma} \begin{proof} - TODO + Part (ii) is also clear. \end{proof} @@ -69,21 +194,55 @@ \chapter{Some irrational numbers} \label{lem_aux_iii} \lean{book.irrational.lem_aux_iii} \leanok - TODO + \begin{enumerate} + \item[(iii)] The derivatives $f^{(k)}(0)$ and $f^{(k)}(1)$ are integers for all $k \ge 0$. + \end{enumerate} \end{lemma} \begin{proof} - TODO + For (iii) note that by (i) the $k$-th derivative $f^{(k)}$ vanishes at $x=0$ unless $n \le k \le 2n$, + and in this range $f^{(k)}(0) = \frac{k!}{n!} c_k$ is an integer. + From $f(x) = f(1-x)$ we get $f'(x) = (-1) f'(1-x)$ for all $x$, + and hence $f^{(k)}(1) = (-1)^k f^{(k)}(0)$, which is an integer. \end{proof} \begin{theorem} \label{e_pow_irrational} \label{book.irrational.Theorem_1} \leanok - \(e^r\) is irrational for every \(r\in\mathbb{Q}\setminus\{0\}\). + $e^r$ is irrational for every $r \in \mathbb{Q} \setminus \{0\}$. \end{theorem} \begin{proof} \uses{lem_aux_i, lem_aux_ii, lem_aux_iii} - TODO + It suffices to show that $e^s$ cannot be rational for a positive integer $s$ (if $e^{\frac{s}{t}}$ were rational, then $\left(e^{\frac{s}{t}}\right)^t = e^s$ would be rational, too). + Assume that $e^s = \frac{a}{b}$ for integers $a, b > 0$, and let $n$ be so large that $n! > a s^{2n+1}$. + Put + \[ + F(x) := s^{2n} f(x) - s^{2n-1} f'(x) + s^{2n-2} f''(x) \mp \dots + f^{(2n)}(x), + \] + where $f(x)$ is the function of the lemma. + $F(x)$ may also be written as an infinite sum + \[ + F(x) = s^{2n} f(x) - s^{2n-1} f'(x) + s^{2n-2} f''(x) \mp \cdots, + \] + since the higher derivatives $f^{(k)}(x)$, for $k > 2n$, vanish. + From this we see that the polynomial $F(x)$ satisfies the identity + \[ + F'(x) = -s F(x) + s^{2n+1} f(x). + \] + Thus differentiation yields + \[ + \frac{d}{dx} [e^{sx} F(x)] = s e^{sx} F(x) + e^{sx} F'(x) = s^{2n+1} e^{sx} f(x) + \] + and hence + \[ + N := b \int_0^1 s^{2n+1} e^{sx} f(x) dx = b [e^{sx} F(x)]_0^1 = a F(1) - b F(0). + \] + This is an integer, since part (iii) of the lemma implies that $F(0)$ and $F(1)$ are integers. + However, part (ii) of the lemma yields estimates for the size of $N$ from below and from above, + \[ + 0 < N = b \int_0^1 s^{2n+1} e^{sx} f(x) dx < b s^{2n+1} e^s \frac{1}{n!} = \frac{a s^{2n+1}}{n!} < 1, + \] + which shows that $N$ cannot be an integer: contradiction. \end{proof} @@ -91,11 +250,36 @@ \chapter{Some irrational numbers} \label{pi_pow_2_irrational} \label{book.irrational.Theorem_2} \leanok - \(\pi^2\) is irrational. + $\pi^2$ is irrational. \end{theorem} \begin{proof} \uses{lem_aux_ii, lem_aux_iii} - TODO + Assume that $\pi^2 = \frac{a}{b}$ for integers $a, b > 0$. + We now use the polynomial + \[ + F(x) := b^n \left( \pi^{2n} f(x) - \pi^{2n-2} f^{(2)}(x) + \pi^{2n-4} f^{(4)}(x) \mp \cdots \right), + \] + which satisfies $F''(x) = -\pi^2 F(x) + b^n \pi^{2n+2} f(x)$. + + From part (iii) of the lemma we get that $F(0)$ and $F(1)$ are integers. + Elementary differentiation rules yield + \begin{align*} + \frac{d}{dx} [F'(x) \sin \pi x - \pi F(x) \cos \pi x] &= \left(F''(x) + \pi^2 F(x)\right) \sin \pi x \\ + &= b^n \pi^{2n+2} f(x) \sin \pi x \\ + &= \pi^2 a^n f(x) \sin \pi x, + \end{align*} + and thus we obtain + \begin{align*} + N := \pi \int_0^1 a^n f(x) \sin \pi x \, dx &= \left[ \frac{1}{\pi} F'(x) \sin \pi x - F(x) \cos \pi x \right]_0^1 \\ + &= F(0) + F(1), + \end{align*} + which is an integer. + Furthermore $N$ is positive since it is defined as the integral of a function that is positive (except on the boundary). + However, if we choose $n$ so large that $\frac{\pi a^n}{n!} < 1$, then from part (ii) of the lemma we obtain + \[ + 0 < N = \pi \int_0^1 a^n f(x) \sin \pi x dx < \frac{\pi a^n}{n!} < 1, + \] + a contradiction. \end{proof} \begin{theorem} @@ -104,11 +288,42 @@ \chapter{Some irrational numbers} \leanok For every odd integer \(n \ge 3\), the number \[ - A(n) := \frac{1}{\pi}\arccos\left(\frac{1}{\sqrt{n}}\right) + A(n) := \frac{1}{\pi} \arccos \left(\frac{1}{\sqrt{n}}\right) \] is irrational. \end{theorem} \begin{proof} \uses{pi_pow_2_irrational} - TODO + We use the addition theorem + \[ + \cos \alpha + \cos \beta = 2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2} + \] + from elementary trigonometry, which for $\alpha = (k+1)\varphi$ and $\beta = (k-1)\varphi$ yields + \begin{equation} \label{eq:cos_add} + \cos (k+1)\varphi = 2 \cos \varphi \cos k\varphi - \cos (k-1)\varphi. + \end{equation} + For the angle $\varphi_n = \arccos \left(\frac{1}{\sqrt{n}}\right)$, which is defined by $\cos \varphi_n = \frac{1}{\sqrt{n}}$ + and $0 \le \varphi_n \le \pi$, this yields representations of the form + \[ + \cos k\varphi_n = \frac{A_k}{\sqrt{n}^k}, + \] + where $A_k$ is an integer that is not divisible by $n$, for all $k \ge 0$. + In fact, we have such a representation for $k=0, 1$ with $A_0 = A_1 = 1$, and by induction on $k$ using \eqref{eq:cos_add} we get for $k \ge 1$ + \[ + \cos (k+1)\varphi_n = 2 \frac{1}{\sqrt{n}} \frac{A_k}{\sqrt{n}^k} - \frac{A_{k-1}}{\sqrt{n}^{k-1}} = \frac{2A_k - n A_{k-1}}{\sqrt{n}^{k+1}}. + \] + Thus we obtain $A_{k+1} = 2A_k - n A_{k-1}$. + If $n \ge 3$ is odd, and $A_k$ is not divisible by $n$, then we find that $A_{k+1}$ cannot be divisible by $n$, either. + + Now assume that + \[ + A(n) = \frac{1}{\pi} \varphi_n = \frac{k}{\ell} + \] + is rational (with integers $k, \ell > 0$). + Then $\ell \varphi_n = k \pi$ yields + \[ + \pm 1 = \cos k \pi = \frac{A_\ell}{\sqrt{n}^\ell}. + \] + Thus $\sqrt{n}^\ell = \pm A_\ell$ is an integer, with $\ell \ge 2$, and hence $n | \sqrt{n}^\ell$. + With $\sqrt{n}^\ell | A_\ell$ we find that $n$ divides $A_\ell$, a contradiction. \end{proof} diff --git a/blueprint/src/chapter/chapter09.tex b/blueprint/src/chapter/chapter09.tex index 6109bc4..b09103f 100644 --- a/blueprint/src/chapter/chapter09.tex +++ b/blueprint/src/chapter/chapter09.tex @@ -5,11 +5,54 @@ \chapter{Four times $π^2/6$} \lean{euler_series} \leanok \[ - \sum_{n\ge 1}\frac{1}{n^2} = \frac{\pi^2}{6} + \sum_{n \ge 1} \frac{1}{n^2} = \frac{\pi^2}{6} \] \end{theorem} \begin{proof} - TODO + The proof consists in two different evaluations of the double integral + \[ + I := \int_0^1 \int_0^1 \frac{1}{1-xy} dx dy. + \] + For the first one, we expand $\frac{1}{1-xy}$ as a geometric series, decompose the summands as products, and integrate effortlessly: + \begin{align*} + I &= \int_0^1 \int_0^1 \sum_{n \ge 0} (xy)^n \, dx \, dy = \sum_{n \ge 0} \int_0^1 \int_0^1 x^n y^n \, dx \, dy \\ + &= \sum_{n \ge 0} \left(\int_0^1 x^n \, dx\right) \left(\int_0^1 y^n \, dy\right) = \sum_{n \ge 0} \frac{1}{n+1} \frac{1}{n+1} \\ + &= \sum_{n \ge 0} \frac{1}{(n+1)^2} = \sum_{n \ge 1} \frac{1}{n^2} = \zeta(2). + \end{align*} + This evaluation also shows that the double integral (over a positive function with a pole at $x=y=1$) is finite. + Note that the computation is also easy and straightforward if we read it backwards + --- thus the evaluation of $\zeta(2)$ leads one to the double integral $I$. + + The second way to evaluate $I$ comes from a change of coordinates: + in the new coordinates given by $u := \frac{y+x}{2}$ and $v := \frac{y-x}{2}$ the domain of integration is a square of side length $\frac{1}{2}\sqrt{2}$, + which we get from the old domain by first rotating it by $45^\circ$ and then shrinking it by a factor of $\sqrt{2}$. + Substitution of $x = u-v$ and $y = u+v$ yields + \[ + \frac{1}{1-xy} = \frac{1}{1-u^2+v^2}. + \] + To transform the integral, we have to replace $dx \, dy$ by $2 \, du \, dv$, to compensate for the fact that our coordinate transformation + reduces areas by a constant factor of $2$ (which is the Jacobi determinant of the transformation). + The new domain of integration, and the function to be integrated, are symmetric with respect to the $u$-axis, + so we just need to compute two times (another factor of $2$ arises here!) the integral over the upper half domain, + which we split into two parts in the most natural way: + \[ + I = 4 \int_0^{1/2} \left(\int_0^u \frac{dv}{1-u^2+v^2}\right) du + 4 \int_{1/2}^1 \left(\int_0^{1-u} \frac{dv}{1-u^2+v^2}\right) du. + \] + Using $\int \frac{dx}{a^2+x^2} = \frac{1}{a} \arctan \frac{x}{a} + C$, this becomes + \begin{align*} + I &= 4 \int_0^{1/2} \frac{1}{\sqrt{1-u^2}} \arctan \left(\frac{u}{\sqrt{1-u^2}}\right) du \\ + &\quad + 4 \int_{1/2}^1 \frac{1}{\sqrt{1-u^2}} \arctan \left(\frac{1-u}{\sqrt{1-u^2}}\right) du. + \end{align*} + These integrals can be simplified and finally evaluated by substituting $u = \sin \theta$ resp. $u = \cos \theta$. + But we proceed more directly, by computing that the derivative of $g(u) := \arctan \left(\frac{u}{\sqrt{1-u^2}}\right)$ is $g'(u) = \frac{1}{\sqrt{1-u^2}}$, + while the derivative of $h(u) := \arctan \left(\frac{1-u}{\sqrt{1-u^2}}\right) = \arctan \left(\sqrt{\frac{1-u}{1+u}}\right)$ is $h'(u) = -\frac{1}{2} \frac{1}{\sqrt{1-u^2}}$. + So we may use $\int_a^b f'(x) f(x) dx = \left[ \frac{1}{2} f(x)^2 \right]_a^b = \frac{1}{2} f(b)^2 - \frac{1}{2} f(a)^2$ and get + \begin{align*} + I &= 4 \int_0^{1/2} g'(u) g(u) du + 4 \int_{1/2}^1 -2 h'(u) h(u) du \\ + &= 2 \left[ g(u)^2 \right]_0^{1/2} - 4 \left[ h(u)^2 \right]_{1/2}^1 \\ + &= 2 g\left(\frac{1}{2}\right)^2 - 2 g(0)^2 - 4 h(1)^2 + 4 h\left(\frac{1}{2}\right)^2 \\ + &= 2 \left(\frac{\pi}{6}\right)^2 - 0 - 0 + 4 \left(\frac{\pi}{6}\right)^2 = \frac{\pi^2}{6}. \qedhere + \end{align*} \end{proof} @@ -18,15 +61,38 @@ \chapter{Four times $π^2/6$} \lean{euler_series'} \leanok \[ - \sum_{k\ge 0}\frac{1}{(2 * k + 1)^2} = \frac{\pi^2}{8} + \sum_{k \ge 0}\frac{1}{(2k+1)^2} = \frac{\pi^2}{8} \] \end{theorem} \begin{proof} - TODO + As above, we may express this as a double integral, namely + \[ + J = \int_0^1 \int_0^1 \frac{1}{1-x^2 y^2} dx \, dy = \sum_{k \ge 0} \frac{1}{(2k+1)^2}. + \] + So we have to compute this integral $J$. + And for this Beukers, Calabi and Kolk proposed the new coordinates + \[ + u := \arccos \sqrt{\frac{1-x^2}{1-x^2 y^2}} \quad v := \arccos \sqrt{\frac{1-y^2}{1-x^2 y^2}}. + \] + To compute the double integral, we may ignore the boundary of the domain, and consider $x, y$ in the range $0 < x < 1$ and $0 < y < 1$. + Then $u, v$ will lie in the triangle $u > 0, v > 0, u+v < \pi/2$. + The coordinate transformation can be inverted explicitly, which leads one to the substitution + \[ + x = \frac{\sin u}{\cos v} \quad \text{and} \quad y = \frac{\sin v}{\cos u}. + \] + It is easy to check that these formulas define a bijective coordinate transformation between the interior of the unit square $S = \{(x, y) : 0 \le x, y \le 1\}$ and the interior of the triangle $T = \{(u, v) : u, v \ge 0, u+v \le \pi/2\}$. + Now we have to compute the Jacobi determinant of the coordinate transformation, and magically it turns out to be + \[ + \det \begin{pmatrix} \frac{\cos u}{\cos v} & \frac{\sin u \sin v}{\cos^2 v} \\ \frac{\sin u \sin v}{\cos^2 u} & \frac{\cos v}{\cos u} \end{pmatrix} = 1 - \frac{\sin^2 u \sin^2 v}{\cos^2 u \cos^2 v} = 1 - x^2 y^2. + \] + But this means that the integral that we want to compute is transformed into + \[ + J = \int_0^{\pi/2} \int_0^{\pi/2-u} 1 dv du, + \] + which is just the area $\frac{1}{2} (\frac{\pi}{2})^2 = \frac{\pi^2}{8}$ of the triangle $T$. \end{proof} - \begin{theorem}[Euler's series: Proof 3] \label{euler_series_3} \lean{euler_series_3} @@ -36,7 +102,87 @@ \chapter{Four times $π^2/6$} \] \end{theorem} \begin{proof} - TODO + The first step is to establish a remarkable relation between values of the (squared) cotangent function. + Namely, for all $m \ge 1$ one has + \begin{equation} \label{eq:cot_sum} + \cot^2 \left(\frac{\pi}{2m+1}\right) + \cot^2 \left(\frac{2\pi}{2m+1}\right) + \dots + \cot^2 \left(\frac{m\pi}{2m+1}\right) = \frac{2m(2m-1)}{6}. + \end{equation} + To establish this, we start with the relation $e^{ix} = \cos x + i \sin x$. + Taking the $n$-th power $e^{inx} = (e^{ix})^n$, we get + \[ + \cos nx + i \sin nx = (\cos x + i \sin x)^n. + \] + The imaginary part of this is + \begin{equation} \label{eq:sin_nx} + \sin nx = \binom{n}{1} \sin x \cos^{n-1} x - \binom{n}{3} \sin^3 x \cos^{n-3} x \pm \cdots + \end{equation} + Now we let $n = 2m+1$, while for $x$ we will consider the $m$ different values $x = \frac{r\pi}{2m+1}$, for $r = 1, 2, \dots, m$. + For each of these values we have $nx = r\pi$, and thus $\sin nx = 0$, while $0 < x < \frac{\pi}{2}$ implies that for $\sin x$ we get $m$ distinct positive values. + + In particular, we can divide \eqref{eq:sin_nx} by $\sin^n x$, which yields + \[ + 0 = \binom{n}{1} \cot^{n-1} x - \binom{n}{3} \cot^{n-3} x \pm \cdots, + \] + that is, + \[ + 0 = \binom{2m+1}{1} \cot^{2m} x - \binom{2m+1}{3} \cot^{2m-2} x \pm \cdots + \] + for each of the $m$ distinct values of $x$. + Thus for the polynomial of degree $m$ + \[ + p(t) := \binom{2m+1}{1} t^m - \binom{2m+1}{3} t^{m-1} \pm \dots + (-1)^m \binom{2m+1}{2m+1} + \] + we know $m$ \emph{distinct} roots + \[ + a_r = \cot^2 \left(\frac{r\pi}{2m+1}\right) \quad \text{for } r = 1, 2, \dots, m. + \] + The roots are distinct because $\cot^2 x = \cot^2 y$ implies $\sin^2 x = \sin^2 y$ + and thus $x = y$ for $x, y \in \{ \frac{r\pi}{2m+1} : 1 \le r \le m \}$. + + Hence the polynomial coincides with + \[ + p(t) = \binom{2m+1}{1} \left(t - \cot^2 \left(\frac{\pi}{2m+1}\right) \right) \dots \left(t - \cot^2 \left(\frac{m\pi}{2m+1}\right) \right). + \] + Comparison of the coefficients of $t^{m-1}$ in $p(t)$ now yields that the sum of the roots is + \[ + a_1 + \dots + a_m = \frac{\binom{2m+1}{3}}{\binom{2m+1}{1}} = \frac{2m(2m-1)}{6}, + \] + which proves \eqref{eq:cot_sum}. + + We also need a second identity, of the same type, + \begin{equation} \label{eq:csc_sum} + \csc^2 \left(\frac{\pi}{2m+1}\right) + \csc^2 \left(\frac{2\pi}{2m+1}\right) + \dots + \csc^2 \left(\frac{m\pi}{2m+1}\right) = \frac{2m(2m+2)}{6}, + \end{equation} + for the cosecant function $\csc x = \frac{1}{\sin x}$. + But + \[ + \csc^2 x = \frac{1}{\sin^2 x} = \frac{\cos^2 x + \sin^2 x}{\sin^2 x} = \cot^2 x + 1, + \] + so we can derive \eqref{eq:csc_sum} from \eqref{eq:cot_sum} by adding $m$ to both sides of the equation. + + Now the stage is set, and everything falls into place. + We use that in the range $0 < y < \frac{\pi}{2}$ we have + \[ + 0 < \sin y < y < \tan y, + \] + and thus + \[ + 0 < \cot y < \frac{1}{y} < \csc y, + \] + which implies + \[ + \cot^2 y < \frac{1}{y^2} < \csc^2 y. + \] + Now we take this double inequality, apply it to each of the $m$ distinct values of $x$, and add the results. + Using \eqref{eq:cot_sum} for the left-hand side, and \eqref{eq:csc_sum} for the right-hand side, we obtain + \[ + \frac{2m(2m-1)}{6} < \left(\frac{2m+1}{\pi}\right)^2 + \left(\frac{2m+1}{2\pi}\right)^2 + \dots + \left(\frac{2m+1}{m\pi}\right)^2 < \frac{2m(2m+2)}{6}, + \] + that is, + \[ + \frac{\pi^2}{6} \frac{2m}{2m+1} \frac{2m-1}{2m+1} < \frac{1}{1^2} + \frac{1}{2^2} + \dots + \frac{1}{m^2} < \frac{\pi^2}{6} \frac{2m}{2m+1} \frac{2m+2}{2m+1}. + \] + Both the left-hand and the right-hand side converge to $\frac{\pi^2}{6}$ for $m \to \infty$: end of proof. \end{proof} @@ -49,15 +195,61 @@ \chapter{Four times $π^2/6$} \] \end{theorem} \begin{proof} - TODO + The first trick in this proof is to consider the Gregory–Leibniz series in doubly-infinite form $\sum_{n=-\infty}^\infty \frac{(-1)^n}{2n+1}$. + As for negative $n = -k < 0$ we get the same terms as for $n = k-1 \ge 0$, since $\frac{(-1)^{-k}}{2(-k)+1} = \frac{(-1)^k}{-(2k-1)} = \frac{(-1)^{k-1}}{2(k-1)+1}$, we infer that $\sum_{n=-N}^N \frac{(-1)^n}{2n+1}$ converges to $\pi/2$ with $N \to \infty$, and thus the square of this sum converges to $\pi^2/4$. + You may write this as + \[ + \lim_{N \to \infty} \sum_{m,n=-N}^N \frac{(-1)^m}{2m+1} \frac{(-1)^n}{2n+1} = \frac{\pi^2}{4}. + \] + The double sum may be interpreted as the sum of all entries of a square matrix of size $(2N+1) \times (2N+1)$, and we know that for $N \to \infty$ this sum of all entries tends to $\pi^2/4$. + We want to know, however, that the sum of only the \emph{diagonal} entries, for $m=n$, also tends to $\pi^2/4$, + \[ + \lim_{N \to \infty} \sum_{n=-N}^N \frac{1}{(2n+1)^2} = \frac{\pi^2}{4}, + \] + because then $\sum_{n=0}^\infty \frac{1}{(2n+1)^2} = \pi^2/8$ will follow, and this, as we know, is equivalent to Euler's theorem. + So let's show that the sum of all off-diagonal terms tends to $0$! + We write $\delta_N$ for this sum, and use a prime to denote that the diagonal terms with $m=n$ are deleted, so + \begin{align*} + \delta_N &= \sideset{}{'}\sum_{m,n=-N}^N \frac{(-1)^{m+n}}{(2m+1)(2n+1)} \\ + &= \sideset{}{'}\sum_{m,n=-N}^N (-1)^{m+n} \left(\frac{1}{2m-2n}\frac{1}{2m+1} - \frac{1}{2m-2n}\frac{1}{2n+1}\right) \\ + &= \sideset{}{'}\sum_{m,n=-N}^N (-1)^{m+n} \left(\frac{1}{2m-2n}\frac{1}{2m+1} - \frac{1}{2n-2m}\frac{1}{2m+1}\right) \\ + &= \sideset{}{'}\sum_{m,n=-N}^N (-1)^{m+n} \frac{1}{m-n} \frac{1}{2m+1} \\ + &= \sum_{m=-N}^N \frac{1}{2m+1} \left(\sideset{}{'}\sum_{n=-N}^N \frac{(-1)^{m-n}}{m-n}\right). + \end{align*} + We only need to show that the terms + \[ + c_{m,N} := \sideset{}{'}\sum_{n=-N}^N \frac{(-1)^{m-n}}{m-n} + \] + are small enough in absolute value. + What do we know about them? + It is easy to see that $c_{-m,N} = -c_{m,N}$, so in particular $c_{0,N} = 0$. + Thus we may assume that $m > 0$, and note that the summands for $n = m+k$ and $n = m-k$ cancel as long as they are in the range between $-N$ and $N$, that is, for $1 \le k \le N-m$. + Thus $c_{m,N}$ equals the alternating sum of fractions of decreasing size given by the remaining terms, where the largest one occurs for $n = m-(N-m)-1 = 2m-N-1$, that is $m-n = N-m+1$. + Hence + \[ + c_{m,N} = (-1)^{N-m+1} \left( \frac{1}{N-m+1} - \frac{1}{N-m+2} \pm \dots \pm \frac{1}{m+N} \right), + \] + which implies that + \[ + |c_{m,N}| \le \frac{1}{N-m+1}. + \] + This finally yields + \begin{align*} + |\delta_N| &\le \sum_{m=-N}^N \left|\frac{1}{2m+1}\right| |c_{m,N}| \le \sum_{m=-N}^N \frac{1}{2|m|-1} |c_{m,N}| \\ + &\le 2 \sum_{m=1}^N \frac{1}{m} |c_{m,N}| \le 2 \sum_{m=1}^N \frac{1}{m} \frac{1}{N-m+1} \\ + &= 2 \sum_{m=1}^N \frac{1}{N+1} \left( \frac{1}{m} + \frac{1}{N-m+1} \right) \\ + &= 2 \frac{1}{N+1} (H_N + H_N) < 4 \frac{\log N + 1}{N+1}, + \end{align*} + and this goes to $0$ as $N$ goes to infinity. \end{proof} \begin{theorem}[Four proofs of Euler's series] \label{four_proofs_euler_series} - Collecting the proofs from the chapter + Collecting the proofs from the chapter. \end{theorem} \begin{proof} \uses{euler_series, euler_series_2, euler_series_3, euler_series_4} + See theorems in this chapter. \end{proof} % Appendix does not contain proofs.. diff --git a/blueprint/src/chapter/chapter10.tex b/blueprint/src/chapter/chapter10.tex index 8178f83..0c93bc8 100644 --- a/blueprint/src/chapter/chapter10.tex +++ b/blueprint/src/chapter/chapter10.tex @@ -39,43 +39,43 @@ \chapter{Hilbert's third problem: decomposing polyhedra} \begin{proof} The name of this lemma stems from the interpretation that the set \[ - C = \{x \in \mathbb{R}^N : Ax = 0, x > 0\} + C = \{\mathbf{x} \in \mathbb{R}^N : A\mathbf{x} = \mathbf{0}, \mathbf{x} > \mathbf{0}\} \] given by an integer matrix \( A \in \mathbb{Z}^{M \times N} \) describes a (relatively open) rational cone. We have to show that if this is nonempty, then it also contains integer points: \( C \cap \mathbb{N}^N \neq \emptyset \). - If \( C \) is nonempty, then so is \(\overline{C} := \{x \in \mathbb{R}^N : Ax = 0, x \geq 1\}\), since + If \( C \) is nonempty, then so is \(\overline{C} := \{\mathbf{x} \in \mathbb{R}^N : A\mathbf{x} = \mathbf{0}, \mathbf{x} \ge \mathbf{1}\}\), since for any positive vector a suitable multiple will have all coordinates equal to or - larger than $1$. (Here $1$ denotes the vector with all coordinates equal to $1$.) + larger than $1$. (Here $\mathbf{1}$ denotes the vector with all coordinates equal to $1$.) It suffices to verify that \( \overline{C} \subseteq C \) contains a point with \emph{rational} coordinates, since then multiplication with a common denominator for all coordinates will yield an integer point in \( \overline{C} \subseteq C \). There are many ways to prove this. We follow a well-trodden path that was first explored by Fourier and Motzkin \([8, \text{Lecture 1}]\): By ``Fourier-Motzkin - elimination" we show that the lexicographically smallest solution to the + elimination'' we show that the lexicographically smallest solution to the system \[ - Ax = 0, x \geq 1 + A\mathbf{x} = \mathbf{0}, \mathbf{x} \ge \mathbf{1} \] exists, and that it is rational if the matrix \( A \) is integral. - Indeed, any linear equation \( a^T x = 0 \) can be equivalently enforced by two - inequalities \( a^T x \geq 0, -a^T x \geq 0 \). (Here \( a \) denotes a column vector and - \( a^T \) its transpose.) Thus it suffices to prove that any system of the type + Indeed, any linear equation \( \mathbf{a}^T \mathbf{x} = 0 \) can be equivalently enforced by two + inequalities \( \mathbf{a}^T \mathbf{x} \ge 0, -\mathbf{a}^T \mathbf{x} \ge 0 \). (Here $\mathbf{a}$ denotes a column vector and + $\mathbf{a}^T$ its transpose.) Thus it suffices to prove that any system of the type \[ - Ax \geq b, x \geq 1 + A\mathbf{x} \ge \mathbf{b}, \mathbf{x} \ge \mathbf{1} \] - with integral \( A \) and \( b \) has a lexicographically smallest solution, which is + with integral \( A \) and \( \mathbf{b} \) has a lexicographically smallest solution, which is rational, provided that the system has any real solution at all. For this we argue with induction on \( N \). The case \( N = 1 \) is clear. For \( N > 1 \) - look at all the inequalities that involve \( x_N \). If \( x' = (x_1, \ldots, x_{N-1}) \) is fixed, + look at all the inequalities that involve \( x_N \). If \( \mathbf{x}' = (x_1, \ldots, x_{N-1}) \) is fixed, these inequalities give lower bounds on \( x_N \) (among them \( x_N \geq 1 \)) and - possibly also upper bounds. So we form a new system \( A' x' \geq b \), \( x' \geq 1 \) + possibly also upper bounds. So we form a new system \( A' \mathbf{x}' \geq \mathbf{b} \), \( \mathbf{x}' \geq 1 \) in \( N-1 \) variables, which contains all the inequalities from the system - \( Ax \geq b \) that do not involve \( x_N \), as well as all the inequalities obtained + \( A\mathbf{x} \geq \mathbf{b} \) that do not involve \( x_N \), as well as all the inequalities obtained by requiring that all upper bounds on \( x_N \) (if there are any) are larger or equal to all the lower bounds on \( x_N \) (which include \( x_N \geq 1 \)). This system in \( N-1 \) variables has a solution, and thus by induction it has a lexicographically diff --git a/blueprint/src/macros/common.tex b/blueprint/src/macros/common.tex index 44f13f0..8084142 100644 --- a/blueprint/src/macros/common.tex +++ b/blueprint/src/macros/common.tex @@ -17,12 +17,15 @@ \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} -\newcommand{\Od}{\operatorname{Od}} -\newcommand{\conv}{\operatorname{conv}} -\newcommand{\per}{\operatorname{per}} +\DeclareMathOperator{\Od}{Od} +\DeclareMathOperator{\conv}{conv} +\DeclareMathOperator{\per}{per} % \cap is already taken, so we name the command `\mycap` instead -\newcommand{\mycap}{\operatorname{cap}} -\newcommand{\supp}{\operatorname{supp}} -\newcommand{\Prop}{\operatorname{Prop}} +\DeclareMathOperator{\mycap}{cap} +\DeclareMathOperator{\supp}{supp} +\DeclareMathOperator{\Prop}{Prop} % \cr is already taken, so we name the command `\mycr` instead -\newcommand{\mycr}{\operatorname{cr}} +\DeclareMathOperator{\mycr}{cr} +\DeclareMathOperator{\ord}{ord} +\DeclareMathOperator{\trace}{trace} +\DeclareMathOperator{\sign}{sign} diff --git a/blueprint/src/print.pdf b/blueprint/src/print.pdf new file mode 100644 index 0000000..cdefad3 Binary files /dev/null and b/blueprint/src/print.pdf differ