Question Number 1937 by Rasheed Soomro last updated on 25/Oct/15 | ||
$$\bullet{Is}\:\:\:'\Leftrightarrow'\:\:{necessary}\:{and}\:{suficient}\:{for}\:{two} \\ $$ $${inequalities}\:{to}\:{be}\:{equivalent}? \\ $$ $$\bullet{If}\:\:\boldsymbol{\mathrm{a}}>\boldsymbol{\mathrm{b}}\:: \\ $$ $${Are}\:\:\boldsymbol{\mathrm{A}}>\boldsymbol{\mathrm{B}}\:{and}\:\boldsymbol{\mathrm{A}}+\boldsymbol{\mathrm{a}}\:>\:\boldsymbol{\mathrm{B}}+\boldsymbol{\mathrm{b}}\:{equivalent}? \\ $$ | ||
Answered by 123456 last updated on 25/Oct/15 | ||
$$\left({a},{b},\mathrm{A},\mathrm{B}\right)\in\mathbb{R}^{\mathrm{4}} \\ $$ $${a}>{b}\wedge\mathrm{A}>\mathrm{B}\Rightarrow\mathrm{A}+{a}>\mathrm{B}+{b} \\ $$ $$−−−−−−−−−− \\ $$ $${a}>{b}\: \\ $$ $$\mathrm{A}+{a}>\mathrm{A}+{b}\:\left(+\mathrm{A}\right) \\ $$ $$\mathrm{B}+{a}>\mathrm{B}+{b}\:\left(+\mathrm{B}\right) \\ $$ $$\mathrm{A}>\mathrm{B} \\ $$ $$\mathrm{A}+{a}>\mathrm{B}+{a}\:\left(+{a}\right) \\ $$ $$\mathrm{A}+{b}>\mathrm{B}+{b}\:\left(+{b}\right) \\ $$ $$\mathrm{A}+{a}>\mathrm{B}+{a}>\mathrm{B}+{b} \\ $$ $$−−−−− \\ $$ $${a}>{b}\wedge\mathrm{A}+{a}>\mathrm{B}+{b}\nRightarrow\mathrm{A}>\mathrm{B} \\ $$ $$−−−−−− \\ $$ $$\mathrm{A}=\mathrm{B}=\mathrm{0}\Rightarrow{a}>{b}\Rightarrow\mathrm{A}+{a}={a}>{b}=\mathrm{B}+{b} \\ $$ | ||
Commented byRasheed Soomro last updated on 25/Oct/15 | ||
$${That}\:{means}\:{adding}\:\boldsymbol{\mathrm{same}}−\boldsymbol{\mathrm{sense}}\:{inequality}\:{to}\:\:{given} \\ $$ $${inequality}\:{doesn}'{t}\:{yeild}\:{equivalent}\:{inequality}. \\ $$ $$ \\ $$ $${Equivalent}\:{inequality}\:{may}\:{be}\:{achieved}\:\boldsymbol{\mathrm{only}}\:\:{by}: \\ $$ $$\bullet{Adding}\:\left({Subtracting}\right){an}\:\underset{−} {{equation}}\:{to}\left({from}\right)\:{both}\:{sides}\:{of}\:{an}\:{inequality}. \\ $$ $$\bullet{Multiplying}/{Dividing}\:\:{an}\:\underset{−} {{equation}}\:{to}\:{both}\:{sides}\:{of}\:{an}\:{inequality}. \\ $$ $${Am}\:{I}\:{correct}? \\ $$ | ||
Commented byprakash jain last updated on 25/Oct/15 | ||
$$\mathrm{Multiplying}\:\mathrm{and}\:\mathrm{dividing}\:\mathrm{an}\:\mathrm{equatily}\:\mathrm{by} \\ $$ $$\mathrm{same}\:\mathrm{value}\:\mathrm{may}\:\mathrm{reverse}\:\mathrm{the}\:\mathrm{sign}\:\mathrm{and}\:\mathrm{equality}. \\ $$ | ||
Commented byRasheed Soomro last updated on 28/Oct/15 | ||
$${Of}\:{course}\:{sir}! \\ $$ | ||