Answer by Kim Jong Un for The floor and ceiling functions are equal for...
With $A=\{k:k\in\mathbb{Z},k\leq x\}$, then $x\in A$ because $x\in\mathbb{Z}$ and $x\leq x$. Moreover $x=\max A$ because (i) $x\in A$, (ii) if $k\in A$, then $k\leq x$. You can do the same for...
View ArticleAnswer by user175994 for The floor and ceiling functions are equal for...
That $x \in \{ k \leq x\}$ follows since $x \leq x$. Similarly for $x \in \{ k \geq x\}$. To show that $x = max\{ k \leq x\}$, suppose not. Then there is some $y \in \{ k \leq x\}$ such that $y >...
View ArticleThe floor and ceiling functions are equal for integer numbers
The exerciseProof the following directly: Let $x \in \mathbb{R}$. If $x \in \mathbb{Z}$, then $\lfloor x \rfloor = \lceil x \rceil$.My problemI mostly fail completely on the formal part of any kind of...
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