BTT Mathematics / Primary Mathematics Learning Hub / Continued Fractions
A simple continued fraction writes a number as a whole-number part followed by repeated reciprocals. For example,
17/7 = 2 + 3/7 = 2 + 1/(7/3) = 2 + 1/(2 + 1/3),
so we may write 17/7=[2;2,3]. The semicolon separates the first whole-number part from the later positive whole-number entries.
For positive rational numbers, the expansion terminates because repeated division eventually reaches remainder zero. This is the same remainder structure that appears in the Euclidean Algorithm.
A convergent is obtained by stopping the continued fraction early. For [2;2,3], the successive convergents are 2, 5/2 and 17/7. These give increasingly detailed rational reconstructions of the target in this example.
This guide is Primary Mathematics enrichment. It connects division with remainder, fractions, reciprocals, approximation and algorithms. Formal approximation theorems are not required here; convergents are treated as structured rational approximations rather than as a universal “closest possible” claim under every notion of closeness.
Use the MOE Primary curriculum page and the learner’s school programme for required content.
Build the expansion · Recover the fraction · Convergents · Remainder chains · Approximation · 24 questions · Worked answers
1. Repeated division builds a continued fraction
Start with a positive fraction or ratio. Divide the larger quantity by the smaller, record the quotient, then continue with the divisor and remainder.
Worked example A: 17/7
17=2×7+3
7=2×3+1
3=3×1
The quotients are 2,2,3, so 17/7=[2;2,3].
Worked example B: 43/19
43=2×19+5
19=3×5+4
5=1×4+1
4=4×1
Hence 43/19=[2;3,1,4].
Worked example C: A proper fraction
For 5/13, the whole-number part is zero. Since 13=2×5+3, 5=1×3+2, 3=1×2+1 and2=2×1,
5/13=[0;2,1,1,2].
Worked example D: Exact division
12/4=3 exactly, so its continued fraction is simply [3].
Termination
A positive rational expansion ends because the Euclidean remainder becomes smaller at every non-zero step and eventually reaches zero.
2. Work from the end to recover the number
To evaluate [a0;a1,a2,…], begin with the last entry, take reciprocals and add as the nesting requires.
Worked example E: [2;2,3]
Start from 3. Then 2+1/3=7/3. Take its reciprocal: 3/7. Add 2: 17/7. Thus [2;2,3]=17/7.
Worked example F: [1;2,2]
2+1/2=5/2. Its reciprocal is2/5. Add1: 7/5.
Worked example G: [0;3,2]
3+1/2=7/2; reciprocal gives 2/7.
Worked example H: [3;4]
3+1/4=13/4.
3. Convergents are obtained by stopping early
For [2;3,1,4], stopping after different numbers of entries gives:
[2]=2
[2;3]=2+1/3=7/3
[2;3,1]=2+1/(3+1)=9/4
[2;3,1,4]=43/19.
Worked example I: Compare errors for 43/19
Target 43/19≈2.26316. The convergents are 2, 7/3≈2.33333, 9/4=2.25 and43/19 exactly. The absolute errors shrink here from about0.26316 to0.07018 to0.01316 to0.
Worked example J: Convergents of [1;2,2]
They are 1, 3/2, 7/5. The target is 7/5=1.4, while 1 and1.5 are coarser approximations.
Worked example K: Proper-fraction convergents
For [0;2,1,1,2], the first two convergents are0 and1/2. Continuing gives1/3,2/5 and finally5/13. Intermediate values may alternate around the target.
Convergents preserve structure
Each convergent uses the same initial quotient decisions as the full expansion. It is not a random rounded fraction.
4. Continued fractions and Euclid record the same division chain
The continued-fraction entries for a rational number are the quotients in the Euclidean algorithm applied to its numerator and denominator.
Worked example L: 84/30
84=2×30+24
30=1×24+6
24=4×6
Therefore 84/30=[2;1,4]. Evaluating [2;1,4] gives 14/5, which equals84/30 after reduction.
Worked example M: Reduction appears automatically
The Euclidean chain for84 and30 ends with GCD6. Dividing gives14/5. The same rational number is represented by the same continued fraction whether we begin from84/30 or14/5.
Worked example N: 252/105
252=2×105+42
105=2×42+21
42=2×21
So 252/105=[2;2,2]=12/5.
Reciprocal connection
If x>1 has expansion [a0;a1,…], then 1/(x−a0) begins with the next quotient structure. Continued fractions repeatedly isolate a whole-number part and invert the leftover fractional part.
5. Rational approximation trades denominator size for accuracy
A simple fraction is often easier to use mentally than a long decimal. Convergents provide a principled sequence of such fractions.
Worked example O: Approximate 43/19
If denominator at most4 is preferred, convergent 9/4=2.25 is a convenient approximation to43/19≈2.26316.
Worked example P: Compare 7/3 and9/4
For target43/19, cross-multiplication gives exact differences:
|43/19−7/3|=|129−133|/57=4/57.
|43/19−9/4|=|172−171|/76=1/76.
Thus 9/4 is closer.
Worked example Q: Bounded-denominator caution
A convergent is often an excellent low-denominator approximation, but this guide does not use “best” without stating the comparison rule. Approximation quality depends on what denominators or error measure are allowed.
Worked example R: Exact target remains the endpoint
For rational targets, the final convergent equals the number exactly. Approximation is useful only when we intentionally stop early.
6. Practice: 24 original questions
Questions 1–8: Build and evaluate
1. Convert 17/7 to a simple continued fraction.
2. Convert 43/19 to a simple continued fraction.
3. Convert 12/4.
4. Convert 7/5.
5. Evaluate [2;2,3].
6. Evaluate [1;2,2].
7. Evaluate [0;3,2].
8. Evaluate [3;4].
Questions 9–16: Convergents and remainder chains
9. List all convergents of [2;2,3].
10. List all convergents of [1;2,2].
11. List all convergents of [2;3,1,4].
12. Convert84/30 to a continued fraction.
13. Reduce84/30 and verify the continued fraction gives the same rational number.
14. Convert252/105 to a continued fraction.
15. Evaluate [2;1,4].
16. Evaluate [2;2,2].
Questions 17–24: Approximation and reasoning
17. For target43/19, which is closer:7/3 or9/4?
18. Find the exact absolute error of9/4 as an approximation to43/19.
19. Find the exact absolute error of7/3 as an approximation to43/19.
20. If a denominator at most4 is desired among the displayed convergents of43/19, which convergent would you choose?
21. Why does a rational continued fraction terminate?
22. What algorithm supplies the quotient sequence used in a rational continued fraction?
23. Does the final convergent of a rational continued fraction approximate or equal the original rational number?
24. Explain why a convergent is not simply an arbitrary rounded fraction.
7. Worked answers
Answers 1–8
1. [2;2,3].
2. [2;3,1,4].
3. [3].
4. [1;2,2]. 7=1×5+2;5=2×2+1;2=2×1.
5. 17/7.
6. 7/5.
7. 2/7.
8. 13/4.
Answers 9–16
9. 2,5/2,17/7.
10. 1,3/2,7/5.
11. 2,7/3,9/4,43/19.
12. [2;1,4].
13. 84/30=14/5, and [2;1,4]=14/5.
14. [2;2,2].
15. 14/5.
16. 12/5.
Answers 17–24
17. 9/4.
18. 1/76. |43/19−9/4|=|172−171|/76.
19. 4/57. |43/19−7/3|=|129−133|/57.
20. 9/4.
21. Repeated Euclidean remainders are positive whole numbers that strictly decrease until zero.
22. The Euclidean algorithm.
23. It equals the original rational number exactly.
24. It is formed by preserving the initial nested quotient structure of the continued fraction and stopping at a definite stage.
8. Teaching and transfer
If a learner sees only unfamiliar brackets, begin with ordinary division. Write17=2×7+3, then ask what fraction remains after removing two wholes. The continued fraction is simply a compact record of repeating that move.
When evaluation order is reversed
Always start from the final entry and work outward. Nested fractions are evaluated from the deepest level first.
When convergents are confused with decimal rounding
Compare 9/4 with2.26. One comes from preserving the continued-fraction structure; the other comes from rounding a decimal representation. They solve different approximation jobs.
When fractions are not reduced first
The Euclidean quotient chain naturally leads to the same rational value after reduction. Use GCD as a check that equivalent input fractions give the same continued fraction under the standard terminating convention used here.
Connect to Euclid
The quotient entries are exactly the repeated-division quotients from the Euclidean Algorithm. This turns one algorithm into a new representation.
Continue through this enrichment collection
For balanced non-crossing counts, use Catalan Numbers, Non-Crossing Paths and Balanced Structures. For polyhedral structure, use Euler Characteristic, Polyhedra, Faces, Edges and Vertices. For repeated scale structure, use Fractals, Self-Similarity and Scaling Patterns.
Return to the BTT Primary Mathematics Learning Hub.
Original enrichment guide with 24 original practice questions and separate worked answers. Continued fractions here are finite simple continued fractions for positive rational numbers unless explicitly stated otherwise.