#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
\fontencoding auto
\font_roman "default" "default"
\font_sans "default" "default"
\font_typewriter "default" "default"
\font_math "auto" "auto"
\font_default_family default
\use_non_tex_fonts false
\font_sc false
\font_roman_osf false
\font_sans_osf false
\font_typewriter_osf false
\font_sf_scale 100 100
\font_tt_scale 100 100
\use_microtype false
\use_dash_ligatures true
\graphics default
\default_output_format default
\output_sync 0
\bibtex_command default
\index_command default
\float_placement class
\float_alignment class
\paperfontsize default
\spacing single
\use_hyperref false
\papersize default
\use_geometry false
\use_package amsmath 1
\use_package amssymb 1
\use_package cancel 1
\use_package esint 1
\use_package mathdots 1
\use_package mathtools 1
\use_package mhchem 1
\use_package stackrel 1
\use_package stmaryrd 1
\use_package undertilde 1
\cite_engine basic
\cite_engine_type default
\biblio_style plain
\use_bibtopic false
\use_indices false
\paperorientation portrait
\suppress_date false
\justification true
\use_refstyle 1
\use_formatted_ref 0
\use_minted 0
\use_lineno 0
\index Index
\shortcut idx
\color #008000
\end_index
\secnumdepth 3
\tocdepth 3
\paragraph_separation indent
\paragraph_indentation default
\is_math_indent 0
\math_numbering_side default
\quotes_style english
\dynamic_quotes 0
\papercolumns 1
\papersides 1
\paperpagestyle default
\tablestyle default
\tracking_changes false
\output_changes false
\change_bars false
\postpone_fragile_content false
\html_math_output 0
\html_css_as_file 0
\html_be_strict false
\docbook_table_output 0
\docbook_mathml_prefix 1
\end_header

\begin_body

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
exerc5[10]
\end_layout

\end_inset

Compute 
\begin_inset Formula $(17/120)+(-27/70)$
\end_inset

 by the method recommended in the text.
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
answer 
\end_layout

\end_inset

We have 
\begin_inset Formula $d_{1}=\gcd\{120,70\}=10$
\end_inset

,
 
\begin_inset Formula $t=17\cdot\frac{70}{10}-27\cdot\frac{120}{10}=119-324=-205$
\end_inset

,
 
\begin_inset Formula $d_{2}=\gcd\{-205,10\}=5$
\end_inset

,
 and so the answer is 
\begin_inset Formula $-\frac{205}{5}\Big/\left(\frac{120}{10}\frac{70}{5}\right)=-\frac{41}{168}$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
rexerc6[M23]
\end_layout

\end_inset

Show that 
\begin_inset Formula $u\bot u'$
\end_inset

 and 
\begin_inset Formula $v\bot v'$
\end_inset

 implies 
\begin_inset Formula $\gcd\{uv'+vu',u'v'\}=d_{1}d_{2}$
\end_inset

,
 where 
\begin_inset Formula $d_{1}=\gcd\{u',v'\}$
\end_inset

 and 
\begin_inset Formula $d_{2}=\gcd\{d_{1},u(v'/d_{1})+v(u'/d_{1})\}$
\end_inset

.
 (Hence if 
\begin_inset Formula $d_{1}=1$
\end_inset

 we have 
\begin_inset Formula $(uv'+vu')\bot u'v'$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
answer 
\end_layout

\end_inset


\begin_inset Note Greyedout
status open

\begin_layout Plain Layout
(I had to look up the solution.) 
\end_layout

\end_inset

Let 
\begin_inset Formula $u''\coloneqq u'/d_{1}$
\end_inset

 and 
\begin_inset Formula $v''\coloneqq v'/d_{1}$
\end_inset

,
 we just need to prove that
\begin_inset Formula 
\[
\gcd\{uv''+vu'',u''v''d_{1}\}=d_{2}=\gcd\{uv''+vu'',d_{1}\},
\]

\end_inset

as multiplying the first equality by 
\begin_inset Formula $d_{1}$
\end_inset

 gives us the required answer.
 Obviously 
\begin_inset Formula $d_{2}\mid uv''+vu'',u''v''d_{1}$
\end_inset

,
 and we have to see that any integer 
\begin_inset Formula $d$
\end_inset

 that divides both 
\begin_inset Formula $uv''+vu''$
\end_inset

 and 
\begin_inset Formula $u''v''d_{1}$
\end_inset

 also divides 
\begin_inset Formula $d_{1}$
\end_inset

 and therefore 
\begin_inset Formula $d_{2}$
\end_inset

.
 Let 
\begin_inset Formula $p$
\end_inset

 be a prime factor of 
\begin_inset Formula $d$
\end_inset

,
 because 
\begin_inset Formula $u\bot u'$
\end_inset

 and therefore 
\begin_inset Formula $u\bot d_{1},u''$
\end_inset

,
 if 
\begin_inset Formula $p\mid u''$
\end_inset

 then 
\begin_inset Formula $p\mid uv''$
\end_inset

 but 
\begin_inset Formula $p\nmid u$
\end_inset

,
 so 
\begin_inset Formula $p\mid v''$
\end_inset

 and 
\begin_inset Formula $u''$
\end_inset

 and 
\begin_inset Formula $v''$
\end_inset

 are not coprime,
\begin_inset Formula $\#$
\end_inset

 and similarly 
\begin_inset Formula $p\nmid v''$
\end_inset

.
 That means that 
\begin_inset Formula $d$
\end_inset

 doesn't have any common factors with either 
\begin_inset Formula $u''$
\end_inset

 and 
\begin_inset Formula $v''$
\end_inset

,
 so 
\begin_inset Formula $d\mid d_{1}$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
rexerc8[22]
\end_layout

\end_inset

Discuss using 
\begin_inset Formula $(1/0)$
\end_inset

 and 
\begin_inset Formula $(-1/0)$
\end_inset

 as representations for 
\begin_inset Formula $\infty$
\end_inset

 and 
\begin_inset Formula $-\infty$
\end_inset

,
 and/or as representations of overflow.
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
answer 
\end_layout

\end_inset

Mathematically they are not that different,
 since 
\begin_inset Formula $\infty$
\end_inset

,
 when treated as a number,
 represents the concept of a number larger than any other,
 or an arbitrarily large number in an asymptotic way,
 while overflow represents a number larger than any 
\emph on
representable
\emph default
 number.
 Mediant rounding would round to 
\begin_inset Formula $\pm1/0$
\end_inset

 for numbers with 
\begin_inset Formula $|x|\geq2^{p}$
\end_inset

,
 which makes sense.
\end_layout

\begin_layout Standard
Using these representations,
 multiplying a number by 
\begin_inset Formula $\infty$
\end_inset

 gives 
\begin_inset Formula $\infty$
\end_inset

 if the number is positive or 
\begin_inset Formula $-\infty$
\end_inset

 if it's negative,
 or vice versa for 
\begin_inset Formula $-\infty$
\end_inset

,
 and 
\begin_inset Formula $\pm\infty\cdot0=(0/0)$
\end_inset

,
 an indeterminate value.
 Dividing 
\begin_inset Formula $\pm\infty$
\end_inset

 by some number yields a similar result,
 with 
\begin_inset Formula $\pm\infty/\pm\infty$
\end_inset

 being indeterminate and 
\begin_inset Formula $\pm\infty/0=\pm\infty$
\end_inset

,
 which works well as a convention,
 and dividing by 
\begin_inset Formula $\pm\infty$
\end_inset

 is equivalent to multiplying by 0.
\end_layout

\begin_layout Standard
Addition and subtraction require a bit more attention:
 if we do 
\begin_inset Formula $\pm\frac{1}{0}+\frac{a}{b}$
\end_inset

,
 we would have 
\begin_inset Formula $d_{1}=b$
\end_inset

 (if 
\begin_inset Formula $b=1$
\end_inset

 then we shortcut to get 
\begin_inset Formula $\pm\frac{1}{0}$
\end_inset

 as the result) and then 
\begin_inset Formula $t=\pm1$
\end_inset

,
 
\begin_inset Formula $d_{2}=1$
\end_inset

,
 and the result is 
\begin_inset Formula $\pm\frac{1}{0}$
\end_inset

,
 as expected.
 The exception is if 
\begin_inset Formula $\frac{a}{b}=\pm\frac{1}{0}$
\end_inset

;
 then 
\begin_inset Formula $d_{1}=0$
\end_inset

 and calculating 
\begin_inset Formula $t$
\end_inset

 would result in a division by 0,
 and the obvious procedure gives us 
\begin_inset Formula $\frac{0}{0}$
\end_inset

 even if both infinities have the same sign,
 so we have to consider this a special case to get the correct result,
 namely 
\begin_inset Formula $\pm\infty\pm\infty=\pm\infty$
\end_inset

 and 
\begin_inset Formula $\pm\infty\mp\infty=0/0$
\end_inset

.
\end_layout

\begin_layout Standard
We also need a special case so that additions and subtractions involving 
\begin_inset Formula $0/0$
\end_inset

 return 
\begin_inset Formula $0/0$
\end_inset

.
\end_layout

\begin_layout Standard
Note,
 however,
 that having 
\begin_inset Formula $x/(\pm1/0)=0$
\end_inset

 for 
\begin_inset Formula $x\neq0$
\end_inset

 would be a hazard in the case that we are representing overflow,
 as we may inadvertently discard the overflow and get a potentially very inaccurate result,
 so sometimes it may be better to use 
\begin_inset Formula $0/0$
\end_inset

 for overflow.
\end_layout

\end_body
\end_document