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23. The Farey series $F_{n}$ of order $n$ is the set of fractions $h / k$ where $h$ and $k$ are integers, $0 \leqslant h \leqslant k \leqslant n$, and $(h, k)=1$, in ascending order. Here, we include 0 and 1 in the forms $\frac{0}{1}$ and $\frac{1}{1}$ respectively. For instance, the Farey series of order 4 is
$$\frac{... | 23. a) $\frac{0}{1}, \frac{1}{7}, \frac{1}{6}, \frac{1}{5}, \frac{1}{4}, \frac{2}{7}, \frac{1}{3}, \frac{2}{5}, \frac{3}{7}, \frac{1}{2}, \frac{4}{7}, \frac{3}{5}, \frac{2}{3}, \frac{5}{7}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}, \frac{1}{1}$ | not found | Number Theory | proof | Yes | Yes | number_theory | false | 739,281 |
Theorem 10.7. Every finite simple continued fraction represents a rational number. | Proof. We will prove the theorem using mathematical induction. For $n=1$ we have
$$\left[a_{o} ; a_{1}\right]=a_{0}+\frac{1}{a_{1}}=\frac{a_{0} a_{1}+1}{a_{0}}$$
which is rational. Now assume that for the positive integer $k$ the simple continued fraction $\left[a_{0} ; a_{1}, a_{2}, \ldots, a_{k}\right.$ ] is rational... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,283 |
Theorem 10.8. Every rational number can be expressed by a finite simple continued fraction. | Proof. Let $x=a / b$ where $a$ and $b$ are integers with $b>0$. Let $r_{0}=a$ and $r_{1}=b$. Then the Euclidean algorithm produces the following sequence of equations: $\square$
$$\begin{array}{rlrl}
r_{0} & =r_{1} q_{1}+r_{2} & & 0<r_{2}<r_{1} \\
r_{1} & =r_{2} q_{2}+r_{3} & 0<r_{3}<r_{2} \\
r_{2} & =r_{3} q_{3}+r_{4}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,284 |
Theorem 10.9. Let $a_{0}, a_{1}, a_{2}, \ldots, a_{n}$ be real numbers, with $a_{1}, a_{2}, \ldots, a_{n}$ positive. Let the sequences $p_{0}, p_{1}, \ldots, p_{n}$ and $q_{0}, q_{1}, \ldots, q_{n}$ be defined recursively by
$$\begin{array}{ll}
p_{0}=a_{0} & q_{0}=1 \\
p_{1}=a_{0} a_{1}+1 & q_{1}=a_{1}
\end{array}$$
an... | Proof. We will prove this theorem using mathematical induction. For $k=0$ we have
$$C_{0}=\left[a_{0}\right]=a_{0} / 1=p_{0} / q_{0}$$
For $k=1$, we see that
$$C_{1}=\left[a_{0} ; a_{1}\right]=a_{0}+\frac{1}{a_{1}}=\frac{a_{0} a_{1}+1}{a_{1}}=\frac{p_{1}}{q_{1}} .$$
Hence, the theorem is valid for $k=0$ and $k=1$.
No... | proof | Algebra | proof | Yes | Yes | number_theory | false | 739,285 |
Corollary 1.2. If $f_{1}$ and $f_{2}$ are $O(g)$, then $f_{1}+f_{2}$ is $O(g)$. | Proof. Proposition 1.7 tells us that $f_{1}+f_{2}$ is $O(2 g)$. But if $f_{1}+f_{2} \leqslant K(2 g)$, then $f_{1}+f_{2} \leqslant(2 K) g$, so that $f_{1}+f_{2}$ is $O(g)$ | proof | Algebra | proof | Yes | Yes | number_theory | false | 739,286 |
Theorem 10.10. Let $k$ be a positive integer, $k \geqslant 1$. Let the $k$ th convergent of the continued fraction $\left[a_{0} ; a_{1}, \ldots, a_{n}\right]$ be $C_{k}=p_{k} / q_{k}$, where $p_{k}$ and $q_{k}$ are as defined in Theorem 10.9. Then
$$p_{k} q_{k-1}-p_{k-1} q_{k}=(-1)^{k-1}$$ | Proof. We use mathematical induction to prove the theorem. For $k=1$ we have
$$p_{1} q_{0}-p_{0} q_{1}=\left(a_{0} a_{1}+1\right) \cdot 1-a_{0} a_{1}=1$$
Assume the theorem is true for an integer $k$ where $1 \leqslant k<n$, so that
$$p_{k} q_{k-1}-p_{k-1} q_{k}=(-1)^{k-1}$$
Then, we have
$$\begin{aligned}
p_{k+1} q_... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,287 |
Corollary 10.1. Let $C_{k}=p_{k} / q_{k}$ be the $k$ th convergent of the simple continued fraction $\left[a_{0} ; a_{1}, \ldots, a_{n}\right]$, where the integers $p_{k}$ and $q_{k}$ are as defined in Theorem 10.9. Then the integers $p_{k}$ and $q_{k}$ are relatively prime. | Proof. Let $d=\left(p_{k}, q_{k}\right)$. From Theorem 10.10 , we know that
$$p_{k} q_{k-1}-q_{k} p_{k-1}=(-1)^{k-1}$$
Hence, from Proposition 1.2 we have
$$d \mid(-1)^{k-1}$$
Therefore, $d=1$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,288 |
Corollary 10.2. Let $C_{k}=p_{k} / q_{k}$ be the $k$ th convergent of the simple continued fraction $\left[a_{0} ; a_{1}, a_{2}, \ldots, a_{k}\right]$. Then
$$C_{k}-C_{k-1}=\frac{(-1)^{k-1}}{q_{k} q_{k-1}}$$
for all integers $k$ with $1 \leqslant k \leqslant n$. Also,
$$C_{k}-C_{k-2}=\frac{a_{k}(-1)^{k}}{q_{k} q_{k-2}}... | Proof. From Theorem 10.10 we know that $p_{k} q_{k-1}-q_{k} p_{k-1}=(-1)^{k-1}$.
We obtain the first identity,
$$C_{k}-C_{k-1}=\frac{p_{k}}{q_{k}}-\frac{p_{k-1}}{q_{k-1}}=\frac{(-1)^{k-1}}{q_{k} q_{k-1}}$$
by dividing both sides by $q_{k} q_{k-1}$.
To obtain the second identity, note that
$$C_{k}-C_{k-2}=\frac{p_{k}}{q... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,289 |
1. Find the rational number, expressed in lowest terms, represented by each of the following simple continued fractions
a) $[2 ; 7]$
e) $[1 ; 1]$
b) $[1 ; 2,3]$
f) $[1 ; 1,1]$
c) $[0 ; 5,6]$
g) $[1 ; 1,1,1]$
d) $[3 ; 7,15,1]$
h) $[1 ; 1,1,1,1]$. | 1. a) $15 / 7$
b) $10 / 7$
c) $6 / 31$
d) $355 / 113$
e) 2
f) $3 / 2$
g) $5 / 3$
h) $8 / 5$ | a) \frac{15}{7}, b) \frac{10}{7}, c) \frac{6}{31}, d) \frac{355}{113}, e) 2, f) \frac{3}{2}, g) \frac{5}{3}, h) \frac{8}{5} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,291 |
2. Find the simple continued fraction expansion not terminating with the partial quotient one, of each of the following rational numbers
a) $6 / 5$
d) 5/999
b) $22 / 7$
e) $-43 / 1001$
c) $19 / 29$
f) $873 / 4867$. | 2. a) $[1 ; 5]$ b) $[3 ; 7]$ c) $[0 ; 1,1,1,9]$
d) $[0 ; 199,1,4]$
f) $[0 ; 5,1,1,2,1,4,1,21]$
e) $[-1 ; 1,22,3,1,1,2,2]$ | a) [1 ; 5] \quad b) [3 ; 7] \quad c) [0 ; 1,1,1,9] \quad d) [0 ; 199,1,4] \quad e) [-1 ; 1,22,3,1,1,2,2] \quad f) [0 ; 5,1,1,2,1,4,1,21 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,292 |
Theorem 1.4. Multiplication of two $n$-bit integers can be performed using $O\left(n^{\log _{2} 3}\right)$ bit operations. (Note: $\log _{2} 3$ is approximately 1.585 , which is considerably less than the exponent 2 that occurs in the estimate of the number of bit operations needed for the conventional multiplication a... | Proof. From (1.8) we have
$$\begin{aligned}
M(n)= & M\left(2^{\log _{2} n}\right) \leqslant M\left(2^{\left[\log _{2} n\right]+1}\right) \\
& \leqslant c\left(3^{\left[\log _{2} n\right]+1}-2^{\left[\log _{2} n\right]+1}\right) \\
& \leqslant 3 c \cdot 3^{\left[\log _{2} n\right]} \leqslant 3 c \cdot 3^{\log _{2} n}=3 ... | proof | Algebra | proof | Yes | Yes | number_theory | false | 739,297 |
Theorem 10.13. Let $a_{0}, a_{1}, a_{2}, \ldots$ be an infinite sequence of integers with $a_{1}, a_{2}, \ldots$ positive, and let $C_{k}=\left[a_{0} ; a_{1}, a_{2}, \ldots, a_{k}\right]$. Then the convergents $C_{k}$ tend to a limit $\alpha$, i.e
$$\lim _{k \rightarrow \infty} C_{k}=\alpha$$ | Proof. Let $m$ be an even positive integer. From Theorem 10.11 , we see that
$$\begin{array}{l}
C_{1}>C_{3}>C_{5}>\cdots>C_{m-1} \\
C_{0}C_{2 k+1}$ whenever $2 j \leqslant m$ and $2 k+1C_{3}>C_{5}>\cdots>C_{2 n-1}>C_{2 n+1}>\cdots \\
C_{0}C_{2 k+1}$ for all positive integers $j$ and $k$. We see that the hypotheses of T... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,304 |
Theorem 10.14. Let $a_{0}, a_{1}, a_{2}, \ldots$ be integers with $a_{1}, a_{2}, \ldots$ positive. Then $\left[a_{0} ; a_{1}, a_{2}, \ldots\right]$ is irrational. | Proof. Let $\alpha=\left[a_{0} ; a_{1}, a_{2}, \ldots\right]$ and let
$$C_{k}=p_{k} / q_{k}=\left[a_{0} ; a_{1}, \ldots, a_{k}\right]$$
denote the $k$ th convergent of $\alpha$. When $n$ is a positive integer, Theorem 10.1 shows that $C_{2 n}2 n+1$, there is an integer $n$ such that $q_{2 n+1}>b$, so that $b / q_{2 n+1... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,305 |
Theorem 10.15. Let $\alpha=\alpha_{0}$ be an irrational number and define the sequence $a_{0}, a_{1}, a_{2}, \ldots$ recursively by
$$a_{k}=\left[\alpha_{k}\right], \alpha_{k+1}=1 /\left(\alpha_{k}-a_{k}\right)$$
for $k=0,1,2, \ldots$. Then $\alpha$ is the value of the infinite, simple continued fraction $\left[a_{0} ;... | Proof. From the recursive definition given above, we see that $a_{k}$ is an integer for every $k$. Further, we can easily show using mathematical induction that $\alpha_{k}$ is irrational for every $k$. We first note that $\alpha_{0}=\alpha$ is irrational. Next, if we assume that $\alpha_{k}$ is irrational, then we can... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,306 |
Theorem 10.16. If the two infinite simple continued fractions $\left[a_{0} ; a_{1}, a_{2}, \ldots\right]$ and $\left[b_{0} ; b_{1}, b_{2}, \ldots\right]$ represents the same irrational number, then $a_{k}=b_{k}$ for $k=0,1,2, \ldots$ | Proof. Suppose that $\alpha=\left[a_{0} ; a_{1}, a_{2}, \ldots\right]$. Then, since $C_{0}=a_{0}$ and $C_{1}=a_{0}+1 / a_{1}$, Theorem 10.11 tells us that
$$a_{0}<\alpha<a_{0}+1 / a_{1}$$
so that $a_{0}=[\alpha]$. Further, we note that
$$\left[a_{0} ; a_{1}, a_{2}, \ldots\right]=a_{0}+\frac{1}{\left[a_{1} ; a_{2}, a_{3... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,307 |
Theorem 10.17. Let $\alpha$ be an irrational number and let $p_{j} / q_{j}, j=1,2, \ldots$, be the convergents of the infinite simple continued fraction of $\alpha$. If $r$ and $s$ are integers with $s>0$ such that
$$|s \alpha-r|<\left|q_{k} \alpha-p_{k}\right|$$
then $s \geqslant q_{k+1}$. | Proof. Assume that $|s \alpha-r|0$, because $q_{k} x>0$ and $q_{k}>0$. When $y>0$, since $q_{k+1} y \geqslant q_{k+1}>s$, we see that $q_{k} x=s-q_{k+1} y<0$, so that $x<0$
From Theorem 10.11, we know that either $p_{k} / q_{k}<\alpha<p_{k+1} / q_{k+1}$ or that $p_{k+1} / q_{k+1}<\alpha<p_{k} / q_{k}$. In either case,... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,309 |
1. Find the simple continued fractions of the following real numbers
a) $\sqrt{2}$
c) $\sqrt{5}$
b) $\sqrt{3}$
d) $\frac{1+\sqrt{5}}{2}$. | 1. a) $[1 ; 2,2,2, \ldots]$
b) $[1 ; 1,2,1,2,1,2, \ldots]$
c) $[2 ; 4,4,4, \ldots]$
d) $[1 ; 1,1,1, \ldots]$ | a) [1 ; 2,2,2, \ldots] \\
b) [1 ; 1,2,1,2,1,2, \ldots] \\
c) [2 ; 4,4,4, \ldots] \\
d) [1 ; 1,1,1, \ldots] | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,312 |
2. Find the first five partial quotients of the simple continued fractions of the following real numbers
a) $\sqrt[3]{2}$
c) $(\mathrm{e}-1) /(\mathrm{e}+1)$
b) $2 \pi$
d) $\left(e^{2}-1\right) /\left(e^{2}+1\right)$. | 2. a) $1,3,1,5,1$
b) $6,3,1,1,7$
c) $0,2,6,10,14$
d) $0,1,3,5,7$ | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,313 |
3. Find the best rational approximation to $\pi$ with a denominator less than 10000 . | 3. $\frac{312689}{99532}$ | \frac{312689}{99532} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,314 |
4. The infinite simple continued fraction expansion of the number $e$ is
$$e=[2 ; 1,2,1,1,4,1,1,6,1,1,8, \ldots]$$
a) Find the first eight convergents of the continued fraction of $e$.
b) Find the best rational approximation to $e$ having a denominator less than 100. | 4. a) $\frac{2}{1}, \frac{3}{1}, \frac{8}{3}, \frac{11}{4}, \frac{19}{7}, \frac{87}{32}, \frac{106}{39}, \frac{193}{71}$
b) $\frac{193}{71}$ | \frac{193}{71} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,315 |
11. Let $\alpha$ be an irrational number, and let the simple continued fraction expansion of $\alpha$ be $\alpha=\left[a_{0} ; a_{1}, a_{2}, \ldots\right]$. Let $p_{k} / q_{k}$ denote, as usual, the $k$ th convergent of this continued fraction. We define the pseudoconvergnts of this continued fraction to be
$$p_{k, t} ... | 11. d) $\frac{25}{8}, \frac{47}{15}, \frac{69}{22}, \frac{91}{29}, \frac{113}{36}, \frac{135}{43}, \frac{157}{50}, \frac{179}{57}, \frac{201}{64}, \frac{223}{71}, \frac{245}{78}, \frac{267}{85}, \frac{289}{92}, \frac{311}{99}$ | \frac{311}{99} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,323 |
Lemma 10.1. The real number $\alpha$ is a quadratic irrational if and only if there are integers $a, b$, and $c$ with $b>0$ and $c \neq 0$, such that $b$ is not a perfect square and
$$\alpha=(a+\sqrt{b}) / c$$ | Proof. If $\alpha$ is a quadratic irrational, then $\alpha$ is irrational, and there are integers $A, B$, and $C$ such that $A \alpha^{2}+B \alpha+C=0$. From the quadratic formula, we know that
$$\alpha=\frac{-B \pm \sqrt{B^{2}-4 A C}}{2 A}$$
Since $\alpha$ is a real number, we have $B^{2}-4 A C>0$, and since $\alpha$... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,324 |
Lemma 10.2. If $\alpha$ is a quadratic irrational and if $r, s, t$, and $u$ are integers, then $(r \alpha+s) /(t \alpha+u)$ is either rational or a quadratic irrational. | Proof. From Lemma 10.1, there are integers $a, b$, and $c$ with $b>0, c \neq 0$, and $b$ not a perfect square such that
$$\alpha=(a+\sqrt{b}) / c$$
Thus
$$\begin{aligned}
\frac{r \alpha+s}{t \alpha+u} & =\left[\frac{r(a+\sqrt{b})}{c}+s\right] /\left[\frac{t(a+\sqrt{b})}{c}+u\right] \\
& =\frac{(a r+c s)+r \sqrt{b}}{(a... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,325 |
Lemma 10.3. If the quadratic irrational $\alpha$ is a root of the polynomial $A x^{2}+B x+C=0$, then the other root of this polynomial is $\alpha^{\prime}$, the conjugate of $\alpha$. | Proof. From the quadratic formula, we see that the two roots of $A x^{2}+B x+C=0$ are
$$\frac{-B \pm \sqrt{B^{2}-4 A C}}{2 A}$$
If $\alpha$ is one of these roots, then $\alpha^{\prime}$ is the other root, because the sign of $\sqrt{B^{2}-4 A C}$ is reversed to obtain $\alpha^{\prime}$ from $\alpha$. | proof | Algebra | proof | Yes | Yes | number_theory | false | 739,326 |
Lemma 10.4. If $\alpha_{1}=\left(a_{1}+b_{1} \sqrt{d}\right) / c_{1}$ and $\alpha_{2}=\left(a_{2}+b_{2} \sqrt{d}\right) / c_{2}$ are quadratic irrationals, then
(i) $\left(\alpha_{1}+\alpha_{2}\right)^{\prime}=\alpha_{1}^{\prime}+\alpha_{2}^{\prime}$
(ii) $\left(\alpha_{1}-\alpha_{2}\right)^{\prime}=\alpha_{1}^{\prime}... | Proof of (iv). Note that
$$\begin{aligned}
\alpha_{1} / \alpha_{2} & =\frac{\left(a_{1}+b_{1} \sqrt{d}\right) / c_{1}}{\left(a_{2}+b_{2} \sqrt{d}\right) / c_{2}} \\
& =\frac{c_{2}\left(a_{1}+b_{1} \sqrt{d}\right)\left(a_{2}-b_{2} \sqrt{d}\right)}{c_{1}\left(a_{2}+b_{2} \sqrt{d}\right)\left(a_{2}-b_{2} \sqrt{d}\right)} ... | proof | Algebra | proof | Yes | Yes | number_theory | false | 739,327 |
Lagrange's Theorem. The infinite simple continued fraction of an irrational number is periodic if and only if this number is a quadratic irrational. | Proof. Let the simple continued fraction of $\alpha$ be periodic, so that
$$\alpha=\left[a_{0} ; a_{1}, a_{2}, \ldots, a_{N-1}, \overline{a_{N}, a_{N+1}, \ldots, a_{N+k}}\right]$$
Now let
$$\beta=\left[\overline{a_{N} ; a_{N+1}, \ldots, a_{N+k}}\right]$$
Then
$$\beta=\left[a_{N} ; a_{N+1}, \ldots, a_{N+k}, \beta\righ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,328 |
Lemma 10.5. If $\alpha$ is a quadratic irrational, then $\alpha$ can be written as
$$\alpha=(P+\sqrt{d}) / Q$$
where $P, Q$, and $d$ are integers, $Q \neq 0, d>0, d$ is not a perfect square, and $Q \mid\left(d-P^{2}\right)$ | Proof. Since $\alpha$ is a quadratic irrational, Lemma 10.1 tells us that
$$\alpha=(a+\sqrt{b}) / c$$
where $a, b$, and $c$ are integers, $b>0$, and $c \neq 0$. We multiply both the numerator and denominator of this expression for $\alpha$ by $|c|$ to obtain
$$\alpha=\frac{a|c|+\sqrt{b c^{2}}}{c|c|}$$
(where we have us... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,329 |
Theorem 10.20 . The simple continued fraction of the quadratic irrational $\alpha$ is purely periodic if and only if $\alpha$ is reduced. Further, if $\alpha$ is reduced and $\alpha=\left[\overline{a_{0} ; a_{1}, a_{2}, \ldots, a_{n}}\right]$ then the continued fraction of $-1 / \alpha^{\prime}$ is $\left[\overline{a_{... | Proof. First, assume that $\alpha$ is a reduced quadratic irrational. Recall from Theorem 10.15 that the partial fractions of the simple continued fraction of $\alpha$ are given by
$$a_{k}=\left[\alpha_{k}\right], \alpha_{k+1}=1 /\left(\alpha_{k}-a_{k}\right)$$
for $k=0,1,2, \ldots$, where $\alpha_{0}=\alpha$. We see t... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,332 |
1. Find the simple continued fractions of
a) $\sqrt{7}$
d) $\sqrt{47}$
b) $\sqrt{11}$
e) $\sqrt{59}$
c) $\sqrt{23}$
f) $\sqrt{94}$. | 1. a) $[2 ; \overline{1,1,1,4}]$
b) $[3 ; \overline{3,6}]$
c) $[4 ; \overline{1,3,1,8}]$
d) $[6 ; \overline{1,5,1,12}]$ | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,333 |
2. Find the simple continued fractions of
a) $(1+\sqrt{3}) / 2$
b) $(14+\sqrt{37}) / 3$
c) $(13-\sqrt{2}) 7$. | 2. a) $[1 ; 2]$ | 1 ; 2 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,334 |
3. Find the quadratic irrational with simple continued fraction expansion
a) $[2 ; 1, \overline{5}]$
b) $[2 ; \overline{1,5}]$
c) $[\overline{2 ; 1,5}]$. | 3. a) $(23 \pm \sqrt{29}) / 10$
b) $(-1+\sqrt{45}) / 2$
c) $(8+\sqrt{82}) / 6$ | a) \frac{23 \pm \sqrt{29}}{10}, b) \frac{-1 + \sqrt{45}}{2}, c) \frac{8 + \sqrt{82}}{6} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,335 |
4. a) Let $d$ be a positive integer. Show that the simple continued fraction of $\sqrt{d^{2}+1}$ is $[d ; 2 d]$.
b) Use part (a) to find the simple continued fractions of $\sqrt{101}, \sqrt{290}$, and $\sqrt{2210}$. | 4. b) $[10 ; \overline{20}],[17 ; \overline{34}],[47 ; \overline{94}]$ | [10 ; \overline{20}],[17 ; \overline{34}],[47 ; \overline{94}] | Number Theory | proof | Yes | Yes | number_theory | false | 739,336 |
5. Let d be a integer, $d \geqslant 2$.
a) Show that the simple continued fraction of $\sqrt{d^{2}-1}$ is $[d-1 ; \overline{1,2 d-2}]$.
b) Show that the simple continued fraction of $\sqrt{d^{2}-d}$ is $[d-1 ; \overline{2,2 d-2}]$.
c) Use parts (a) and (b) to find the simple continued fractions of $\sqrt{99}, \sqrt{110... | 5. с) $[9 ; \overline{1,18}],[10 ; \overline{2,20}],[16 ; \overline{2,32}],[24 ; \overline{2,48}]$ | [9 ; \overline{1,18}],[10 ; \overline{2,20}],[16 ; \overline{2,32}],[24 ; \overline{2,48}] | Number Theory | proof | Yes | Yes | number_theory | false | 739,337 |
6. a) Show that if d is an integer, $d \geqslant 3$, then the simple continued fraction of $\sqrt{d^{2}-2}$ is $[d-1 ; \overline{1, d-2,1,2 d-2}]$.
b) Show that if $d$ is a positive integer, then the simple continued fraction of $\sqrt{d^{2}+2}$ is $[d ; \overline{d, 2 d}]$.
c) Find the simple continued fraction expans... | 6. c) $[6 ; 1,5,1,12],[7 ; 7,14],[16 ; 1,15,1,32]$ | [6 ; 1,5,1,12],[7 ; 7,14],[16 ; 1,15,1,32] | Number Theory | proof | Yes | Yes | number_theory | false | 739,338 |
1. Add $(101111011)_{2}$ and $(1100111011)_{2}$. | 1. $(10010110110)_{2}$ | (10010110110)_{2} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,341 |
Lemma 11.1. If $x, y, z$ is a primitive Pythagorean triple, then $(x, y)=(x, z)=(y, z)=1$ | Proof. Suppose $x, y, z$ is a primitive Pythagorean triple and $(x, y)>1$. Then, there is a prime $p$ such that $p \mid(x, y)$, so that $p \mid x$ and $p \mid y$. Since $p \mid x$ and $p \mid y$, we know that $p \mid\left(x^{2}+y^{2}\right)=z^{2}$. Because $p \mid z^{2}$, we can conclude that $p \mid z$ (using problem ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,349 |
Lemma 11.2. If $x, y, z$ is a primitive Pythagorean triple, then $x$ is even and $y$ is odd or $x$ is odd and $y$ is even. | Proof. Let $x, y, z$ be a primitive Pythagorean triple. By Lemma 11.1 , we know that $(x, y)=1$, so that $x$ and $y$ cannot both be even. Also $x$ and $y$ cannot both be odd. If $x$ and $y$ were both odd, then (from problem 2 of Section 2.1) we would have
$$x^{2} \equiv y^{2} \equiv 1(\bmod 4)$$
so that
$$z^{2}=x^{2}+y... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,350 |
Lemma 11.3. If $r, s$, and $t$ are positive integers such that $(r, s)=1$ and $r s=t^{2}$, then there are integers $m$ and $n$ such that $r=m^{2}$ and $s=n^{2}$ | Proof. If $r=1$ or $s=1$, then the lemma is obviously true, so we may suppose that $r>1$ and $s>1$. Let the prime-power factorizations of $r, s$, and $t$ be
$$\begin{array}{l}
r=p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots p_{u}^{a_{*}} \\
s=p_{u+1}^{a_{u+1}} p_{u+2}^{a_{*+2}} \cdots \cdots p_{v}^{a_{v}^{v}}
\end{array}$$
and
$$... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,351 |
Theorem 11.1. The positive integers $x, y, z$ form a primitive Pythagorean triple, with $y$ even, if and only if there are relatively prime positive integers $m$ and $n, m>n$, with $m$ odd and $n$ even or $m$ even and $n$ odd, such that
$$\begin{array}{l}
x=m^{2}-n^{2} \\
y=2 m n \\
z=m^{2}+n^{2}
\end{array}$$ | Proof. Let $x, y, z$ be a primitive Pythagorean triple. Lemma 11.2 tells us that $x$ is odd and $y$ is even, or vice versa. Since we have assumed that $y$ is even, $x$ and $z$ are both odd. Hence, $z+x$ and $z-x$ are both even, so that there are positive integers $r$ and $s$ with $r=(z+x) / 2$ and $s=(z-x) / 2$.
Since... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,353 |
1. Find all
a) primitive Pythagorean triples $x, y, z$ with $z \leqslant 40$.
b) Pythagorean triples $x, y, z$ with $z \leqslant 40$. | $$\begin{array}{l}
\text { 1. а) } 3,4,5 ; 5,12,13 ; 15,8,17 ; 7,24,25 ; 21,20,29 ; 35,12,37 \text { b) } 3,4,5 ; 6,8,10 ; 5,12,13 ; 9 \text {, } \\
12,15 ; 15,8,17 ; 12,16,20 ; 7,24,25 ; 15,20,25 ; 10,24,26 ; 21,20,29 ; 18,24,30 ; 30,16,34 ; \\
21,28,35 ; 35,12,37 ; 15,36,39 ; 24,32,40
\end{array}$$ | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,354 |
8. Find all solutions in positive integers of the diophantine equation $x^{2}+2 y^{2}=z^{2}$. | 8. $x=\frac{1}{2}\left(m^{2}-2 n^{2}\right), y=m n, z=\frac{1}{2}\left(m^{2}+2 n^{2}\right)$ where $m$ and $n$ are positive integers, $x=\frac{1}{2}\left(2 m^{2}-n^{2}\right), y=m n, z=\frac{1}{2}\left(2 m^{2}+n^{2}\right)$ where $m$ and $n$ are positive integers, $m>n / \sqrt{2}$, and $n$ is even | x=\frac{1}{2}\left(m^{2}-2 n^{2}\right), y=m n, z=\frac{1}{2}\left(m^{2}+2 n^{2}\right) \text{ or } x=\frac{1}{2}\left(2 m^{2}-n^{2}\right), y=m n, z=\frac{1}{2}\left(2 m^{2}+n^{2}\right) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,361 |
9. Find all solutions in positive integers of the diophantine equation $x^{2}+3 y^{2}=z^{2}$. | 9. $x=\frac{1}{2}\left(m^{2}-3 n^{2}\right), y=m n, z=\frac{1}{2}\left(m^{2}+3 n^{2}\right)$ where $m$ and $n$ are positive integers, $m>\sqrt{3} n$, and $m \equiv n(\bmod 2)$ | x=\frac{1}{2}\left(m^{2}-3 n^{2}\right), y=m n, z=\frac{1}{2}\left(m^{2}+3 n^{2}\right) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,362 |
3. Multiply $(11101)_{2}$ and $(110001)_{2}$. | 3. $(10110001101)_{2}$ | (10110001101)_{2} | Algebra | math-word-problem | Yes | Yes | number_theory | false | 739,364 |
Theorem 11.2. The diophantine equation
$$x^{4}+y^{4}=z^{2}$$
has no solutions in nonzero integers $x, y, z$. | Proof. Assume that the above equation has a solution in nonzero integers $x, y, z$. Since we may replace any number of the variables with their negatives without changing the validity of the equation, we may assume that $x, y, z$ are positive integers.
We may also suppose that $(x, y)=1$. To see this, let $(x, y)=d$. ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,372 |
4. Find the quotient and remainder when $(110100111)_{2}$ is divided by $(11101)_{2}$. | 4. $(1110)_{2},(10001)_{2}$ | (1110)_{2},(10001)_{2} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,375 |
Theorem 11.3. Let $d$ and $n$ be integers such that $d>0, d$ is not a perfect square, and $|n|<\sqrt{d}$. If $x^{2}-d y^{2}=n$, then $x / y$ is a convergent of the simple continued fraction of $\sqrt{d}$. | Proof. First consider the case where $n>0$. Since $x^{2}-d y^{2}=n$, we see that $(11.3)$
$$(x+y \sqrt{d})(x-y \sqrt{d})=n$$
From (11.3), we see that $x-y \sqrt{d}>0$, so that $x>y \sqrt{d}$. Consequently,
$$\frac{x}{y}-\sqrt{d}>0$$
and since $00$, we see that $y / x$ is a convergent of the simple continued fraction e... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,384 |
Theorem 11.4. Let $d$ be a positive integer that is not a perfect square. Define $\quad \alpha_{k}=\left(P_{k}+\sqrt{d}\right) / Q_{k}, \quad a_{k}=\left[\alpha_{k}\right], \quad P_{k+1}=a_{k} Q_{k}-P_{k}, \quad$ and $Q_{k+1}=\left(d-P_{k+1}^{2}\right) / Q_{k}, \quad$ for $k=0,1,2, \ldots$ where $\alpha_{0}=\sqrt{d}$. ... | Proof. Since $\sqrt{d}=\alpha_{0}=\left[a_{0} ; a_{1}, a_{2}, \ldots, a_{k}, \alpha_{k+1}\right]$, Theorem 10.9 tells us that
$$\sqrt{d}=\frac{\alpha_{k+1} p_{k}+p_{k-1}}{\alpha_{k+1} q_{k}+q_{k-1}}$$
Since $\alpha_{k+1}=\left(P_{k+1}+\sqrt{d}\right) / Q_{k+1}$ we have
$$\sqrt{d}=\frac{\left(P_{k+1}+\sqrt{d}\right) p_... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,385 |
Lemma 11.4. Let $r+s \sqrt{d}=t+u \sqrt{d}$ where $r, s, t$, and $u$ are rational numbers and $d$ is a positive integer that is not a perfect square. Then $r=t$ and $s=u$ | Proof. Since $r+s \sqrt{d}=t+u \sqrt{d}$, we see that if $s \neq u$ then
$$\sqrt{d}=\frac{r-t}{u-s} .$$
By Theorem 10.1, $(r-t) /(u-s)$ is rational, and by Theorem $10.2 \quad \sqrt{d}$ is irrational. Hence, $s=u$, and consequently $r=t$. | proof | Algebra | proof | Yes | Yes | number_theory | false | 739,387 |
Theorem 11.6. Let $x_{1}, y_{1}$ be the least positive solution of the diophantine equation $x^{2}-d y^{2}=1$, where $d$ is a positive integer that is not a perfect square. Then all positive solutions $x_{k}, y_{k}$ are given by
$$x_{k}+y_{k} \sqrt{d}=\left(x_{1}+y_{1} \sqrt{d}\right)^{k}$$
for $k=1,2,3, \ldots$. (Note... | Proof. We need to show that $x_{k}, y_{k}$ is a solution for $k=1,2,3, \ldots$ and that every solution is of this form.
To show that $x_{k}, y_{k}$ is a solution, first note that by taking conjugates, it follows that $x_{k}-y_{k} \sqrt{d}=\left(x_{1}-y_{1} \sqrt{d}\right)^{k}$, because from Lemma 10.4, the conjugate o... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,389 |
1. Find all the solutions of each of the following diophantine equations
a) $x^{2}+3 y^{2}=4$
b) $x^{2}+5 y^{2}=7$
c) $2 x^{2}+7 y^{2}=30$. | 1. a) $x= \pm 2, y=0 ; x= \pm 1, y= \pm 1$ b) no solution c) $x= \pm 1, y= \pm 2$ | a) x= \pm 2, y=0 ; x= \pm 1, y= \pm 1 \quad b) \text{no solution} \quad c) x= \pm 1, y= \pm 2 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,390 |
2. Find all the solutions of each of the following diophantine equations
a) $x^{2}-y^{2}=8$
b) $x^{2}-4 y^{2}=40$
c) $4 x^{2}-9 y^{2}=100$. | 2. a) $x= \pm 3, y= \pm 1$
b) no solution
c) $x= \pm 5, y=0 ; x= \pm 13, y= \pm 8$ | a) x= \pm 3, y= \pm 1 \\ b) \text{no solution} \\ c) x= \pm 5, y=0 ; x= \pm 13, y= \pm 8 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,391 |
5. Find the three smallest positive solutions of the diophantine equation $x^{2}-37 y^{2}=1$. | 5. $\begin{aligned} x & =1520, y=273 ; x=4620799, y=829920 ; x=42703566796801, \\ y & =766987012160\end{aligned}$ | x =1520, y=273 ; x=4620799, y=829920 ; x=42703566796801, y=766987012160 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,394 |
6. For each of the following values of $d$ determine whether the diophantine equation $x^{2}-d y^{2}=-1$ has solutions
a) 2
e) 17
b) 3
f) 31
c) 6
g) 41
d) 13
h) 50 . | 6. a) , d), e), g), h) yes b), c), f) no | a), d), e), g), h) yes; b), c), f) no | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,395 |
7. The least positive solution of the diophantine equation $x^{2}-61 y^{2}=1$ is $x_{1}=1766319049, y_{1}=226153980$. Find the least positive solution other than $x_{1}, y_{1}$. | 7. $x=6239765965720528801, y=79892016576262330040$ | x=6239765965720528801, y=79892016576262330040 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,396 |
3. Let $T$ be a set composed of integers. If $T$ contains positive integers, then $T$ must have a smallest positive integer. | 3. Let $T^{*}$ be the subset of $T$ consisting of all positive integers. Apply Theorem 2 of $\S 1$ to $T^{*}$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,411 |
12. Under the notation of problem 10, find $s$ and $j_{1}, \cdots, j_{s}$, such that
$1 \bmod 3=j_{1} \bmod 21 \cup \cdots \cup j_{s} \bmod 21$.
In general, let $a \mid b$, and $j$ be a given integer, find $s$ and $j_{1}, \cdots, j_{s}$, such that
$j \bmod a=j_{1} \bmod b \cup \cdots \cup j_{s} \bmod b$.
Explain the me... | 12. $s=7, j_{i}=1+(i-1) 3,1 \leqslant i \leqslant 7$. Generally,
$$s=b / a, \quad j_{i}=j+(i-1) a, \quad 1 \leqslant i \leqslant b / a$$ | s=b / a, \quad j_{i}=j+(i-1) a, \quad 1 \leqslant i \leqslant b / a | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,412 |
Theorem 3 Let $\xi_{0}=\sqrt{d}$, with the period of its continued fraction being $l$, and the convergents being $h_{n} / k_{n}$ $(n \geqslant 0)$, then we have
$$h_{l j-1}+\sqrt{d} k_{l j-1}=\left(h_{l-1}+\sqrt{d} k_{l-1}\right)^{j}, \quad j=1,2, \cdots$$ | Let $\rho_{j}=h_{l j-1}+\sqrt{d} k_{l j-1}$, its conjugate number is
$$\rho_{j}^{\prime}=h_{l j-1}-\sqrt{d} k_{l j-1}$$
Theorem 1 proves that all positive solutions of the indefinite equations (1) and (2) (if they have solutions) are given by $\rho_{j}(j \geqslant 1)$. We have
$$\begin{array}{c}
\rho_{j} \rho_{j}^{\pr... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,414 |
Example 1 Find all solutions to the indefinite equations
$$x^{2}-73 y^{2}=-1$$
and
$$x^{2}-73 y^{2}=1$$ | From Example 3 of §5, we know that $\sqrt{73}=\langle 8, \overline{1,1,5,5,1,1,16}\rangle$. The period is 7. Therefore, by Theorem 1 and Theorem 3, the smallest positive solution to the indeterminate equation (19) is $x=h_{6}, y=k_{6}$. It is not difficult to find that
$$h_{6} / k_{6}=\langle 8,1,1,5,5,1,1\rangle=1068 ... | x+y \sqrt{73}= \pm(1068 \pm 125 \sqrt{73})^{j}, \quad j=1,3,5,7, \cdots \text{ for } x^{2}-73 y^{2}=-1 \text{ and } x+y \sqrt{73}= \pm(1068 \pm 125 \sqrt{73})^{j}, \ | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,416 |
Example 2 Find all solutions to the indeterminate equations
$$x^{2}-8 y^{2}=-1$$
and
$$x^{2}-8 y^{2}=1$$ | From Example 4 in §3, we know that $\sqrt{8}=\langle 2, \overline{1,4}\rangle$. The period is 2. Therefore, by Theorem 1 and Theorem 3, the indeterminate equation (21) has no solution, and the smallest positive solution of the indeterminate equation (22) is $x=h_{1}, y=k_{1}$. It is easy to find that
$$h_{1} / k_{1}=\l... | x+y \sqrt{8}= \pm(3 \pm \sqrt{8})^{j}, \quad j=0,1,2,3, \cdots | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,417 |
Example 3 Solve the indeterminate equations
$$x^{2}-29 y^{2}=-1$$
and
$$x^{2}-29 y^{2}=1$$
for all solutions. | For $y=1,2, \cdots, 12$, it is known by calculation that $\pm 1+29 y^{2}$ are not perfect squares. When $y=13$, $-1+29 \cdot 13^{2}=70^{2}$. Therefore, $x=70, y=13$ is the smallest positive solution of the indeterminate equation (23). By
$$(70+13 \sqrt{29})^{2}=9801+1820 \sqrt{29}$$
it is known that the smallest posit... | x+y \sqrt{29}= \pm(70 \pm 13 \sqrt{29})^{j}, \quad j=1,3,5,7, \cdots \text{ and } x+y \sqrt{29}= \pm(70 \pm 13 \sqrt{29})^{j}, \quad j=0,2,4,6, \cdots | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,418 |
4. Find all solutions to the indeterminate equation $x^{2}+(x+1)^{2}=y^{2}$, and explain the geometric meaning of this problem
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 4. The original equation can be written as $(2 x+1)^{2}-2 y^{2}=-1$. This problem is to find a Pythagorean triangle with two legs being consecutive integers.
保留源文本的换行和格式，应该是这样的：
4. The original equation can be written as $(2 x+1)^{2}-2 y^{2}=-1$. This problem is to find a Pythagorean triangle with two legs being cons... | not found | Logic and Puzzles | math-word-problem | Yes | Yes | number_theory | false | 739,422 |
5. Prove: There exist infinitely many positive integers $n$, such that $1+2+\cdots+n$ is a perfect square.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 5. Solve the indeterminate equation $n(n+1)=2 y^{2}$, i.e., $(2 n+1)^{2}-8 y^{2}=1$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,424 |
6. Let $x_{n}+y_{n} \sqrt{2}=(1+\sqrt{2})^{n}$. Prove:
(i) $y_{n+1}=x_{n}+y_{n}, x_{n+1}=y_{n+1}+y_{n}, \quad n \geqslant 1$;
(ii) $y_{2 n+1}=y_{n+1}^{2}+y_{n}^{2}, \quad n \geqslant 1$;
(iii) $y_{2 n+1}^{2}$ is the sum of the squares of two consecutive natural numbers, find these two natural numbers;
(iv) If $x_{0}=1,... | 6. (i) It can be deduced from \(x_{n+1}+y_{n+1} \sqrt{2}=\left(x_{n}+y_{n} \sqrt{2}\right)(1+\sqrt{2})\).
(ii) It follows from \(x_{2 n+1}+y_{2 n+1} \sqrt{2}=\left(x_{n}+y_{n} \sqrt{2}\right)\left(x_{n+1}+y_{n+1} \sqrt{2}\right)\) and (i).
(iii) \(1+\sqrt{2}\) is the smallest positive solution of \(u^{2}-2 v^{2}=-1\), ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,425 |
7. Let prime $p \equiv 1(\bmod 4)$. Prove: $x^{2}-p y^{2}=-1$ must have a solution.
The text has been translated while preserving the original formatting and line breaks. | 7. Let $x_{2}+y_{2} \sqrt{p}$ be the smallest positive solution of $x^{2}-p y^{2}=1$. From $p y_{2}^{2}=\left(x_{2}^{2}-1\right)$, it follows that $2 \nmid x_{2}$, $2 \mid y_{2}$, and $p$ can only divide one of $x_{2}+1, x_{2}-1$. Therefore, we have $x_{2} \pm 1=2 x_{1}^{2}, x_{2} \mp 1=2 p y_{1}^{2}$. This leads to $x... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,426 |
8. Let $d>1$ be a non-square number, and $u, v$ be the smallest positive solution of $x^{2}-d y^{2}=1$. Prove: $x^{2}-d y^{2}=-1$ has a solution if and only if
$$s^{2}+d t^{2}=u, \quad 2 s t=v$$
has positive integer solutions $s, t$, and $s, t$ are the smallest positive integer solutions of $x^{2}-d y^{2}=-1$.
The fol... | 8. $s, t$ are the minimal positive solutions of $x^{2}-d y^{2}=-1$ if and only if
$$(s+t \sqrt{d})^{2}=u+v \sqrt{d} .$$ | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,427 |
9. Prove: There exist infinitely many pairs of positive integers $x, y$ satisfying $\left|x^{2}-d y^{2}\right|<1+2 \sqrt{d}$ (using Theorem 8 from §4). | 9. By Theorem 8 of §4, we know that there are infinitely many pairs of positive integers satisfying $|x-y \sqrt{d}|<1 / y$. From this and $|x+y \sqrt{d}| \leqslant|x-y \sqrt{d}|+2 y \sqrt{d}$, the conclusion follows. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,428 |
10. Prove: There must exist an integer $m$, such that $x^{2}-d y^{2}=m$ has infinitely many solutions $x_{j}, y_{j}$, where $x_{j}, y_{j}$ are natural numbers, and for any $j_{1}, j_{2}$, it satisfies
$$x_{j_{1}} \equiv x_{j_{2}}(\bmod m), \quad y_{j_{1}} \equiv y_{j_{2}}(\bmod m)$$ | 10. From the previous problem, we know that there are infinitely many pairs of $x, y$ such that $\left|x^{2}-d y^{2}\right|$ takes on only a finite number of positive integer values less than $1+2 \sqrt{d}$. Therefore, there must be infinitely many pairs of $x, y$ such that $x^{2}-d y^{2}=m,|m|<1+2 \sqrt{d}$, and since... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,429 |
11. Prove: (i) There is at least one pair of positive integers $x, y$ that satisfies the equation $x^{2}-d y^{2}=1$. Let $x_{1}, y_{1}$ and $x_{2}, y_{2}$ be two pairs of positive integer solutions, then the necessary and sufficient condition for $x_{1} \leqslant x_{2}$ is $y_{1} \leqslant$ $y_{2}$. If the positive int... | 11. (i) From the previous problem, we know there must be different pairs of positive integers $x_{1}, y_{1}; x_{2}, y_{2}$ satisfying: $x_{1} \equiv x_{2}(\bmod m)$, $y_{1} \equiv y_{2}(\bmod m), x_{1}^{2}-d y_{1}^{2}=x_{2}^{2}-d y_{2}^{2}=m$. It is evident that $x_{1} \neq x_{2}, y_{1} \neq y_{2}$. From the relation
$... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,430 |
12. Prove: If $x^{2}-d y^{2}=c$ has one solution, then it must have infinitely many solutions, where $d>1$ is a non-square number, and $c$ is an integer. | 12. Let $x_{1}+y_{1} \sqrt{d}$ be a positive solution of $x^{2}-d y^{2}=1$, and $u_{1}, v_{1}$ be solutions of the original indeterminate equation. Then $\left(x_{1}+y_{1} \sqrt{d}\right)^{n}(u+v \sqrt{d})=\left(u_{n}+v_{n} \sqrt{d}\right)(n=1,2, \cdots)$ gives $u_{n}, v_{n}$ which are all solutions of the original equ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,431 |
14. Let $d>1$ be a non-square number, $c$ an integer, $|c|<\sqrt{d}$. If positive integers $h, k$ are a solution to $x^{2}-d y^{2}=c$, and $(h, k)=1$, then $h / k$ must be a convergent of $\sqrt{d}$. | 14. At this point, we have $|h-k \sqrt{d}|=|c| /|h+k \sqrt{d}|$. First, assume $c$ is a positive real number, satisfying $0 < k \sqrt{d} < h$, thus $|h-k \sqrt{d}| < 1 / (2 \sqrt{d})$. Since $h > k \sqrt{d}$, it follows that $|h-k \sqrt{d}| < h \sqrt{1 / d}$, leading to $|k-h \sqrt{1 / d}| < 1 / (2 h)$. Therefore, by T... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,433 |
15. Let $d>1$ be a non-square number, and $\xi, \eta$ be two positive integers, satisfying $\xi^{2}-d \eta^{2}=1$. Prove: If $\xi>\eta^{2} / 2-1$, then $\xi+\eta \sqrt{d}$ is the smallest positive solution of the indefinite equation $x^{2}-d y^{2}=1$. | 15. When $\eta=1$, the conclusion holds. Let $\eta>1$. Use proof by contradiction. Suppose $x_{0}+y_{0} \sqrt{d}$ is the smallest positive solution, $1 \leqslant y_{1}0$, and thus we get
$$x_{1} \eta+y_{1} \xi=c_{1}>0, \quad x_{1} \eta-y_{1} \xi=c_{2}>0, \quad c_{1} c_{2}=\eta^{2}-y_{1}^{2}$$
This implies $\xi \leqsla... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,435 |
Theorem 2 Let $A$ be a given finite sequence of integers, and $K$ a given positive integer; let $A_{d}$ denote the subsequence of $A$ consisting of all integers divisible by the positive integer $d$, and let $p_{1}, \cdots, p_{s}$ be all the distinct prime factors of $K$, and $\left|A_{d}\right|$ denote the number of i... | Theorem 2 Proof: By Lemma 3, we know
$$\sum_{\substack{a \in A \\(a, K)=1}} 1=\sum_{a \in A d \mid(a, K)} \sum_{d \mid} \mu(d)=\sum_{d \mid K} \mu(d) \sum_{\substack{a \in A \\ d \mid a}} 1=\sum_{d \mid K} \mu(d)\left|A_{d}\right| .$$
This proves equation (24), which is also equation (21). Q.E.D. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,438 |
Example 1 Find the number of prime numbers not exceeding 100. | We take $N=100$, the prime numbers not exceeding $\sqrt{100}=10$ are $2,3,5,7$, so by formula (20) we get
$$\begin{aligned}
\pi(100)= & 4-1+100-\left\{\left[\frac{100}{2}\right]+\left[\frac{100}{3}\right]+\left[\frac{100}{5}\right]+\left[\frac{100}{7}\right]\right\} \\
& +\left\{\left[\frac{100}{2 \cdot 3}\right]+\left... | 25 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,439 |
Example 2 Let $N$ be a positive integer, and $\varphi(N)$ be the number of positive integers in $1,2, \cdots, N$ that are coprime to $N$, then
$$\varphi(N)=N \prod_{p \mid N}\left(1-\frac{1}{p}\right),$$
where the meaning of the product symbol is given in Chapter 1, §5, Equation (18). | Let $p_{1}, p_{2}, \cdots, p_{m}$ be all the distinct prime divisors of $N$. In Theorem 2, take the sequence $A$ to be $1, 2, \cdots, N, K=p_{1} \cdots p_{m}$. Thus, $\varphi(N)$ is the number of integers in $A$ that are coprime to $K$, i.e., $S(A ; K)$. Note that in this case we have (why)
$$\left[p_{i_{1}}, \cdots, p... | \varphi(N)=N\left(1-\frac{1}{p_{1}}\right)\left(1-\frac{1}{p_{2}}\right) \cdots\left(1-\frac{1}{p_{m}}\right) | Number Theory | proof | Yes | Yes | number_theory | false | 739,440 |
Lemma 3 Let $n$ be a positive integer. We have
$$\sum_{d \mid n} \mu(d)=\left[\frac{1}{n}\right]=\left\{\begin{array}{ll}
1, & n=1 \\
0, & n>1
\end{array}\right.$$ | Proof: When $n=1$, equation (25) obviously holds. Now suppose $n=p_{1}^{q_{1}} \cdots p_{s}^{s_{s}^{s}}, \alpha_{j} \geqslant 1$. By definition (22), we have
$$\begin{aligned}
\sum_{d \mid n} \mu(d)=\sum_{d \mid p_{1} \cdots p_{s}} \mu(d) & =1-\binom{s}{1}+\binom{s}{2}-\cdots+(-1)^{s}\binom{s}{s} \\
= & (1-1)^{s}=0
\en... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,441 |
Theorem 4 Let $x \geqslant y \geqslant 2, \Phi(x ; y)$ denote the number of positive integers not exceeding $x$, all of whose prime factors are greater than $y$, then
$$x \prod_{p \leqslant y}\left(1-\frac{1}{p}\right)-2^{\pi(y)} \leqslant 1+\Phi(x ; y) \leqslant x \prod_{p \leq y}\left(1-\frac{1}{p}\right)+2^{\pi(y)} ... | Let $A$ be the sequence $1,2, \cdots,[x]$,
$$K=P(y)=\prod_{p \leqslant y} p$$
We have
$$\begin{aligned}
1+\Phi(x ; y) & =S(A ; P(y))=\sum_{\substack{1 \leqslant a \leqslant x \\
(a, P(y))=1}} 1 \\
& =\sum_{d \mid P(y)} \mu(d)\left[\frac{x}{d}\right] \\
& =x \sum_{d \mid P(y)} \frac{\mu(d)}{d}-\sum_{d \mid P(y)} \mu(d)... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,442 |
Theorem 5 Let $x \geqslant 10$. There exists a positive constant $c_{1}$ such that
$$\begin{array}{l}
\pi(x) \leqslant c_{1} x(\ln \ln x)^{-1} \\
p_{n} \geqslant c_{1}^{-1} n \ln \ln n, \quad n \geqslant 5
\end{array}$$
Here $p_{n}$ denotes the $n$-th prime number. | It is easy to see that
$$\pi(x)-\pi(y) \leqslant \Phi(x ; y), \quad 2 \leqslant y \leqslant x.$$
From this and equation (28), we get
$$\pi(x) \leqslant \pi(y)+x \prod_{p \leqslant y}\left(1-\frac{1}{p}\right)+2^{\pi(y)}, \quad 2 \leqslant y \leqslant x.$$
Notice that
$$\begin{aligned}
\prod_{p \leq y}\left(1-\frac{1}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,443 |
Theorem 6 Let the sequence of all prime numbers arranged in increasing order be
$$p_{1}=2, p_{2}=3, p_{3}, p_{4}, \cdots$$
We have
$$p_{n} \leqslant 2^{2^{n-1}}, \quad n=1,2, \cdots$$
and
$$\pi(x)>\log _{2} \log _{2} x, \quad x \geqslant 2$$
where $\log _{2} y$ denotes the logarithm to the base 2. | Proof From the proof of Theorem 7 in Chapter 1, §2, we know that
$$p_{n} \leqslant p_{1} p_{2} \cdots p_{n-1}+1, \quad n>1$$
We prove formula (39) by induction. When $n=1$, formula (39) clearly holds. Assuming that formula (39) holds for $n \leqslant k (\geqslant 1)$, when $n=k+1$, by formula (41) and the induction hy... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,444 |
Theorem 7 (Principle of Inclusion-Exclusion) Under the above notation and conventions, let $A$ be a finite sequence, and $P_{1}, P_{2}, \cdots, P_{m}$ be $m$ properties related to the elements of $A$. Then, the number of elements in $A$ that do not have any property $P_{j}(1 \leqslant j \leqslant m)$ is
$$\begin{aligne... | Theorem 7's proof: $B^{(0)}$ is easy to solve. Taking the algebraic sum of the above $m+1$ equations as follows: the first equation minus the second equation, plus the third equation, minus the fourth equation, $\cdots$, we get
$$\begin{aligned}
A^{(0)}- & A^{(1)}+A^{(2)}-\cdots+(-1)^{r} A^{(r)}+\cdots+(-1)^{m-1} A^{(m... | proof | Combinatorics | proof | Yes | Yes | number_theory | false | 739,446 |
Example 3 (i) Find the number of integers between 1 and 500 that are not divisible by any of 5, 6, 8;
(ii) Find the number of integers between 1 and 500 that are divisible by 5, or 6, or 8. | Consider the sequence $A$ composed of integers from 1 to 500. Property $P_{1}$: divisible by 5, property $P_{2}$: divisible by 6, property $P_{3}$: divisible by 8. Thus, (i) is to find the number of $B^{(0)}$, and (ii) is to find the number of $A^{(0)}-B^{(0)}$.
The number of $A(1)$ is: 100; the number of $A(2)$ is: 83... | 299 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,449 |
Example 4 Let sequence $A$ be $1,3,2,5,4,3,5,7,6,11$. Property $P_{1}$ is that the element is divisible by 3, property $P_{2}$ is that the element leaves a remainder of 1 when divided by 3, and property $P_{3}$ is that the element is divisible by 2. Try to verify whether Theorem 7 holds. | $A(1)$ is $3,3,6; A(2)$ is $1,4,7; A(3)$ is $2,4,6. A(1,2)$ has no elements; $A(1,3)$ is $6; A(2,3)$ is $4. A(1,2,3)$ has no elements.$
$$B^{(0)}=B(0) \text { is } 5,5,11; B(1) \text { is } 3,3; B(2) \text { is } 1,7; B(3) \text { is } 2 \text {. }$$
$B(1,2)$ has no elements; $B(1,3)$ is $6; B(2,3)$ is $4. B(1,2,3)$ ha... | 3 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,450 |
Example 5 Let sequence $A$ be $1,3,2,5,4,3,5,7,6,11$. Property $P_{1}$ is that the element is divisible by 3, Property $P_{2}$ is that the element leaves a remainder of 1 when divided by 3, Property $P_{3}$ is that the element is divisible by 2, Property $P_{4}$ is that the element is divisible by 3. Try to verify whet... | Solve $A(1)$ is $3,3,6 ; A(2)$ is $1,4,7 ; A(3)$ is $2,4,6 ; A(4)$ is $3,3,6. A(1,2)$ has no elements; $A(1,3)$ is $6 ; A(1,4)$ is $3,3,6 ; A(2,3)$ is $4 ; A(2,4)$ has no elements; $A(3,4)$ is $6. A(1,2,3)$ has no elements; $A(1,2,4)$ has no elements; $A(1,3,4)$ is $6 ; A(2,3,4)$ has no elements; $A(1,2,3,4)$ has no el... | 3 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,451 |
Example 6 Let $a_{1}, a_{2}, \cdots, a_{m}$ be $m$ non-negative integers, then the maximum value among them is
$$\begin{aligned}
\max \left(a_{1}, a_{2}, \cdots, a_{m}\right)= & \sum_{1 \leqslant i_{1} \leqslant m} a_{i_{1}}-\sum_{1 \leqslant i_{1}<i_{2} \leqslant m} \min \left(a_{i_{1}}, a_{i_{2}}\right)+\cdots \\
& +... | Let $N$ be a positive integer not less than all $a_{j}(1 \leqslant j \leqslant m)$. In Theorem 7, take the sequence $A$ to be $1,2, \cdots, N$, and let property $P_{j}$ be not greater than $a_{j}(1 \leqslant j \leqslant m)$. Thus, we have
$$\begin{array}{c}
\left|A\left(i_{1}, \cdots, i_{k}\right)\right|=\min \left(a_{... | proof | Algebra | proof | Yes | Yes | number_theory | false | 739,452 |
1. (i) Find the number of integers from 1 to 2000 that are not divisible by 10, 14, or 21;
(ii) Find the number of integers from 1 to 1000 that are divisible by 3 and 7, but not by 5;
(iii) Find the number of integers from 1 to 1000 that are divisible by 3 or 7, but not by 5;
(iv) Find $\pi(N)$, for $N=200,300,400,500,... | $\begin{array}{l}\text { 1. (i) } 1638 \text {; (ii) } 47-9=38 \text {; (iii) } 333+142-66-28+38=419 \text {; (iv) } 46,62 \text {, } \\ 78,95,109,125,139,154,168 \text {; (v) } 168-7+7+5+1=174 \text {; (vi) } 11 \text {. }\end{array}$ | 1638; 38; 419; 46, 62, 78, 95, 109, 125, 139, 154, 168; 174; 11 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,453 |
16. (I) Let $n=c_{k} \cdot 10^{k}+\cdots+c_{1} \cdot 10+c_{0}$. Prove:
(i) $2|n \Longleftrightarrow 2| c_{0}$;
(ii) $5|n \Longleftrightarrow 5| c_{0}$;
(iii) $3|n \Longleftrightarrow 3|\left(c_{k}+\cdots+c_{0}\right)$;
(iv) $9|n \Longleftrightarrow 9|\left(c_{k}+\cdots+c_{0}\right)$;
(v) $11|n \Longleftrightarrow 11|\l... | $\begin{array}{l}\text { 16. (IV) } 1535625=3^{3} \cdot 5^{4} \cdot 7 \cdot 13 ; \quad 1158066=2 \cdot 3^{2} \cdot 7^{2} \cdot 13 \cdot 101 \text {; } \\ 82798848=2^{8} \cdot 3^{5} \cdot 11^{3} ; \\ 81057226635000=2^{3} \cdot 3^{3} \cdot 5^{4} \cdot 7^{3} \cdot 11^{2} \cdot 17 \cdot 23 \cdot 37\end{array}$ | null | Number Theory | proof | Yes | Yes | number_theory | false | 739,456 |
4. Let $n$ be any given positive integer. Find the value of $\mu(n) \mu(n+1) \mu(n+2) \mu(n+3)$. | 4. 0 , because $n, n+1, n+2, n+3$ must include one number that is divisible by 4. | 0 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,457 |
6. Find the positive integer $k$ such that
$$\mu(k)+\mu(k+1)+\mu(k+2)=0, \pm 1, \pm 2, \pm 3$$ | 6.
\begin{tabular}{|c|rrrrrrr|}
\hline$k$ & 4 & 8 & 1 & 14 & 2 & 33 & 29 \\
\hline$\mu(k)+\mu(k+1)+\mu(k+2)$ & 0 & 1 & -1 & 2 & -2 & 3 & -3 \\
\hline
\end{tabular} | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,459 |
7. Prove: $\sum_{d^{2} \mid n} \mu(d)=u^{2}(n)=|\mu(n)|$, here the summation sign indicates the sum over all positive integers $d$ such that $d^{2} \mid n$. | 7. $n=m^{2} n_{1}, \mu\left(n_{1}\right) \neq 0 . \sum_{d^{2} \mid n} f(d)=\sum_{d \mid m} f(d)$. Then use Lemma 3. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,460 |
8. Prove: (i) $\sum_{d \mid n} \mu^{2}(d)=2^{\omega(n)}, \omega(n)$ is the number of distinct prime factors of $n$, $\omega(1)=0$;
(ii) $\sum_{d \mid n} \mu(d) \tau(d)=(-1)^{\alpha(n)}$. | 8. (i) Similarly to (27), we can prove:
$$\sum_{d \mid n} \mu^{2}(d)=\left(\sum_{i_{1}=0}^{a_{1}} \mu^{2}\left(p_{1}^{i_{1}}\right)\right) \cdots\left(\sum_{i_{r}=0}^{a_{r}} \mu^{2}\left(p_{r}^{i_{r}}\right)\right),$$
where $n=p_{1}^{q_{1}} \cdots p_{r}^{\alpha_{r}}$;
(ii) $\sum_{d \mid n} \mu(d) \tau(d)=\left(\sum_{i... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,461 |
9. Let $k$ be a given positive integer. Prove:
$$\sum_{d^{k} \mid n} \mu(d)=\left\{\begin{array}{ll}
0, & \text { if there exists } m>1 \text { such that } m^{k} \mid n, \\
1 & \text { otherwise, }
\end{array}\right.$$
Here the summation sign indicates the sum over all positive integers $d$ such that $d^{k} \mid n$. | 9. $n=m^{k} n_{1}$, for any prime $p, p^{k} \nmid n_{1} . d^{k}|n \Longleftrightarrow d| m$.
| proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,462 |
10. Let $2 \mid n$. Prove: $\sum_{d \mid n} \mu(d) \varphi(d)=0$. | 10. Use the method from question 8, and note that $\varphi(p)=1$ if and only if $p=2$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,463 |
11. Let $f(n)$ be a function defined on the set of positive integers (called a number-theoretic function), and let $F(n)=\sum_{d \mid n} f(d)$. Prove: If for any $\left(n_{1}, n_{2}\right)=1$ there must be $f\left(n_{1} n_{2}\right)=$ $f\left(n_{1}\right) f\left(n_{2}\right)$, then for any $\left(n_{1}, n_{2}\right)=1$... | 11. $d \mid n=n_{1} n_{2},\left(n_{1}, n_{2}\right)=1$ is a sufficient and necessary condition for
$$d=d_{1} d_{2}, \quad d_{1}=\left(d, n_{1}\right), \quad d_{2}=\left(d, n_{2}\right) .$$ | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,464 |
13. (i) Let $k \mid n$. Prove: $\sum_{\substack{d=1 \\(d, n)=k}} 1=\varphi(n / k)$.
(ii) Let $f(n)$ be an arithmetic function. Prove:
$$\sum_{d=1}^{n} f((d, n))=\sum_{d \mid n} f(d) \varphi(n / d)$$
(iii) Prove: $\sum_{d=1}^{n}(d, n) \mu((d, n))=\mu(n)$. | 13. (i) Deduce from the definition of $\varphi(n)$.
(ii) $\sum_{d=1}^{n} f((d, n))=\sum_{k \mid n} \sum_{\substack{d=1 \\(d, n)=k}}^{n} f((d, n))$.
(iii) From (ii), the sum is $=\sum_{d \mid n} d \mu(d) \varphi(n / d)$. When $n=1$, it equals 1; when $n>1$, using the method of proving problem 11, we can show: when $\lef... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,466 |
14. Prove: $\mu(n)=\sum_{\substack{d=1 \\(d, n)=1}}^{n} \mathrm{e}^{2 \pi i d / n}$. | 14. The right-hand side $=\sum_{d=1}^{n} \mathrm{e}^{2 \pi i d / n} \sum_{k \mid (d, n)} \mu(k)$, then interchange the summation signs. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,468 |
15. Prove: $\sum_{d \leqslant x} \mu(d)\left[\frac{x}{d}\right]=1$. | 15. Left-hand side $=\sum_{d \leqslant x} \mu(d) \sum_{\substack{k \leqslant x \\ d k}} 1$, then interchange the summation signs. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,469 |
16. Find the values of $\Phi(x ; y)$ for the following:
(i) $x=400, y=3,5,7,11$;
(ii) $x=1000, y=5,7,11,17,29$.
Compare
$$\Phi(x ; y) \quad \text { and } \quad x \prod_{p \leqslant y}\left(1-\frac{1}{p}\right)$$
for the values of $(x, y)$ given in (i) and (ii). | 16.
\begin{tabular}{|c|ccccc|}
\hline$y$ & 3 & 5 & 7 & 11 \\
\hline$\Phi(400 ; y)$ & 132 & 105 & 90 & 81 \\
\hline $400 \prod_{p \leqslant y}\left(1-\frac{1}{p}\right) \approx$ & 133 & 106 & 91 & 83 \\
\hline$y$ & 5 & 7 & 11 & 17 & 29 \\
\hline$\Phi(1000 ; y)$ & 331 & 293 & 263 & 225 & 212 \\
\hline $1000 \prod_{p \leq... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 739,470 |
17. Let $k>l \geqslant 0, (k, l)=1$ and $x>k$; let $A$ denote the sequence of integers: $1 \leqslant a \leqslant x, a \equiv l(\bmod k)$, $\pi(x ; k, l)$ denote the number of primes in $A$, $K$ denote the product of all primes not exceeding $\sqrt{x}$ and not dividing $k$.
(i) Prove:
$$\pi(x ; k, l)=S(A, K)-\varepsilon... | 17. (ii)
\begin{tabular}{|c|ccccc|}
\hline$x$ & 100 & 300 & 500 & 700 & 1000 \\
\hline$\pi(x ; 4,1)$ & 11 & 29 & 44 & 59 & 80 \\
\hline$\pi(x ; 4,3)$ & 13 & 32 & 50 & 65 & 87 \\
\hline
\end{tabular} | not found | Number Theory | proof | Yes | Yes | number_theory | false | 739,471 |
18. Let the integer $N \geqslant 2$. Prove:
(i) $\sum_{p \leqslant N} \frac{1}{p-1}>\ln \ln (N+1)$, where the summation is over all primes $p$ not exceeding $N$;
(ii) $\sum_{p \leqslant N} \frac{1}{p}>\ln \ln (N+1)-1$. | 18. (i) Use $1 / n>\ln (1+1 / n)$ and
$$\ln (1-1 / p)^{-1}=\ln (1+1 /(p-1))<1 /(p-1)$$
(ii) $\sum_{p \leqslant N}\{1 /(p-1)-1 / p\}<1$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,472 |
19. Prove: (i) any positive integer $a$ can definitely be expressed as $a=k^{2} l$, where $k$ is a positive integer, and $l=1$ or is the product of distinct primes;
(ii) for any integer $N \geqslant 2$, then $N \leqslant \sqrt{N} \cdot 2^{\pi^{(N)}}$;
(iii) $\pi(N) \geqslant\left(\log _{2} N\right) / 2$;
(iv) the $n$-t... | 19. (i) Use Theorem 5 of Chapter 1, § 2; (ii) when $1 \leqslant a \leqslant N$, the $k, l$ in (i) satisfy: $1 \leqslant k$ $\leqslant \sqrt{N}$, and the number of possible values for $l$ $\leqslant 1+\binom{\pi(N)}{1}+\binom{\pi(N)}{2}+\cdots+\binom{\pi(N)}{\pi(N)}=2^{2(N)}$; from (ii) deduce (iii) and (iv). | proof | Number Theory | proof | Yes | Yes | number_theory | false | 739,473 |
Lemma 2 Let $y \geqslant 2$. We have
$$\begin{array}{c}
\ln \ln ([y]+1)-\ln \ln 2<\sum_{2 \leqslant k \leqslant y} \frac{1}{k \ln k} \\
\quad<\ln \ln [y]+\frac{1}{2 \ln 2}-\ln \ln 2
\end{array}$$
and
$$\begin{array}{l}
{[y]\{\ln [y]-1\}+1<\sum_{1 \leqslant k \leqslant y} \ln k} \\
\quad<([y]+1)\{\ln ([y]+1)-1\}+2-2 \l... | Prove that we have
$$\int_{k}^{k+1} \frac{\mathrm{d} t}{t \ln t}\int_{2}^{[y]+1} \frac{\mathrm{d} t}{t \ln t}=\ln \ln ([y]+1)-\ln \ln 2 .
\end{array}$$
From the above two equations, we obtain equation (15). Similarly, from
$$\int_{k-1}^{k} \ln t \mathrm{~d} t\int_{1}^{[y]} \ln t \mathrm{~d} t=[y] \ln [y]-[y]+1
\end{ar... | proof | Inequalities | proof | Yes | Yes | number_theory | false | 739,480 |
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