BTT Mathematics / Primary Mathematics Learning Hub / Farey Sequences and Mediants
The Farey sequence of order n, written Fn, is the increasing list of all reduced fractions between 0 and 1 whose denominators are at most n. In this guide, 0/1 and 1/1 are included.
For example:
F1: 0/1, 1/1
F2: 0/1, 1/2, 1/1
F3: 0/1, 1/3, 1/2, 2/3, 1/1
F4: 0/1, 1/4, 1/3, 1/2, 2/3, 3/4, 1/1
Only fractions in lowest terms appear. Thus 2/4 is not listed separately because it is the same number as 1/2.
This guide is Primary Mathematics enrichment. It connects fraction equivalence, order, GCD, systematic listing and structured approximation. It extends Fractions of a Quantity, Equivalence and Operations, Euclidean Algorithm, GCD and Remainder Reduction and Systematic Listing.
Farey sequences and neighbour identities are optional enrichment rather than universal Primary syllabus requirements. Use the MOE Primary curriculum page and the learner’s school programme for required content.
Build Farey sequences · Reduction and order · Mediants · Farey neighbours · Insertion structure · 24 questions · Worked answers · Teaching and transfer
1. Build the sequence by denominator limit and reduction
To build Fn, list fractions a/b with 0≤a≤b≤n, reduce duplicates to lowest terms, then sort by size.
Worked example A: F2
Allowed reduced fractions are 0/1,1/2,1/1. Thus F2=0/1,1/2,1/1.
Worked example B: F3
New reduced denominator-three fractions are 1/3 and2/3. Therefore F3=0/1,1/3,1/2,2/3,1/1.
Worked example C: F4
Reduced denominator-four candidates are 1/4 and3/4; 2/4 reduces to1/2 and is already present. So F4=0/1,1/4,1/3,1/2,2/3,3/4,1/1.
Worked example D: F5
All numerators 1,2,3,4 are coprime to5, so four new terms appear:
0/1,1/5,1/4,1/3,2/5,1/2,3/5,2/3,3/4,4/5,1/1.
F5 therefore has 11 terms.
Worked example E: F6
Among denominator-six fractions between zero and one, only 1/6 and5/6 are reduced. Fractions 2/6,3/6 and4/6 reduce to earlier terms. Hence F6 is F5 with 1/6 inserted near zero and5/6 near one, for 13 terms.
2. Lowest terms prevent duplicate values
A Farey sequence is a list of rational values, not every written numerator-denominator pair. Equivalent fractions must collapse to one reduced representative.
Worked example F: 2/4
gcd(2,4)=2, so 2/4=1/2. The value appears once as 1/2.
Worked example G: 3/6
3/6 reduces to1/2 and creates no new term in F6.
Worked example H: Compare 2/5 and1/2
Cross-products give 2×2=4 and1×5=5, so 2/5<1/2.
Worked example I: Compare 3/5 and2/3
3×3=9 and2×5=10, so 3/5<2/3.
Exact ordering avoids decimal-rounding ambiguity.
3. The mediant of two positive-denominator fractions lies between them
The mediant of a/b and c/d is
(a+c)/(b+d).
If a/b<c/d and b,d are positive, then the mediant lies strictly between the two fractions.
Worked example J: 1/3 and1/2
Mediant=(1+1)/(3+2)=2/5. Check 1/3<2/5<1/2.
Worked example K: 2/5 and1/2
Mediant=3/7. It lies between 0.4 and0.5 without needing decimal approximation: compare by cross-products.
Worked example L: 1/4 and1/3
Mediant=2/7. Indeed 1/4<2/7<1/3.
Worked example M: 2/3 and3/4
Mediant=5/7, and 2/3<5/7<3/4.
Why the mediant lies between
From a/b<c/d we have ad<bc. To show a/b<(a+c)/(b+d), cross-multiply: a(b+d)<b(a+c) reduces to ad<bc. The other side follows similarly.
The mediant is not the arithmetic mean
For 1/3 and1/2, arithmetic mean is5/12, while mediant is2/5. They are different constructions even though both lie between the endpoints.
4. Farey neighbours have a determinant-one relationship
If a/b<c/d are adjacent terms in a Farey sequence, then
bc−ad=1.
This is a powerful exact check for Farey neighbours.
Worked example N: 1/3 and2/5
3×2−1×5=6−5=1. They are adjacent in F5.
Worked example O: 2/5 and1/2
5×1−2×2=5−4=1. They are adjacent in F5.
Worked example P: 1/2 and3/5
2×3−1×5=6−5=1.
Worked example Q: Denominator-sum test
Adjacent Farey terms a/b and c/d in Fn satisfy b+d>n. If b+d≤n, their mediant would have denominator at most n and would lie strictly between them, contradicting adjacency.
Worked example R: F5 neighbours around 1/2
In F5, the immediate neighbours of1/2 are 2/5 and3/5. Denominator sums are5+2=7>5 on both sides.
5. Mediant insertion explains how new Farey terms appear
If a/b and c/d are Farey neighbours and n reaches b+d, their mediant (a+c)/(b+d) is the natural new fraction inserted between them. For Farey neighbours, determinant one ensures the mediant is already reduced.
Worked example S: From F4 to F5
In F4, 1/3 and1/2 are neighbours. Their denominator sum is5, so at order5 the mediant 2/5 appears between them. Similarly 1/2 and2/3 produce3/5.
Worked example T: A later mediant
2/5 and1/2 are neighbours in F5. Their mediant is3/7, so it appears between them when order reaches seven.
Building from 0/1 and1/1
The mediant is1/2. Taking mediants again with the outer neighbours produces 1/3 between0/1 and1/2, and2/3 between1/2 and1/1. This reconstructs F3:
0/1,1/3,1/2,2/3,1/1.
Neighbour does not mean “numerically closest rational number”
There are infinitely many rational numbers between any two distinct rational numbers. “Farey neighbours” means adjacent under a denominator limit, not that no rational lies between them at all.
Fractions near a target
Farey sequences provide systematically ordered low-denominator fractions. This makes them useful for studying rational approximation without claiming that the nearest fraction under every possible metric is automatically a Farey neighbour.
6. Practice: 24 original questions
Use the definition: Fn contains reduced fractions between0 and1 with denominator at most n, in increasing order.
Questions 1–8: Build the sequences
1. Write F1.
2. Write F2.
3. Write F3.
4. Write F4.
5. How many terms are in F5?
6. Which four reduced denominator-five fractions are new when moving from F4 to F5?
7. Which two reduced denominator-six fractions are new when moving from F5 to F6?
8. Is 2/4 listed separately in F4? Explain.
Questions 9–16: Mediants and neighbour checks
9. Find the mediant of1/3 and1/2.
10. Find the mediant of2/5 and1/2.
11. Find the mediant of1/4 and1/3.
12. Find the mediant of2/3 and3/4.
13. Verify exactly that1/3<2/5<1/2.
14. Compute bc−ad for1/3<2/5.
15. Compute bc−ad for2/5<1/2.
16. Are1/3 and1/2 adjacent in F4? If so, what mediant appears between them in F5?
Questions 17–24: Fraction neighbours and insertion
17. What are the immediate neighbours of1/2 in F5?
18. What are the immediate neighbours of2/3 in F5?
19. Which F5 fraction lies between1/3 and1/2?
20. Which F5 fraction lies between1/2 and2/3?
21. Why does3/6 not create a new F6 term?
22. How many terms are in F6?
23. For Farey neighbours2/5 and1/2 in F5, verify that the denominator sum is greater than5.
24. Starting with0/1 and1/1, insert1/2, then the mediants with the two outer intervals. Write the resulting ordered list.
7. Worked answers
Answers 1–8
1. 0/1,1/1.
2. 0/1,1/2,1/1.
3. 0/1,1/3,1/2,2/3,1/1.
4. 0/1,1/4,1/3,1/2,2/3,3/4,1/1.
5. 11 terms.
6. 1/5,2/5,3/5,4/5.
7. 1/6 and5/6. The other denominator-six fractions reduce.
8. No. 2/4 reduces to1/2, which is already present.
Answers 9–16
9. 2/5.
10. 3/7.
11. 2/7.
12. 5/7.
13. 1/3<2/5 because1×5<2×3; and2/5<1/2 because2×2<1×5.
14. 1. 3×2−1×5=1.
15. 1. 5×1−2×2=1.
16. Yes; 2/5. Their denominator sum is5, and their mediant is2/5.
Answers 17–24
17. 2/5 and3/5.
18. 3/5 and3/4.
19. 2/5.
20. 3/5.
21. Because3/6 reduces to1/2, so it represents an existing Farey value.
22. 13 terms. F6 adds only1/6 and5/6 to the eleven F5 terms.
23. 5+2=7>5.
24. 0/1,1/3,1/2,2/3,1/1.
8. Teaching and transfer
If a learner lists every denominator form, require reduction before insertion. Farey sequences order rational values, so equivalent forms must collapse to one representative.
When decimal approximations determine order
Use cross-products instead. Comparing2/5 and1/2 by4<5 is exact and keeps the reasoning inside fractions.
When the mediant is confused with an average
Calculate both for1/3 and1/2. The mediant2/5 and arithmetic mean5/12 are different, even though both lie between the endpoints.
When “neighbour” is misunderstood
Emphasise the denominator limit. Two Farey neighbours still have infinitely many rational numbers between them; those intervening rationals simply require larger denominators than the current order allows.
When determinant one is memorised without meaning
Pair it with mediant insertion. If neighbours had a reduced mediant whose denominator were already within the limit, they would not be adjacent. The arithmetic and ordering conditions support each other.
Connect to GCD
Lowest terms are central to the construction. The Euclidean Algorithm guide provides an efficient way to decide whether numerator and denominator are coprime.
Continue through this enrichment collection
For lattice-area counting, use Pick’s Theorem, Lattice Points and Coordinate Area. For folding-as-reflection, use Paper Folding, Crease Patterns and Symmetry. For repeated-remainder reduction, use Euclidean Algorithm, GCD and Remainder Reduction.
Return to the BTT Primary Mathematics Learning Hub.
Original enrichment guide with 24 original practice questions and separate worked answers. Farey-order conventions are stated explicitly; the guide uses reduced fractions between0 and1 inclusive.