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Question Number 1937 by Rasheed Soomro last updated on 25/Oct/15

$$\bullet Is{}^{\prime }{\iff }^{\prime }necessaryandsuficientfortwo$$ $$inequalitiestobeequivalent?$$ $$\bullet If\mathbf{a}>\mathbf{b}:$$ $$Are\mathbf{A}>\mathbf{B}and\mathbf{A}+\mathbf{a}>\mathbf{B}+\mathbf{b}equivalent?$$

Answered by 123456 last updated on 25/Oct/15

$$(a,b,\mathrm{A},\mathrm{B})\in {\mathbb{R}}^4$$ $$a>b\wedge \mathrm{A}>\mathrm{B}\Rightarrow \mathrm{A}+a>\mathrm{B}+b$$ $$----------$$ $$a>b$$ $$\mathrm{A}+a>\mathrm{A}+b(+\mathrm{A})$$ $$\mathrm{B}+a>\mathrm{B}+b(+\mathrm{B})$$ $$\mathrm{A}>\mathrm{B}$$ $$\mathrm{A}+a>\mathrm{B}+a(+a)$$ $$\mathrm{A}+b>\mathrm{B}+b(+b)$$ $$\mathrm{A}+a>\mathrm{B}+a>\mathrm{B}+b$$ $$-----$$ $$a>b\wedge \mathrm{A}+a>\mathrm{B}+b\nRightarrow \mathrm{A}>\mathrm{B}$$ $$------$$ $$\mathrm{A}=\mathrm{B}=0\Rightarrow a>b\Rightarrow \mathrm{A}+a=a>b=\mathrm{B}+b$$

Commented byRasheed Soomro last updated on 25/Oct/15

$$Thatmeansadding\mathbf{same}-\mathbf{sense}inequalitytogiven$$ $$inequality{doesn}^{\prime }tyeildequivalentinequality.$$ $$$$ $$Equivalentinequalitymaybeachieved\mathbf{only}by:$$ $$\bullet Adding\left(Subtracting\right)an\underset{-}{equation}to\left(from\right)bothsidesofaninequality.$$ $$\bullet Multiplying/Dividingan\underset{-}{equation}tobothsidesofaninequality.$$ $$AmIcorrect?$$

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

$$Ofcoursesir!$$

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