Cardinality of Integer Interval
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Theorem
Let $a, b \in \Z$ be integers.
Let $\left[{a \,.\,.\, b}\right]$ denote the integer interval between $a$ and $b$.
Then $\left[{a \,.\,.\, b}\right]$ is finite and its cardinality equals:
- $\begin{cases} b - a + 1 & : b \ge a - 1 \\ 0 & : b \le a - 1 \end{cases}$
Proof
Let $b < a$.
Then $\left[{a \,.\,.\, b}\right]$ is empty.
By Empty Set is Finite, $\left[{a \,.\,.\, b}\right]$ is finite.
By Cardinality of Empty Set, $\left[{a \,.\,.\, b}\right]$ has cardinality $0$.
Let $b \ge a$.
By Translation of Integer Interval is Bijection, there exists a bijection between $\left[{a \,.\,.\, b}\right]$ and $\left[{0 \,.\,.\, b - a}\right]$.
Thus $\left[{a \,.\,.\, b}\right]$ is finite of cardinality $b - a + 1$.
$\blacksquare$