From 4f670b750af5c11e1eac16d9cd8556455f89f46a Mon Sep 17 00:00:00 2001
From: Juan Marín Noguera <juan@mnpi.eu>
Date: Fri, 16 May 2025 22:18:44 +0200
Subject: Changed layout for more manageable volumes

---
 1.2.4.lyx | 2433 -------------------------------------------------------------
 1 file changed, 2433 deletions(-)
 delete mode 100644 1.2.4.lyx

(limited to '1.2.4.lyx')

diff --git a/1.2.4.lyx b/1.2.4.lyx
deleted file mode 100644
index ed3730a..0000000
--- a/1.2.4.lyx
+++ /dev/null
@@ -1,2433 +0,0 @@
-#LyX 2.4 created this file. For more info see https://www.lyx.org/
-\lyxformat 620
-\begin_document
-\begin_header
-\save_transient_properties true
-\origin unavailable
-\textclass book
-\begin_preamble
-\input defs
-\end_preamble
-\use_default_options true
-\maintain_unincluded_children no
-\language english
-\language_package default
-\inputencoding utf8
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-\use_indices false
-\paperorientation portrait
-\suppress_date false
-\justification true
-\use_refstyle 1
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-\index Index
-\shortcut idx
-\color #008000
-\end_index
-\secnumdepth 3
-\tocdepth 3
-\paragraph_separation indent
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-\end_header
-
-\begin_body
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc1[00]
-\end_layout
-
-\end_inset
-
-What are 
-\begin_inset Formula $\lfloor1.1\rfloor$
-\end_inset
-
-,
- 
-\begin_inset Formula $\lfloor-1.1\rfloor$
-\end_inset
-
-,
- 
-\begin_inset Formula $\lceil-1.1\rceil$
-\end_inset
-
-,
- 
-\begin_inset Formula $\lfloor0.99999\rfloor$
-\end_inset
-
-,
- and 
-\begin_inset Formula $\lfloor\lg35\rfloor$
-\end_inset
-
-?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-
-\begin_inset Formula $\lfloor1.1\rfloor=1$
-\end_inset
-
-,
- 
-\begin_inset Formula $\lfloor-1.1\rfloor=-2$
-\end_inset
-
-,
- 
-\begin_inset Formula $\lceil-1.1\rceil=-1$
-\end_inset
-
-,
- 
-\begin_inset Formula $\lfloor0.99999\rfloor=0$
-\end_inset
-
-,
- 
-\begin_inset Formula $\lfloor\lg35\rfloor=5$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc2[01]
-\end_layout
-
-\end_inset
-
-What is 
-\begin_inset Formula $\lceil\lfloor x\rfloor\rceil$
-\end_inset
-
-?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-The same as 
-\begin_inset Formula $\lfloor x\rfloor$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc3[10]
-\end_layout
-
-\end_inset
-
-Let 
-\begin_inset Formula $n$
-\end_inset
-
- be an integer,
- and let 
-\begin_inset Formula $x$
-\end_inset
-
- be a real number.
- Prove that
-\end_layout
-
-\begin_layout Enumerate
-\begin_inset Formula $\lfloor x\rfloor<n$
-\end_inset
-
- if and only if 
-\begin_inset Formula $x<n$
-\end_inset
-
-;
-\end_layout
-
-\begin_layout Enumerate
-\begin_inset Formula $n\leq\lfloor x\rfloor$
-\end_inset
-
- if and only if 
-\begin_inset Formula $n\leq x$
-\end_inset
-
-;
-\end_layout
-
-\begin_layout Enumerate
-\begin_inset Formula $\lceil x\rceil\leq n$
-\end_inset
-
- if and only if 
-\begin_inset Formula $x\leq n$
-\end_inset
-
-;
-\end_layout
-
-\begin_layout Enumerate
-\begin_inset Formula $n<\lceil x\rceil$
-\end_inset
-
- if and only if 
-\begin_inset Formula $n<x$
-\end_inset
-
-;
-\end_layout
-
-\begin_layout Enumerate
-\begin_inset Formula $\lfloor x\rfloor=n$
-\end_inset
-
- if and only if 
-\begin_inset Formula $x-1<n\leq x$
-\end_inset
-
-,
- and if and only if 
-\begin_inset Formula $n\leq x<n+1$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Enumerate
-\begin_inset Formula $\lceil x\rceil=n$
-\end_inset
-
- if and only if 
-\begin_inset Formula $x\leq n<x+1$
-\end_inset
-
-,
- and if and only if 
-\begin_inset Formula $n-1<x\leq n$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer
-\end_layout
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Enumerate
-\begin_inset Note Note
-status open
-
-\begin_layout Plain Layout
-
-\end_layout
-
-\end_inset
-
-
-\end_layout
-
-\begin_deeper
-\begin_layout Enumerate
-\begin_inset Argument item:1
-status open
-
-\begin_layout Plain Layout
-\begin_inset Formula $\implies]$
-\end_inset
-
-
-\end_layout
-
-\end_inset
-
-By contraposition,
- if 
-\begin_inset Formula $x\geq n$
-\end_inset
-
-,
- then 
-\begin_inset Formula $n$
-\end_inset
-
- is an integer less than or equal to 
-\begin_inset Formula $x$
-\end_inset
-
- and so 
-\begin_inset Formula $\lfloor x\rfloor\geq n$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Enumerate
-\begin_inset Argument item:1
-status open
-
-\begin_layout Plain Layout
-\begin_inset Formula $\impliedby]$
-\end_inset
-
-
-\end_layout
-
-\end_inset
-
-
-\begin_inset Formula $\lfloor x\rfloor\leq x<n$
-\end_inset
-
-.
-\end_layout
-
-\end_deeper
-\begin_layout Enumerate
-It's the contrapositive of the previous statement.
-\end_layout
-
-\begin_layout Enumerate
-Derived from the previous statement replacing 
-\begin_inset Formula $x$
-\end_inset
-
- and 
-\begin_inset Formula $n$
-\end_inset
-
- with 
-\begin_inset Formula $-x$
-\end_inset
-
- and 
-\begin_inset Formula $-n$
-\end_inset
-
- and using that 
-\begin_inset Formula $\lfloor-x\rfloor=-\lceil x\rceil$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Enumerate
-It's the contrapositive of the previous statement.
-\end_layout
-
-\begin_layout Enumerate
-\begin_inset Note Note
-status open
-
-\begin_layout Plain Layout
-
-\end_layout
-
-\end_inset
-
-
-\end_layout
-
-\begin_deeper
-\begin_layout Description
-\begin_inset Formula $1\implies2]$
-\end_inset
-
- Obviously 
-\begin_inset Formula $n=\lfloor x\rfloor\leq x$
-\end_inset
-
-,
- and if 
-\begin_inset Formula $x-1\geq n$
-\end_inset
-
- then it would be 
-\begin_inset Formula $x\geq n+1\in\mathbb{Z}$
-\end_inset
-
- and 
-\begin_inset Formula $\lfloor x\rfloor\geq n+1>n\#$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Description
-\begin_inset Formula $2\implies3]$
-\end_inset
-
- Multiply the inequality by 
-\begin_inset Formula $-1$
-\end_inset
-
- and add 
-\begin_inset Formula $x+n$
-\end_inset
-
- to each member.
-\end_layout
-
-\begin_layout Description
-\begin_inset Formula $3\implies1]$
-\end_inset
-
- 
-\begin_inset Formula $n\leq x$
-\end_inset
-
- and any 
-\begin_inset Formula $m\in\mathbb{Z}$
-\end_inset
-
- with 
-\begin_inset Formula $m\leq x$
-\end_inset
-
- has 
-\begin_inset Formula $m<n+1$
-\end_inset
-
- and therefore 
-\begin_inset Formula $m\leq n$
-\end_inset
-
-,
- so 
-\begin_inset Formula $n=\max\{m\in\mathbb{Z}\mid m\leq x\}=\lfloor x\rfloor$
-\end_inset
-
-.
-\end_layout
-
-\end_deeper
-\begin_layout Enumerate
-Derived from the previous statement by replacing 
-\begin_inset Formula $x$
-\end_inset
-
- and 
-\begin_inset Formula $n$
-\end_inset
-
- with 
-\begin_inset Formula $-x$
-\end_inset
-
- and 
-\begin_inset Formula $-n$
-\end_inset
-
- and using that 
-\begin_inset Formula $-\lceil x\rceil=\lfloor-x\rfloor$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc4[M10]
-\end_layout
-
-\end_inset
-
-Using the previous exercise,
- prove that 
-\begin_inset Formula $\lfloor-x\rfloor=-\lceil x\rceil$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-Since that would be circular reasoning (I did the previous exercise before reading the statement of this one),
- we'll prove it directly.
-\end_layout
-
-\begin_layout Standard
-\begin_inset Formula 
-\begin{align*}
-\lfloor-x\rfloor=n & \iff\forall m\in\mathbb{Z},(m\leq-x\implies m\leq n)\\
- & \iff\forall m\in\mathbb{Z},(-m\geq x\implies-m\geq-n)\\
- & \iff\forall m\in\mathbb{Z},(m\geq x\implies m\geq-n)\iff-n=\lceil x\rceil\iff n=-\lceil x\rceil.
-\end{align*}
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc6[20]
-\end_layout
-
-\end_inset
-
-Which of the following equations are true for all positive real numbers 
-\begin_inset Formula $x$
-\end_inset
-
-?
-\end_layout
-
-\begin_layout Enumerate
-\begin_inset Formula $\left\lfloor \sqrt{\lfloor x\rfloor}\right\rfloor =\lfloor\sqrt{x}\rfloor$
-\end_inset
-
-;
-\end_layout
-
-\begin_layout Enumerate
-\begin_inset Formula $\left\lceil \sqrt{\lceil x\rceil}\right\rceil =\lceil\sqrt{x}\rceil$
-\end_inset
-
-;
-\end_layout
-
-\begin_layout Enumerate
-\begin_inset Formula $\left\lceil \sqrt{\lfloor x\rfloor}\right\rceil =\lceil\sqrt{x}\rceil$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer
-\end_layout
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Enumerate
-True,
- if 
-\begin_inset Formula $n=\lfloor\sqrt{x}\rfloor$
-\end_inset
-
-,
- then 
-\begin_inset Formula $n\leq\sqrt{x}<n+1$
-\end_inset
-
-,
- so 
-\begin_inset Formula $n^{2}\leq\lfloor x\rfloor\leq x<(n+1)^{2}$
-\end_inset
-
- and so 
-\begin_inset Formula $n\leq\sqrt{\lfloor x\rfloor}<n+1$
-\end_inset
-
- and 
-\begin_inset Formula $n=\left\lfloor \sqrt{\lfloor x\rfloor}\right\rfloor $
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Enumerate
-True,
- if 
-\begin_inset Formula $n=\lceil\sqrt{x}\rceil$
-\end_inset
-
-,
- then 
-\begin_inset Formula $n-1<\sqrt{x}\leq n$
-\end_inset
-
- and 
-\begin_inset Formula $(n-1)^{2}<x\leq\lceil x\rceil\leq n$
-\end_inset
-
-,
- so 
-\begin_inset Formula $n-1<\sqrt{\lceil x\rceil}\leq n$
-\end_inset
-
- and therefore 
-\begin_inset Formula $\left\lceil \sqrt{\lceil x\rceil}\right\rceil =n$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Enumerate
-False,
- for example 
-\begin_inset Formula $\left\lceil \sqrt{\lfloor\frac{9}{2}\rfloor}\right\rceil =\lceil\sqrt{4}\rceil=2$
-\end_inset
-
- but 
-\begin_inset Formula $\lceil\sqrt{\frac{9}{2}}\rceil=3$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc8[00]
-\end_layout
-
-\end_inset
-
-What are 
-\begin_inset Formula $100\bmod3$
-\end_inset
-
-,
- 
-\begin_inset Formula $100\bmod7$
-\end_inset
-
-,
- 
-\begin_inset Formula $-100\bmod7$
-\end_inset
-
-,
- 
-\begin_inset Formula $-100\bmod0$
-\end_inset
-
-?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-
-\begin_inset Formula $100\bmod3=1$
-\end_inset
-
-,
- 
-\begin_inset Formula $100\bmod7=2$
-\end_inset
-
-,
- 
-\begin_inset Formula $-100\bmod7=-2\bmod7=5$
-\end_inset
-
-,
- 
-\begin_inset Formula $-100\bmod0=-100$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc9[05]
-\end_layout
-
-\end_inset
-
-What are 
-\begin_inset Formula $5\bmod-3$
-\end_inset
-
-,
- 
-\begin_inset Formula $18\bmod-3$
-\end_inset
-
-,
- 
-\begin_inset Formula $-2\bmod-3$
-\end_inset
-
-?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-
-\begin_inset Formula $5\bmod-3=5-(-3)\lfloor\frac{5}{-3}\rfloor=5+3(-2)=5-6=-1$
-\end_inset
-
-,
- 
-\begin_inset Formula $18\bmod-3=18-(-3)(-6)=0$
-\end_inset
-
-,
- 
-\begin_inset Formula $-2\bmod-3=-2-(-3)\lfloor\frac{-2}{-3}\rfloor=-2$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc10[10]
-\end_layout
-
-\end_inset
-
-What are 
-\begin_inset Formula $1.1\bmod1$
-\end_inset
-
-,
- 
-\begin_inset Formula $0.11\bmod.1$
-\end_inset
-
-,
- 
-\begin_inset Formula $0.11\bmod-.1$
-\end_inset
-
-?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-
-\begin_inset Formula 
-\begin{eqnarray*}
-1.1\bmod1 & = & 1.1-\lfloor1.1\rfloor=.1,\\
-0.11\bmod.1 & = & .11-.1\lfloor1.1\rfloor=.01,\\
-0.11\bmod-.1 & = & .11-(-.1)\lfloor-1.1\rfloor=-.09.
-\end{eqnarray*}
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc11[00]
-\end_layout
-
-\end_inset
-
-What does 
-\begin_inset Quotes eld
-\end_inset
-
-
-\begin_inset Formula $x\equiv y\bmod0$
-\end_inset
-
-
-\begin_inset Quotes erd
-\end_inset
-
- mean by our conventions?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-It means 
-\begin_inset Formula $x=x\bmod0=y\bmod0=y$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc12[00]
-\end_layout
-
-\end_inset
-
-What integers are relatively prime to 1?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-All of them,
- since 
-\begin_inset Formula $\gcd\{a,1\}=1$
-\end_inset
-
- for any 
-\begin_inset Formula $a\in\mathbb{Z}$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc13[M00]
-\end_layout
-
-\end_inset
-
-By convention,
- we say that the greatest common divisor of 0 and 
-\begin_inset Formula $n$
-\end_inset
-
- is 
-\begin_inset Formula $|n|$
-\end_inset
-
-.
- What integers are relatively prime to 0?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-Only 1 and 
-\begin_inset Formula $-1$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc14[12]
-\end_layout
-
-\end_inset
-
-If 
-\begin_inset Formula $x\bmod3=2$
-\end_inset
-
- and 
-\begin_inset Formula $x\bmod5=3$
-\end_inset
-
-,
- what is 
-\begin_inset Formula $x\bmod15$
-\end_inset
-
-?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-This is a simple Diophantine equation,
- there exist 
-\begin_inset Formula $n,m\in\mathbb{Z}$
-\end_inset
-
- such that
-\begin_inset Formula 
-\[
-x=3n+2=5m+3,
-\]
-
-\end_inset
-
-so 
-\begin_inset Formula $3n-5m=1$
-\end_inset
-
- and one solution is 
-\begin_inset Formula $n=2$
-\end_inset
-
- and 
-\begin_inset Formula $m=1$
-\end_inset
-
-.
- In this case 
-\begin_inset Formula $x\bmod15=(3n+2)\bmod15=8$
-\end_inset
-
-,
- and the Chinese remainder theorem tells us that this is the only possibility.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc15[10]
-\end_layout
-
-\end_inset
-
-Prove that 
-\begin_inset Formula $z(x\bmod y)=(zx)\bmod(zy)$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-If 
-\begin_inset Formula $z=0$
-\end_inset
-
-,
- then 
-\begin_inset Formula $z(x\bmod y)=0=0\bmod0=(zx)\bmod(zy)$
-\end_inset
-
-.
- If 
-\begin_inset Formula $y=0$
-\end_inset
-
-,
- 
-\begin_inset Formula $z(x\bmod y)=zx=(zx)\bmod(zy)$
-\end_inset
-
-.
- If 
-\begin_inset Formula $y,z\neq0$
-\end_inset
-
-,
-\begin_inset Formula 
-\[
-(zx)\bmod(zy)=zx-zy\left\lfloor \frac{zx}{zy}\right\rfloor =zx-zy\left\lfloor \frac{x}{y}\right\rfloor =z\left(x-y\left\lfloor \frac{x}{y}\right\rfloor \right)=z(x\bmod y).
-\]
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc16[M10]
-\end_layout
-
-\end_inset
-
-Assume that 
-\begin_inset Formula $y>0$
-\end_inset
-
-.
- Show that if 
-\begin_inset Formula $(x-z)/y$
-\end_inset
-
- is an integer and if 
-\begin_inset Formula $0\leq z<y$
-\end_inset
-
-,
- then 
-\begin_inset Formula $z=x\bmod y$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-In this case,
- there exists an integer 
-\begin_inset Formula $k$
-\end_inset
-
- such that 
-\begin_inset Formula $x-z=ky$
-\end_inset
-
-,
- so 
-\begin_inset Formula $z=x-ky$
-\end_inset
-
- and we just have to show that 
-\begin_inset Formula $k=\lfloor\frac{x}{y}\rfloor$
-\end_inset
-
-.
- But since 
-\begin_inset Formula $0\leq z<y$
-\end_inset
-
-,
- 
-\begin_inset Formula $0\leq\frac{z}{y}<1$
-\end_inset
-
- and,
- multiplying this formula by 
-\begin_inset Formula $-1$
-\end_inset
-
- and adding 
-\begin_inset Formula $\frac{x}{y}$
-\end_inset
-
-,
- 
-\begin_inset Formula $\frac{x}{y}-1<\frac{x}{y}-\frac{z}{y}=k\leq\frac{x}{y}$
-\end_inset
-
-,
- which means that 
-\begin_inset Formula $k=\lfloor\frac{x}{y}\rfloor$
-\end_inset
-
- by Exercise 3.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc19[M10]
-\end_layout
-
-\end_inset
-
-(
-\emph on
-Law of inverses.
-\emph default
-) If 
-\begin_inset Formula $n\bot m$
-\end_inset
-
-,
- there is an integer 
-\begin_inset Formula $n'$
-\end_inset
-
- such that 
-\begin_inset Formula $nn'\equiv1$
-\end_inset
-
- (modulo 
-\begin_inset Formula $m$
-\end_inset
-
-).
- Prove this,
- using the extension of Euclid's algorithm (Algorithm 1.2.1E).
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-We already know that the algorithm terminates and that returns 
-\begin_inset Formula $a,b\in\mathbb{Z}$
-\end_inset
-
- such that 
-\begin_inset Formula $am+bn=d\coloneqq\gcd\{n,m\}$
-\end_inset
-
-.
- But in this case 
-\begin_inset Formula $d=1$
-\end_inset
-
-,
- so 
-\begin_inset Formula $bn=d-am=1-am=1$
-\end_inset
-
- (modulo 
-\begin_inset Formula $m$
-\end_inset
-
-).
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc22[M10]
-\end_layout
-
-\end_inset
-
-Give an example to show that Law B is not always true if 
-\begin_inset Formula $a$
-\end_inset
-
- is not relatively prime to 
-\begin_inset Formula $m$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-Let 
-\begin_inset Formula $a=b=3$
-\end_inset
-
-,
- 
-\begin_inset Formula $x=2$
-\end_inset
-
-,
- 
-\begin_inset Formula $y=4$
-\end_inset
-
-,
- and 
-\begin_inset Formula $m=6$
-\end_inset
-
- in Law B,
- then 
-\begin_inset Formula $ax=6\equiv12=by$
-\end_inset
-
- and 
-\begin_inset Formula $a=3=b$
-\end_inset
-
- but 
-\begin_inset Formula $x=2\not\equiv4=y$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc23[M10]
-\end_layout
-
-\end_inset
-
-Give an example to show that Law D is not always true if 
-\begin_inset Formula $r$
-\end_inset
-
- is not relatively prime to 
-\begin_inset Formula $s$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-If 
-\begin_inset Formula $a=6$
-\end_inset
-
-,
- 
-\begin_inset Formula $b=12$
-\end_inset
-
-,
- 
-\begin_inset Formula $r=3$
-\end_inset
-
-,
- and 
-\begin_inset Formula $s=6$
-\end_inset
-
-,
- in Law D,
- then 
-\begin_inset Formula $a=6\equiv12=b$
-\end_inset
-
- modulo 
-\begin_inset Formula $r=3$
-\end_inset
-
- and modulo 
-\begin_inset Formula $s=6$
-\end_inset
-
- but not modulo 
-\begin_inset Formula $rs=18$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc24[M20]
-\end_layout
-
-\end_inset
-
-To what extent can Laws A,
- B,
- C,
- and D be generalized to apply to arbitrary real numbers instead of integers?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-Let's check the laws one by one.
-\end_layout
-
-\begin_layout Description
-Law
-\begin_inset space ~
-\end_inset
-
-A If 
-\begin_inset Formula $a,b,x,y,m\in\mathbb{R}$
-\end_inset
-
-,
- if 
-\begin_inset Formula $m=0$
-\end_inset
-
- the law is trivial.
- If not,
- we have 
-\begin_inset Formula $a-m\lfloor\frac{a}{m}\rfloor=b-m\lfloor\frac{b}{m}\rfloor$
-\end_inset
-
- and 
-\begin_inset Formula $x-m\lfloor\frac{x}{m}\rfloor=y-m\lfloor\frac{y}{m}\rfloor$
-\end_inset
-
-,
- and for addition
-\begin_inset Formula 
-\begin{multline*}
-a+x-m\left\lfloor \frac{a+x}{m}\right\rfloor =b+y-m\left\lfloor \frac{b+y}{m}\right\rfloor \iff\\
-\iff m\left(\left\lfloor \frac{a+x}{m}\right\rfloor -\left\lfloor \frac{b+y}{m}\right\rfloor \right)=\\
-=a+x-b-y=m\left(\left\lfloor \frac{a}{m}\right\rfloor +\left\lfloor \frac{x}{m}\right\rfloor -\left\lfloor \frac{b}{m}\right\rfloor -\left\lfloor \frac{y}{m}\right\rfloor \right)\iff\\
-\iff\left\lfloor \frac{a+x}{m}\right\rfloor -\left\lfloor \frac{a}{m}\right\rfloor -\left\lfloor \frac{x}{m}\right\rfloor =\left\lfloor \frac{b+y}{m}\right\rfloor -\left\lfloor \frac{b}{m}\right\rfloor -\left\lfloor \frac{y}{m}\right\rfloor .
-\end{multline*}
-
-\end_inset
-
-Now,
-\begin_inset Formula 
-\[
-\left\lfloor \frac{a}{m}\right\rfloor +\left\lfloor \frac{x}{m}\right\rfloor \leq\frac{a+x}{m}<\left\lfloor \frac{a}{m}\right\rfloor +\left\lfloor \frac{x}{m}\right\rfloor +2,
-\]
-
-\end_inset
-
-so 
-\begin_inset Formula $\left\lfloor \frac{a+x}{m}\right\rfloor $
-\end_inset
-
- is either 
-\begin_inset Formula $\left\lfloor \frac{a}{m}\right\rfloor +\left\lfloor \frac{x}{m}\right\rfloor $
-\end_inset
-
- or 
-\begin_inset Formula $\left\lfloor \frac{a}{m}\right\rfloor +\left\lfloor \frac{x}{m}\right\rfloor +1$
-\end_inset
-
-,
- and similarly for 
-\begin_inset Formula $b$
-\end_inset
-
- and 
-\begin_inset Formula $y$
-\end_inset
-
-.
- But
-\begin_inset Formula 
-\begin{multline*}
-\left\lfloor \frac{a+x}{m}\right\rfloor =\left\lfloor \frac{a}{m}\right\rfloor +\left\lfloor \frac{x}{m}\right\rfloor \iff\left\lfloor \frac{a}{m}\right\rfloor +\left\lfloor \frac{x}{m}\right\rfloor \leq\frac{a+x}{m}<\left\lfloor \frac{a}{m}\right\rfloor +\left\lfloor \frac{x}{m}\right\rfloor +1\iff\\
-\iff a+x<m\left\lfloor \frac{a}{m}\right\rfloor +m\left\lfloor \frac{x}{m}\right\rfloor +m\iff\\
-\iff(a\bmod m)+(x\bmod m)=(b\bmod m)+(x\bmod m)<m\iff\\
-\iff\left\lfloor \frac{b+y}{m}\right\rfloor =\left\lfloor \frac{b}{m}\right\rfloor +\left\lfloor \frac{y}{m}\right\rfloor ,
-\end{multline*}
-
-\end_inset
-
-which proves the equality in the last line of the first formula.
- For subtraction,
- we have to see that 
-\begin_inset Formula $-x\equiv-y\bmod m$
-\end_inset
-
- and so 
-\begin_inset Formula $a-x=a+(-x)\equiv a+(-y)=a-y\pmod m$
-\end_inset
-
-.
- If 
-\begin_inset Formula $x\bmod m=0$
-\end_inset
-
-,
- then 
-\begin_inset Formula $\frac{x}{m}\in\mathbb{Z}$
-\end_inset
-
-,
- so 
-\begin_inset Formula $-\frac{x}{m}\in\mathbb{Z}$
-\end_inset
-
- and 
-\begin_inset Formula $-x\bmod m=0$
-\end_inset
-
-,
- and similarly 
-\begin_inset Formula $y\bmod m=0$
-\end_inset
-
- and so 
-\begin_inset Formula $-y\bmod m=0$
-\end_inset
-
-.
- Otherwise 
-\begin_inset Formula $\frac{x}{m}\notin\mathbb{Z}$
-\end_inset
-
-,
- so 
-\begin_inset Formula $\lceil\frac{x}{m}\rceil=\lfloor\frac{x}{m}\rfloor+1$
-\end_inset
-
-,
- but by Exercise 4,
- 
-\begin_inset Formula 
-\begin{align*}
--x\bmod m & =-x-m\left\lfloor \frac{-x}{m}\right\rfloor =-x+m\left\lceil \frac{x}{m}\right\rceil =-x+m\left\lfloor \frac{x}{m}\right\rfloor +m=\\
- & =m-(x\bmod m)=m-(y\bmod m)=-y\bmod m.
-\end{align*}
-
-\end_inset
-
-For multiplication this doesn't hold;
- for example,
- if 
-\begin_inset Formula $m=2.5$
-\end_inset
-
-,
- 
-\begin_inset Formula $a=0.5$
-\end_inset
-
-,
- 
-\begin_inset Formula $b=-2$
-\end_inset
-
-,
- and 
-\begin_inset Formula $x=y=2.2$
-\end_inset
-
-,
- then 
-\begin_inset Formula $ax\bmod m=1.1\bmod2.5=1.1$
-\end_inset
-
- but 
-\begin_inset Formula $by\bmod m=-4.4\bmod2.5=0.6$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Description
-Law
-\begin_inset space ~
-\end_inset
-
-B Obviously 
-\begin_inset Formula $a$
-\end_inset
-
- and 
-\begin_inset Formula $m$
-\end_inset
-
- must be integers,
- otherwise the statement wouldn't make sense.
- Even then,
- however,
- if 
-\begin_inset Formula $a=1$
-\end_inset
-
-,
- 
-\begin_inset Formula $b=-2$
-\end_inset
-
-,
- 
-\begin_inset Formula $x=\frac{4}{3}$
-\end_inset
-
-,
- 
-\begin_inset Formula $y=\frac{7}{3}$
-\end_inset
-
-,
- and 
-\begin_inset Formula $m=3$
-\end_inset
-
-,
- then 
-\begin_inset Formula $ax=\frac{4}{3}\equiv-\frac{14}{3}=by$
-\end_inset
-
- and 
-\begin_inset Formula $a=1\equiv-2=b$
-\end_inset
-
-,
- but 
-\begin_inset Formula $\frac{4}{3}\not\equiv\frac{7}{3}$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Description
-Law
-\begin_inset space ~
-\end_inset
-
-C Using Exercise 15,
- for 
-\begin_inset Formula $a,b,m,n\in\mathbb{R}$
-\end_inset
-
- with 
-\begin_inset Formula $n\neq0$
-\end_inset
-
-,
-\begin_inset Formula 
-\begin{multline*}
-a\equiv b\bmod m\iff a\bmod m=b\bmod m\iff\\
-\iff an\bmod mn=n(a\bmod m)=n(b\bmod m)=bn\bmod mn\iff\\
-\iff an\equiv bn\bmod mn.
-\end{multline*}
-
-\end_inset
-
-Note that the second double implication would not hold in the left direction for 
-\begin_inset Formula $n=0$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Description
-Law
-\begin_inset space ~
-\end_inset
-
-D Here 
-\begin_inset Formula $r$
-\end_inset
-
- and 
-\begin_inset Formula $s$
-\end_inset
-
- must be integers.
- Then,
- if 
-\begin_inset Formula $a,b\in\mathbb{R}$
-\end_inset
-
- with 
-\begin_inset Formula $a\equiv b$
-\end_inset
-
- modulo 
-\begin_inset Formula $r$
-\end_inset
-
- and 
-\begin_inset Formula $s$
-\end_inset
-
-,
- if 
-\begin_inset Formula $r=0$
-\end_inset
-
- or 
-\begin_inset Formula $s=0$
-\end_inset
-
-,
- the statement is obvious,
- and otherwise 
-\begin_inset Formula $a-r\lfloor\frac{a}{r}\rfloor=b-r\lfloor\frac{b}{r}\rfloor$
-\end_inset
-
- and so 
-\begin_inset Formula $a-b=r(\lfloor\frac{a}{r}\rfloor-\lfloor\frac{b}{r}\rfloor)\eqqcolon d\in\mathbb{Z}$
-\end_inset
-
-.
- By Law A,
- since 
-\begin_inset Formula $a\equiv b$
-\end_inset
-
- and 
-\begin_inset Formula $b\equiv b$
-\end_inset
-
-,
- 
-\begin_inset Formula $d=a-b\equiv b-b=0$
-\end_inset
-
- modulo both 
-\begin_inset Formula $r$
-\end_inset
-
- and 
-\begin_inset Formula $s$
-\end_inset
-
-,
- so from Law D for integers we have 
-\begin_inset Formula $a-b\equiv0$
-\end_inset
-
- modulo 
-\begin_inset Formula $rs$
-\end_inset
-
- (assuming 
-\begin_inset Formula $r\bot s$
-\end_inset
-
-) and so 
-\begin_inset Formula $a\equiv b$
-\end_inset
-
- modulo 
-\begin_inset Formula $rs$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-Thus,
- Law A holds for addition and subtraction,
- Law C always holds,
- and Law D holds if we still maintain 
-\begin_inset Formula $r,s\in\mathbb{Z}$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc25[M02]
-\end_layout
-
-\end_inset
-
-Show that,
- according to Theorem F,
- 
-\begin_inset Formula $a^{p-1}\bmod p=[a\text{ is not a multiple of }p]$
-\end_inset
-
-,
- whenever 
-\begin_inset Formula $p$
-\end_inset
-
- is a prime number.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-If 
-\begin_inset Formula $a$
-\end_inset
-
- is a multiple of 
-\begin_inset Formula $p$
-\end_inset
-
-,
- then so is 
-\begin_inset Formula $a^{p-1}$
-\end_inset
-
- and 
-\begin_inset Formula $a^{p-1}\bmod p=0$
-\end_inset
-
-.
- Otherwise 
-\begin_inset Formula $a\bot p$
-\end_inset
-
-,
- so we can cancel out in 
-\begin_inset Formula $a^{p}\equiv a\pmod p$
-\end_inset
-
- to get 
-\begin_inset Formula $a^{p-1}\equiv1\pmod p$
-\end_inset
-
- and so 
-\begin_inset Formula $a^{p-1}\bmod p=1\bmod p=1$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc28[M25]
-\end_layout
-
-\end_inset
-
-Show that the method used to prove Theorem F can be used to prove the following extension,
- called 
-\emph on
-Euler's theorem
-\emph default
-:
- 
-\begin_inset Formula $a^{\varphi(m)}\equiv1\pmod m$
-\end_inset
-
-,
- for 
-\emph on
-any
-\emph default
- positive integer 
-\begin_inset Formula $m$
-\end_inset
-
-,
- when 
-\begin_inset Formula $a\bot m$
-\end_inset
-
-.
- (In particular,
- the number 
-\begin_inset Formula $n'$
-\end_inset
-
- in exercise 19 may be taken to be 
-\begin_inset Formula $n^{\varphi(m)-1}\bmod m$
-\end_inset
-
-.)
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-First,
- 
-\begin_inset Formula $\gcd\{x,y\}\equiv\gcd\{x+ky,y\}$
-\end_inset
-
- for any 
-\begin_inset Formula $x,y,k\in\mathbb{Z}$
-\end_inset
-
-,
- because if 
-\begin_inset Formula $z\mid x$
-\end_inset
-
- and 
-\begin_inset Formula $z\mid y$
-\end_inset
-
- then obviously 
-\begin_inset Formula $z\mid x+ky$
-\end_inset
-
- and if 
-\begin_inset Formula $z\mid x+ky$
-\end_inset
-
- and 
-\begin_inset Formula $z\mid y$
-\end_inset
-
- then 
-\begin_inset Formula $z\mid x$
-\end_inset
-
-,
- so the set of common divisors is the same.
- It's also clear that if 
-\begin_inset Formula $x\bot z$
-\end_inset
-
- and 
-\begin_inset Formula $y\bot z$
-\end_inset
-
- then 
-\begin_inset Formula $xy\bot z$
-\end_inset
-
-,
- since neither 
-\begin_inset Formula $x$
-\end_inset
-
- nor 
-\begin_inset Formula $y$
-\end_inset
-
- has common divisors with 
-\begin_inset Formula $z$
-\end_inset
-
- and so neither does 
-\begin_inset Formula $xy$
-\end_inset
-
-.
- 
-\end_layout
-
-\begin_layout Standard
-With these lemmas,
- we are ready to prove the statement.
- Let 
-\begin_inset Formula 
-\[
-\{x_{1},\dots,x_{\varphi(m)}\}\coloneqq\{x\in\{0,\dots,m-1\}\mid x\bot m\},
-\]
-
-\end_inset
-
- then the 
-\begin_inset Formula $x_{i}a\bmod m$
-\end_inset
-
- are all distinct,
- for if 
-\begin_inset Formula $x_{i}a\bmod m=x_{j}a\bmod m$
-\end_inset
-
-,
- since 
-\begin_inset Formula $a\bot m$
-\end_inset
-
-,
- then 
-\begin_inset Formula $x_{i}=x_{i}\bmod m=x_{j}\bmod m=x_{j}$
-\end_inset
-
-.
- Since 
-\begin_inset Formula $x_{i},a\bot m$
-\end_inset
-
-,
- 
-\begin_inset Formula $x_{i}a\bot m$
-\end_inset
-
- and also 
-\begin_inset Formula $(x_{i}a\bmod m)\bot m$
-\end_inset
-
-,
- so all the 
-\begin_inset Formula $x_{i}a\bmod m$
-\end_inset
-
- are different and relatively prime to 
-\begin_inset Formula $m$
-\end_inset
-
- and therefore 
-\begin_inset Formula $\{x_{i}a\bmod m\}_{i}=\{x_{i}\}_{i}$
-\end_inset
-
-.
- Thus
-\begin_inset Formula 
-\[
-\prod_{i}x_{i}\equiv\prod_{i}(x_{i}a\bmod m)\equiv\prod_{i}x_{i}a\equiv a^{\varphi(m)}\prod_{i}x_{i}\pmod m,
-\]
-
-\end_inset
-
-and therefore using Law B with the fact that 
-\begin_inset Formula $\prod_{i}x_{i}\bot m$
-\end_inset
-
- gives us the result.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc34[M21]
-\end_layout
-
-\end_inset
-
-What conditions on the real number 
-\begin_inset Formula $b>1$
-\end_inset
-
- are necessary and sufficient to guarantee that 
-\begin_inset Formula $\lfloor\log_{b}x\rfloor=\left\lfloor \log_{b}\lfloor x\rfloor\right\rfloor $
-\end_inset
-
- for all real 
-\begin_inset Formula $x\geq1$
-\end_inset
-
-?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-It happens if and only if 
-\begin_inset Formula $b\in\mathbb{Z}$
-\end_inset
-
-.
- To prove this,
- observe that,
- since the logarithm in base 
-\begin_inset Formula $b>1$
-\end_inset
-
- is strictly increasing,
- 
-\begin_inset Formula $\log_{b}\lfloor x\rfloor<\log_{b}x$
-\end_inset
-
-,
- so 
-\begin_inset Formula $\lfloor\log_{b}x\rfloor\neq\left\lfloor \log_{b}\lfloor x\rfloor\right\rfloor $
-\end_inset
-
- if and only if 
-\begin_inset Formula $\left\lfloor \log_{b}\lfloor x\rfloor\right\rfloor <\lfloor\log_{b}x\rfloor$
-\end_inset
-
-,
- that is,
- if there exists an integer 
-\begin_inset Formula $k$
-\end_inset
-
- (namely 
-\begin_inset Formula $\lfloor\log_{b}x\rfloor$
-\end_inset
-
-) such that 
-\begin_inset Formula $\log_{b}\lfloor x\rfloor<k\leq\log_{b}x$
-\end_inset
-
-,
- if and only if 
-\begin_inset Formula $\lfloor x\rfloor<b^{k}\leq x$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-If 
-\begin_inset Formula $b\in\mathbb{Z}$
-\end_inset
-
-,
- 
-\begin_inset Formula $b^{k}\in\mathbb{Z}$
-\end_inset
-
- as well,
- so that would mean that 
-\begin_inset Formula $\lfloor x\rfloor+1\leq b^{k}\leq x\#$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-If 
-\begin_inset Formula $b\notin\mathbb{Z}$
-\end_inset
-
-,
- we just have to set 
-\begin_inset Formula $x=b$
-\end_inset
-
- and 
-\begin_inset Formula $k=1$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc35[M20]
-\end_layout
-
-\end_inset
-
-Given that 
-\begin_inset Formula $m$
-\end_inset
-
- and 
-\begin_inset Formula $n$
-\end_inset
-
- are integers and 
-\begin_inset Formula $n>0$
-\end_inset
-
-,
- prove that
-\begin_inset Formula 
-\[
-\lfloor(x+m)/n\rfloor=\left\lfloor (\lfloor x\rfloor+m)/n\right\rfloor 
-\]
-
-\end_inset
-
-for all real 
-\begin_inset Formula $x$
-\end_inset
-
-.
- (When 
-\begin_inset Formula $m=0$
-\end_inset
-
-,
- we have an important special case.) Does an analogous result hold for the ceiling function?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-Clearly 
-\begin_inset Formula $\left\lfloor \frac{\lfloor x\rfloor+m}{n}\right\rfloor \leq\left\lfloor \frac{x+m}{n}\right\rfloor $
-\end_inset
-
-.
- If this inequality were strict,
- however,
- there would be an integer 
-\begin_inset Formula $k$
-\end_inset
-
- such that 
-\begin_inset Formula $\frac{\lfloor x\rfloor+m}{n}<k\leq\frac{x+m}{n}$
-\end_inset
-
-,
- and so 
-\begin_inset Formula $\lfloor x\rfloor<nk-m\leq x$
-\end_inset
-
-,
- but this is impossible because 
-\begin_inset Formula $nk-m\in\mathbb{Z}$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-For the ceiling,
- clearly 
-\begin_inset Formula $\left\lceil \frac{\lceil x\rceil+m}{n}\right\rceil \geq\left\lceil \frac{x+m}{n}\right\rceil $
-\end_inset
-
-,
- but if this inequality were strict,
- there would be an integer 
-\begin_inset Formula $k$
-\end_inset
-
- (namely 
-\begin_inset Formula $\left\lceil \frac{x+m}{n}\right\rceil $
-\end_inset
-
-) such that 
-\begin_inset Formula $\frac{x+m}{n}\leq k<\frac{\lceil x\rceil+m}{n}$
-\end_inset
-
-,
- and so 
-\begin_inset Formula $x\leq nk-m<\lceil x\rceil$
-\end_inset
-
-,
- an absurdity because 
-\begin_inset Formula $nk-m\in\mathbb{Z}$
-\end_inset
-
-.
-\end_layout
-
-\end_body
-\end_document
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