Characteristics of Floor and Ceiling Function

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Theorem

Let $f: \R \to \Z$ be an integer-valued function which satisfies both of the following:

$(1): \quad \map f {x + 1} = \map f x + 1$
$(2): \quad \forall n \in \Z_{> 0}: \map f x = \map f {\dfrac {\map f {n x} } n}$


Then either:

$\forall x \in \Q: \map f x = \floor x$

or:

$\forall x \in \Q: \map f x = \ceiling x$


Real Domain

Let $f: \R \to \Z$ be an integer-valued function which satisfies both of the following:

$(1): \quad \map f {x + 1} = \map f x + 1$
$(2): \quad \forall n \in \Z_{> 0}: \map f x = \map f {\dfrac {\map f {n x} } n}$


Then it is not necessarily the case that either:

$\forall x \in \R: \map f x = \floor x$

or:

$\forall x \in \R: \map f x = \ceiling x$


Proof

From $(1)$, by induction we have:

$\forall n \in \N: \map f {x + n} = \map f x + n$

and

$\forall n \in\N: \map f {x - n} = \map f x - n$

and therefore, in particular:

$(3): \quad \forall n \in \Z: \map f n = \map f 0 + n$


From $(2)$, we get

\(\displaystyle \map f 0\) \(=\) \(\displaystyle \map f {\map f 0}\)
\(\displaystyle \) \(=\) \(\displaystyle \map f 0 + \map f 0\) using $(3)$ and $f$ being integer valued

Hence

$\map f 0 = 0$


Thus from $(3)$ it follows that:

$\forall n \in \Z: \map f n = n$

Suppose that $\map f {\dfrac 1 2} = k \le 0$.

Then:

\(\displaystyle k\) \(=\) \(\displaystyle \map f {\dfrac 1 {1 - 2 k} \map f {\dfrac 1 2 - k} }\)
\(\displaystyle \) \(=\) \(\displaystyle \map f {\dfrac 1 {1 - 2 k} \paren {\map f {\dfrac 1 2} - k} }\)
\(\displaystyle \) \(=\) \(\displaystyle \map f 0\)
\(\displaystyle \) \(=\) \(\displaystyle 0\)


We show that in this case, by induction:

$\map f {\dfrac 1 n} = 0$ for all $n \in \N$


Induction hypothesis:

$\map f {\dfrac 1 {n - 1} } = 0$

Then from $(1)$:

$\map f {\dfrac n {n - 1} } = \map f {\dfrac 1 {n - 1} + 1} = 0 + 1 = 1$

so:

\(\displaystyle \map f {\dfrac 1 n}\) \(=\) \(\displaystyle \map f {\dfrac 1 n \map f {\dfrac n {n - 1} } }\)
\(\displaystyle \) \(=\) \(\displaystyle \map f {\dfrac 1 {n - 1} }\) using $(2)$
\(\displaystyle \) \(=\) \(\displaystyle 0\)

Finally, we show by induction on $m$ that even:

$\map f {\dfrac m n} = 0$

for $m \in \set {1, \ldots, n - 1}$.


Above we have shown this for $m = 1$.

Let $1 \le m < n$.

If $m \divides n$, then

$\map f {\dfrac m n} = \map f {\dfrac 1 {n / m} } = 0$

Otherwise, write:

$n = \paren {k - 1} m + r$

Then:

$k > 1$

and:

$1 \le r \le m - 1$

We therefore have:

$k m = n + m - r$

so by the induction hypothesis:

$\map f {\dfrac {k m} n} = \map f {1 + \dfrac {m - r} n} = 1 + \map f {\dfrac {m - r} n} = 1$

Then:

\(\displaystyle \map f {\dfrac m n}\) \(=\) \(\displaystyle \map f {\dfrac 1 m \map f {\dfrac {m k} n} }\)
\(\displaystyle \) \(=\) \(\displaystyle \map f {\dfrac 1 m}\)
\(\displaystyle \) \(=\) \(\displaystyle 0\)


By $(1)$, this shows that

$\map f {\dfrac 1 2} \le 0 \implies \map f x = \floor x$

for all rational $x$.


Suppose that $\map f {\dfrac 1 2} = k > 0$.

Then the integer-valued function $g: \R \to \Z$ satisfies:

$\map g x = -\map f {-x}$

satisfies $(1)$ and $(2)$, and also:

\(\displaystyle \map g {\dfrac 1 2}\) \(=\) \(\displaystyle 1 - \map f {\dfrac 1 2}\)
\(\displaystyle \) \(\ge\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map f x\) \(=\) \(\displaystyle -\map g {-x}\)
\(\displaystyle \) \(=\) \(\displaystyle -\floor {-x}\)
\(\displaystyle \) \(=\) \(\displaystyle \ceiling x\) Floor of Negative equals Negative of Ceiling


Thus:

$\map f {\dfrac 1 2} > 0 \implies \map f x = \ceiling x$

for all rational $x$.

$\blacksquare$


Historical Note

The result Characteristics of Floor and Ceiling Function was presented by Peter Eisele and Karl Peter Hadeler in $1990$.

The previous work done in $1905$ by Georg Hamel to establish that the result does not necessarily hold in the real domain is to be noted.


Sources