Show that if $n− 1 < x < n$ where $n$ is a natural number then
(a) $⌈x⌉ = n$
(b) $⌊x⌋ = n− 1$
(a) The value of $\lceil x \rceil$ is the smallest integer larger than or equal to $x$.
We are given that $x < n$. We need to establish that there is no other integer between $x$ and $n$. The integer below $n$ is $n-1$, and we are given that $n-1 < x$, so there is no other integer between $x$ and $n$. So $n$ is the smallest integer larger than or equal to $x$.
$$\lceil x \rceil = n $$
(b) The value of $\lfloor x \rfloor$ is the largest integer less than or equal to $x$.
We are given that $n-1 < x$. We need to establish that there is no integer between $n-1$ and $x$. The integer above $n-1$ is $n$, and we are given $x < n$, so there is no integer between $n-1$ and $x$. So $n-1$ is the largest integer less than or equal to $x$.
$$ \lfloor x \rfloor = n $$