Show that if $x− 1 < n < x$ where $n$ is a natural number then
(a) $⌊x⌋ = n$
(b) $⌈x⌉ = n + 1$
(a) $\lfloor x \rfloor$ is the largest integer smaller than or equal to $x$.
We are given $n < x$ where $n$ is a natural number (and so an integer).
We need to show there is no other integer greater than $n$ but less than or equal to $x$.
For the purpose of contradiction, let's assume there is an integer $m = n+1$ such that
$$n \; < \; m \; \le \; x$$
$$n \; < \; n+1 \; \le \; x$$
Subtracting 1 gives us
$$ n \le \; x -1$$
This contradicts the given $x-1 < n$, and so there is no integer larger than $n$ that is smaller than or equal to $x$. And so
$$ \lfloor x \rfloor = n $$
(b) $\lceil x \rceil$ is the smallest integer greater than or equal to $x$.
We are given
$$x -1 \; < \; n$$
Adding 1 gives us
$$x \; < \; n+1$$
We need to show there is no other integer smaller than $n+1$ that is greater than to equal to $x$.
For the purpose of contradiction, let's assume there is an integer $m = (n+1) -1 = n$ such that
$$ x \; \le \; m < \; n + 1 $$
This gives us
$$ x \; \le \; n $$
This contradicts the given $n < x$, and so there is no integer smaller than $n+1$ that is greater than or equal to $x$. And so
$$ \lceil x \rceil = n + 1 $$