Exercise 2: Proving "Same Last Two Digits of Employee ID" Is an Equivalence Relation — Possible Solution
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REFLEXIVE
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For any employee A, does A have the same last two digits of employee
ID as themselves? Trivially yes - A's ID's last two digits are
identical to A's own ID's last two digits, by definition. Reflexive
holds for every employee.

SYMMETRIC
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Suppose employee A has the same last two digits as employee B. Does B
then have the same last two digits as A? Yes - "the same as" is a
statement about two specific two-digit values being equal, and
equality itself doesn't have a direction: if A's last two digits equal
B's last two digits, then B's last two digits equal A's last two
digits too, automatically. Symmetric holds.

TRANSITIVE
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Suppose A has the same last two digits as B, and B has the same last
two digits as C. Does A then have the same last two digits as C?
Yes - if A's value equals B's value, and B's value equals C's value,
then A's value equals C's value, by the ordinary transitivity of
numeric equality itself. Transitive holds.

WHY ALL THREE HOLDING TOGETHER MATTERS
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Per this chapter, "a relation with all three of the first properties
together is an equivalence relation." Since reflexive, symmetric, and
transitive have each been confirmed individually, "has the same last
two digits of employee ID" is a genuine equivalence relation on the
set of employees.

WHAT THE EQUIVALENCE CLASSES ACTUALLY LOOK LIKE
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Per this chapter, an equivalence relation "splits its entire set into
disjoint equivalence classes, where everything inside one class is
considered equivalent to everything else in that same class." Here,
each equivalence class is the group of all employees whose ID happens
to end in one specific two-digit combination - one class for "...00,"
another for "...01," and so on, up to "...99" - as many as 100
possible classes, though not every class needs to actually contain an
employee. Every employee belongs to exactly one class (their own ID's
own last two digits), and this is precisely what a database query
grouping employees by the last two digits of their ID would produce.

WHY THIS WORKS AS AN ANSWER
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It verifies each of the three required properties individually with
its own specific justification, confirms the conclusion follows
directly from having all three, and describes the resulting
equivalence classes concretely, connecting back to this chapter's own
GROUP BY analogy.