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Is my description of the floor function correct?

$$ f = \begin{cases} \mathbb{R} \rightarrow \mathbb{Z} \\ x \mapsto z = \inf(x) \end{cases} $$

Explanation:

The floor function maps a real number $x$ to the smallest whole number less than or equal to $x$. The infimum of is the largest lower bound of a set. The above stated function $f$ maps a real number $x$ to the largest whole number $z$ for which $z \leq x$, which is the definition of the floor function. Hence $f = \operatorname{floor}$.

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    $\begingroup$ $\inf$ should be defined on a set $\endgroup$ – J. W. Tanner Oct 2 '20 at 12:58
  • $\begingroup$ Agree with @J.W.Tanner, $\inf{x}=x$ if $x\in\mathbb{R}$, which makes your statement odd. $\endgroup$ – Weierstraß Ramirez Oct 2 '20 at 13:01
  • $\begingroup$ well but $x$ is not in $\mathbb{Z}$ $\endgroup$ – user2550228 Oct 2 '20 at 13:04
  • $\begingroup$ Doesn't the floor function map a real number $x$ to the largest whole number less than or equal to $x$? How about $f:x\mapsto \sup \{z\in\mathbb Z|z\le x\}$? $\endgroup$ – J. W. Tanner Oct 2 '20 at 13:06
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    $\begingroup$ @J.W.Tanner you are correct. Want to post it as answer, then I can upvote. at All: thank you for your comments, i really appreciate it. $\endgroup$ – user2550228 Oct 2 '20 at 13:14
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Inf should be defined on a set, not one real number.

Furthermore, the floor function maps a real number $x$

to the largest integer less than or equal to $x$,

so it could be defined as floor$(x)=\sup\{z\in\mathbb Z|z\le x\}$.

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