exercise:D2ad6faccc: Difference between revisions

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\newcommand{\mathds}{\mathbb}
\newcommand{\mathds}{\mathbb}
</math></div>
</math></div>
If <math>a  <  b  <  c</math>, prove that
If <math>a  <  b  <  c</math>, prove that


<math display="block">
<math display="block">
\cond{\frac bc  <  \frac ba & \mbox{if $a  >  0$}}
\frac bc  <  \frac ba, \quad  \mbox{if $a  >  0$}
,
,
</math>
</math>


<math display="block">
<math display="block">
\cond{\frac bc >  \frac ba & \mbox{if $c  <  0$}}.
\frac{b}{c} >  \frac{b}{a},\quad  \mbox{if $c  <  0$}.
</math>
</math>

Latest revision as of 23:34, 22 November 2024

[math] \newcommand{\ex}[1]{\item } \newcommand{\sx}{\item} \newcommand{\x}{\sx} \newcommand{\sxlab}[1]{} \newcommand{\xlab}{\sxlab} \newcommand{\prov}[1] {\quad #1} \newcommand{\provx}[1] {\quad \mbox{#1}} \newcommand{\intext}[1]{\quad \mbox{#1} \quad} \newcommand{\R}{\mathrm{\bf R}} \newcommand{\Q}{\mathrm{\bf Q}} \newcommand{\Z}{\mathrm{\bf Z}} \newcommand{\C}{\mathrm{\bf C}} \newcommand{\dt}{\textbf} \newcommand{\goesto}{\rightarrow} \newcommand{\ddxof}[1]{\frac{d #1}{d x}} \newcommand{\ddx}{\frac{d}{dx}} \newcommand{\ddt}{\frac{d}{dt}} \newcommand{\dydx}{\ddxof y} \newcommand{\nxder}[3]{\frac{d^{#1}{#2}}{d{#3}^{#1}}} \newcommand{\deriv}[2]{\frac{d^{#1}{#2}}{dx^{#1}}} \newcommand{\dist}{\mathrm{distance}} \newcommand{\arccot}{\mathrm{arccot\:}} \newcommand{\arccsc}{\mathrm{arccsc\:}} \newcommand{\arcsec}{\mathrm{arcsec\:}} \newcommand{\arctanh}{\mathrm{arctanh\:}} \newcommand{\arcsinh}{\mathrm{arcsinh\:}} \newcommand{\arccosh}{\mathrm{arccosh\:}} \newcommand{\sech}{\mathrm{sech\:}} \newcommand{\csch}{\mathrm{csch\:}} \newcommand{\conj}[1]{\overline{#1}} \newcommand{\mathds}{\mathbb} [/math]

If [math]a \lt b \lt c[/math], prove that

[[math]] \frac bc \lt \frac ba, \quad \mbox{if $a \gt 0$} , [[/math]]

[[math]] \frac{b}{c} \gt \frac{b}{a},\quad \mbox{if $c \lt 0$}. [[/math]]