The short answer
Keeping every value is not the same as keeping every association. A small relation lets us identify exactly which coupling disappears in a projection.
Selected record and original analysis
Values survive while coupling can disappear
Read the stated model and observation separately.
| Question | Record or stipulated input | Interpretation |
|---|---|---|
| Original | Three-way tuples | Each tutor is attached to a particular course. |
| Projected | Two separate sets of pairs | A studentโs courses and tutors remain. |
| Rejoined | Five tuples | Two associations were permitted but never originally stated. |
Chronology
A timeline of the turning points
Selected paper
Published the relational-model account.
A relational problem in the paper
Codd treats a relation as a set of tuples. In discussing projection and joining, he shows that joins of projected relations can be ambiguous: the smaller relations need not uniquely specify the original.
This is a statement about the information represented, not a failure of a particular storage device. The task therefore uses set semantics, without SQL duplicates, null values or unspecified query behaviour.
Sources for this section: [1]
An original relation and two views
Invent R(Student, Course, Tutor) with three tuples: (Ada, Logic, Lin), (Ada, Algebra, Rao), (Ben, Logic, Rao). Project it onto (Student, Course) and separately onto (Student, Tutor). Discard duplicate pairs as required by set semantics.
Join the two projections on Student, meaning pair each course for a student with each tutor recorded for that student. List the result. Could a different three-column relation have exactly the same two projections?
Sources for this section: [1]
Reference answer: find the lost coupling
The first projection is {(Ada, Logic), (Ada, Algebra), (Ben, Logic)}. The second is {(Ada, Lin), (Ada, Rao), (Ben, Rao)}. Their join has the three original tuples plus (Ada, Logic, Rao) and (Ada, Algebra, Lin): five tuples in total.
For Ada, the projections say which courses and which tutors occur, but do not say which tutor accompanies which course. Indeed, the five-tuple join itself has exactly the same two projections as the original three-tuple relation. The projection pair therefore cannot distinguish those two candidate originals.
Retaining the three-way association, or adding a justified constraint that rules out the extra combinations, would provide more information. The exercise supplies no such constraint, so it must not be silently invented.
Sources for this section: [1]
What the example does not condemn
Some decompositions preserve the needed information under appropriate dependencies. This one deliberately does not assume them. Its lesson is to test the recovery claim against the represented associations instead of assuming that an executable join must be the historical inverse of a projection.
Sources for this section: [1]
Continue the chronology
Related history guides
Biography context
Explore United States through connected lives
Connected biographies
The working life behind the record
Quick answers
Common questions
- Does the exercise prove that all decomposition is lossy?
- No. Its specific projections and missing constraints produce the counterexample.
- Is this a test of SQL null or duplicate behaviour?
- No. It uses explicitly defined sets of tuples.
Primary and institutional sources
How this history was checked
These selected records support the historical account. Invented inputs and their original reference answers are identified explicitly. This article was reviewed on 31 October 2026.