I was solving an excercise on calculus when i came up with this problem: Let $a,b\in\mathbb{R}$ if $a>8-b$ and $a>3$ can it be shown that $b<5$? I've given it some tries but it got me nowhere, could someone propose an idea?
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If $a=4$ and $b=8$, then $a>8-b$ and $a>3$. But $b\not<5$. – CY Aries Apr 04 '20 at 07:38
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We can only do this: $$3<a>8-b$$ This is not a chain of inequalities. It is not possible to derive a relation between $3$ and $8-b$, so $b<5$ cannot be concluded (we can e.g. take $b=6$ and $a=10$).
Parcly Taxel
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