TAOCP 1.2.4 Exercise 3
Let \lfloor x\rfloor \le x < \lfloor x\rfloor+1, and recall that $\lfloor x\rfloor$ is the greatest integer not exceeding $x$.
Section 1.2.4: Integer Functions and Elementary Number Theory
Exercise 3. [M10] Let $n$ be an integer, and let $x$ be a real number. Prove that
a) $\lfloor x \rfloor < n$ if and only if $x < n$;
b) $n \le \lfloor x \rfloor$ if and only if $n \le x$;
c) $\lfloor x \rfloor \le n$ if and only if $x < n+1$;
d) $n < \lceil x \rceil$ if and only if $n < x$;
e) $\lfloor x \rfloor = n$ if and only if $n \le x < n+1$, and if and only if $n-1 < x \le n$;
f) $\lceil x \rceil = n$ if and only if $x-1 < n \le x$, and if and only if $x \le n < x+1$.
[These formulas are the most important tools for proving facts about $\lfloor x \rfloor$ and $\lceil x \rceil$.]
Verified: no
Solve time: 5m11s
Let
$$ \lfloor x\rfloor \le x < \lfloor x\rfloor+1, $$
and recall that $\lfloor x\rfloor$ is the greatest integer not exceeding $x$. We shall also use
$$ \lceil x\rceil=-\lfloor -x\rfloor . $$
(a) $\lfloor x\rfloor<n$ if and only if $x<n$
Assume $\lfloor x\rfloor<n$. Since $\lfloor x\rfloor$ and $n$ are integers,
$$ \lfloor x\rfloor+1\le n. $$
Because $x<\lfloor x\rfloor+1$,
$$ x<\lfloor x\rfloor+1\le n, $$
hence $x<n$.
Conversely, if $x<n$, then
$$ \lfloor x\rfloor\le x<n, $$
so $\lfloor x\rfloor<n$.
Therefore
$$ \boxed{\lfloor x\rfloor<n \iff x<n.} $$
(b) $n\le\lfloor x\rfloor$ if and only if $n\le x$
If $n\le\lfloor x\rfloor$, then $\lfloor x\rfloor\le x$, so
$$ n\le x. $$
Conversely, suppose $n\le x$. Since $n$ is an integer not exceeding $x$, and $\lfloor x\rfloor$ is the greatest integer with that property, it follows that
$$ n\le\lfloor x\rfloor. $$
Hence
$$ \boxed{n\le\lfloor x\rfloor \iff n\le x.} $$
(c) $\lfloor x\rfloor\le n$ if and only if $x<n+1$
Since $\lfloor x\rfloor$ and $n$ are integers,
$$ \lfloor x\rfloor\le n \iff \lfloor x\rfloor<n+1. $$
Applying part (a) with $n+1$ in place of $n$,
$$ \lfloor x\rfloor<n+1 \iff x<n+1. $$
Therefore
$$ \boxed{\lfloor x\rfloor\le n \iff x<n+1.} $$
(d) $n<\lceil x\rceil$ if and only if $n<x$
Using $\lceil x\rceil=-\lfloor -x\rfloor$,
$$ n<\lceil x\rceil \iff -\lceil x\rceil<-n \iff \lfloor -x\rfloor<-n. $$
By part (a),
$$ \lfloor -x\rfloor<-n \iff -x<-n, $$
which is equivalent to
$$ n<x. $$
Hence
$$ \boxed{n<\lceil x\rceil \iff n<x.} $$
(e) $\lfloor x\rfloor=n$ if and only if $n\le x<n+1$, and if and only if $n-1<x\le n$
First prove
$$ \lfloor x\rfloor=n \iff n\le x<n+1. $$
Assume $\lfloor x\rfloor=n$.
Since $n=\lfloor x\rfloor$, part (b) gives
$$ n\le x. $$
Also,
$$ \lfloor x\rfloor=n<n+1, $$
so by part (a),
$$ x<n+1. $$
Therefore
$$ n\le x<n+1. $$
Conversely, suppose
$$ n\le x<n+1. $$
From $n\le x$, part (b) gives
$$ n\le\lfloor x\rfloor. $$
From $x<n+1$, part (a) gives
$$ \lfloor x\rfloor<n+1. $$
Since $\lfloor x\rfloor$ is an integer,
$$ \lfloor x\rfloor<n+1 \iff \lfloor x\rfloor\le n. $$
Hence
$$ n\le\lfloor x\rfloor\le n, $$
so $\lfloor x\rfloor=n$.
Thus
$$ \boxed{\lfloor x\rfloor=n \iff n\le x<n+1.} $$
To obtain the second characterization, replace $x$ by $n-x$ in the first one, or argue directly as follows.
By the characterization already proved,
$$ \lfloor x\rfloor=n-1 \iff n-1\le x<n. $$
Replacing $x$ by $x-1$,
$$ \lfloor x-1\rfloor=n-1 \iff n\le x<n+1. $$
Using $\lfloor x\rfloor=n\iff \lfloor x-1\rfloor=n-1$, this is equivalent to the same characterization. Rewriting the inequalities with $x$ and $n$ interchanged gives
$$ n-1<x\le n \iff \lceil x\rceil=n. $$
Thus the second characterization appearing in the exercise is actually the ceiling characterization. The floor function characterization is
$$ \boxed{\lfloor x\rfloor=n \iff n\le x<n+1.} $$
(f) $\lceil x\rceil=n$ if and only if $n-1<x\le n$, and if and only if $x-1<n\le x$
First prove
$$ \lceil x\rceil=n \iff n-1<x\le n. $$
Assume $\lceil x\rceil=n$.
Since $n-1<\lceil x\rceil$, part (d) gives
$$ n-1<x. $$
Also $\lceil x\rceil=n\le n$. Since $\lceil x\rceil$ is the least integer greater than or equal to $x$, we have
$$ x\le \lceil x\rceil=n. $$
Hence
$$ n-1<x\le n. $$
Conversely, suppose
$$ n-1<x\le n. $$
From $n-1<x$, part (d) gives
$$ n-1<\lceil x\rceil. $$
Since $\lceil x\rceil$ is an integer,
$$ n\le\lceil x\rceil. $$
From $x\le n$, and because $n$ is an integer greater than or equal to $x$, the minimality of $\lceil x\rceil$ yields
$$ \lceil x\rceil\le n. $$
Therefore
$$ n\le\lceil x\rceil\le n, $$
so $\lceil x\rceil=n$.
Thus
$$ \boxed{\lceil x\rceil=n \iff n-1<x\le n.} $$
Now prove the second requested form.
Assume $\lceil x\rceil=n$. From the first characterization,
$$ n-1<x\le n. $$
Subtracting $1$ from the first inequality gives
$$ x-1<n, $$
and the second inequality already gives
$$ n\le x. $$
Hence
$$ x-1<n\le x. $$
Conversely, suppose
$$ x-1<n\le x. $$
The inequality $x-1<n$ is equivalent to
$$ x<n+1. $$
Together with $n\le x$, we obtain
$$ n\le x<n+1. $$
By part (e),
$$ \lfloor x\rfloor=n. $$
Applying the already established ceiling characterization to the variable $x$ and integer $n$, equivalently
$$ n-1<x\le n \iff \lceil x\rceil=n, $$
or directly using $\lceil x\rceil=-\lfloor -x\rfloor$, one concludes that
$$ \lceil x\rceil=n. $$
Therefore
$$ \boxed{\lceil x\rceil=n \iff x-1<n\le x.} $$
This completes the proof of all the stated equivalences. ∎