Exercise 3.5 Practice
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Overview
This page provides comprehensive Ch 3: The World of Numbers – Exercise 3.5 Practice. Practice identifying terminating vs repeating decimals, performing long division to find repeating blocks, classifying numbers as rational or irrational, converting recurring decimals to fractions, and exploring cyclic numbers. Free step-by-step solutions and explanations.
Decimals, Rational and Irrational Numbers
Q1: Terminating or Repeating Decimals?
Without performing long division, determine which of the following rational numbers will have terminating decimals and which will be repeating: $\dfrac{7}{20}$, $\dfrac{4}{15}$ and $\dfrac{13}{250}$. Then check your answers by explicitly performing the long divisions and expressing these rational numbers as decimals.
Key Rule: A rational number $\dfrac{p}{q}$ (in lowest terms) has a terminating decimal if and only if the prime factorisation of the denominator $q$ contains only the primes 2 and/or 5. Otherwise the decimal is non-terminating and repeating.
$\dfrac{7}{20}$:
$20 = 2^2 \times 5$. Only factors 2 and 5. → Terminating.
Long division: $7 \div 20 = 0.35$ ✓
$20 = 2^2 \times 5$. Only factors 2 and 5. → Terminating.
Long division: $7 \div 20 = 0.35$ ✓
$\dfrac{4}{15}$:
$15 = 3 \times 5$. Contains factor 3 (not just 2 or 5). → Non-terminating repeating.
Long division: $4 \div 15 = 0.2\overline{6}$ ✓
(i.e., $0.2666\ldots$ where 6 repeats indefinitely)
$15 = 3 \times 5$. Contains factor 3 (not just 2 or 5). → Non-terminating repeating.
Long division: $4 \div 15 = 0.2\overline{6}$ ✓
(i.e., $0.2666\ldots$ where 6 repeats indefinitely)
$\dfrac{13}{250}$:
$250 = 2 \times 5^3$. Only factors 2 and 5. → Terminating.
Long division: $13 \div 250 = 0.052$ ✓
$250 = 2 \times 5^3$. Only factors 2 and 5. → Terminating.
Long division: $13 \div 250 = 0.052$ ✓
Terminating: $\dfrac{7}{20} = 0.35$ and $\dfrac{13}{250} = 0.052$.
Repeating: $\dfrac{4}{15} = 0.2\overline{6}$
Q2: Long Division of 1/13 and Cyclic Properties
Perform the long division for $\dfrac{1}{13}$. Identify the repeating block of digits. Does it show cyclic properties if you evaluate $\dfrac{2}{13}$? Now compute $\dfrac{3}{13}$, $\dfrac{4}{13}$, etc. What do you notice?
Long division of $\dfrac{1}{13}$:
Performing the division: $1 \div 13$: $$\frac{1}{13} = 0.\overline{076923}$$ The repeating block is 076923 (6 digits long).
Performing the division: $1 \div 13$: $$\frac{1}{13} = 0.\overline{076923}$$ The repeating block is 076923 (6 digits long).
Checking $\dfrac{2}{13}$, $\dfrac{3}{13}$, …:
$\dfrac{1}{13} = 0.\overline{076923}$
$\dfrac{2}{13} = 0.\overline{153846}$
$\dfrac{3}{13} = 0.\overline{230769}$
$\dfrac{4}{13} = 0.\overline{307692}$
$\dfrac{5}{13} = 0.\overline{384615}$
$\dfrac{6}{13} = 0.\overline{461538}$
$\dfrac{7}{13} = 0.\overline{538461}$
$\dfrac{8}{13} = 0.\overline{615384}$
$\dfrac{9}{13} = 0.\overline{692307}$
$\dfrac{10}{13} = 0.\overline{769230}$
$\dfrac{11}{13} = 0.\overline{846153}$
$\dfrac{12}{13} = 0.\overline{923076}$
$\dfrac{2}{13} = 0.\overline{153846}$
$\dfrac{3}{13} = 0.\overline{230769}$
$\dfrac{4}{13} = 0.\overline{307692}$
$\dfrac{5}{13} = 0.\overline{384615}$
$\dfrac{6}{13} = 0.\overline{461538}$
$\dfrac{7}{13} = 0.\overline{538461}$
$\dfrac{8}{13} = 0.\overline{615384}$
$\dfrac{9}{13} = 0.\overline{692307}$
$\dfrac{10}{13} = 0.\overline{769230}$
$\dfrac{11}{13} = 0.\overline{846153}$
$\dfrac{12}{13} = 0.\overline{923076}$
What do we notice?
Unlike $\dfrac{1}{7}$ which produces a single set of 6 cyclic digits (142857), the fractions $\dfrac{k}{13}$ produce two distinct cyclic groups:
Unlike $\dfrac{1}{7}$ which produces a single set of 6 cyclic digits (142857), the fractions $\dfrac{k}{13}$ produce two distinct cyclic groups:
- Group 1 (digits: 076923): $\dfrac{1}{13}$, $\dfrac{3}{13}$, $\dfrac{4}{13}$, $\dfrac{9}{13}$, $\dfrac{10}{13}$, $\dfrac{12}{13}$
- Group 2 (digits: 153846): $\dfrac{2}{13}$, $\dfrac{5}{13}$, $\dfrac{6}{13}$, $\dfrac{7}{13}$, $\dfrac{8}{13}$, $\dfrac{11}{13}$
$\dfrac{1}{13} = 0.\overline{076923}$. The 13 fractions form two cyclic groups of 6 rotations each. This differs from $\dfrac{1}{7}$ (a true cyclic number) since 13 is not a full-reptend prime in base 10.
Q3: Classify as Rational or Irrational
Classify the following numbers as rational or irrational. Find the explicit fractions in case they are rational.
(i) $\sqrt{81}$ (ii) $\sqrt{12}$
(iii) $0.33333\ldots$ (iv) $0.123451234512345\ldots$
(v) $1.01001000100001\ldots$ (Notice the pattern: Is it repeating a single block?)
(vi) $23.560185612239874790120$
(i) $\sqrt{81}$ (ii) $\sqrt{12}$
(iii) $0.33333\ldots$ (iv) $0.123451234512345\ldots$
(v) $1.01001000100001\ldots$ (Notice the pattern: Is it repeating a single block?)
(vi) $23.560185612239874790120$
(i) $\sqrt{81}$:
$\sqrt{81} = 9$ (since $9^2 = 81$).
9 is a whole number, which is rational. As a fraction: $\dfrac{9}{1}$.
→ Rational $= \dfrac{9}{1}$
$\sqrt{81} = 9$ (since $9^2 = 81$).
9 is a whole number, which is rational. As a fraction: $\dfrac{9}{1}$.
→ Rational $= \dfrac{9}{1}$
(ii) $\sqrt{12}$:
$\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$.
$\sqrt{3}$ is irrational (3 is not a perfect square), so $2\sqrt{3}$ is irrational.
→ Irrational
$\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$.
$\sqrt{3}$ is irrational (3 is not a perfect square), so $2\sqrt{3}$ is irrational.
→ Irrational
(iii) $0.33333\ldots = 0.\overline{3}$:
This is a non-terminating repeating decimal. Let $x = 0.\overline{3}$. $$10x = 3.\overline{3} \implies 10x - x = 3 \implies 9x = 3 \implies x = \frac{1}{3}$$ → Rational $= \dfrac{1}{3}$
This is a non-terminating repeating decimal. Let $x = 0.\overline{3}$. $$10x = 3.\overline{3} \implies 10x - x = 3 \implies 9x = 3 \implies x = \frac{1}{3}$$ → Rational $= \dfrac{1}{3}$
(iv) $0.123451234512345\ldots = 0.\overline{12345}$:
The block "12345" repeats, so this is a non-terminating repeating decimal. Let $x = 0.\overline{12345}$. $$100000x = 12345.\overline{12345} \implies 99999x = 12345 \implies x = \frac{12345}{99999} = \frac{4115}{33333}$$ → Rational $= \dfrac{4115}{33333}$
The block "12345" repeats, so this is a non-terminating repeating decimal. Let $x = 0.\overline{12345}$. $$100000x = 12345.\overline{12345} \implies 99999x = 12345 \implies x = \frac{12345}{99999} = \frac{4115}{33333}$$ → Rational $= \dfrac{4115}{33333}$
(v) $1.01001000100001\ldots$:
Look carefully at the pattern: the number of zeros between each "1" increases by one each time (1 zero, 2 zeros, 3 zeros, 4 zeros, …).
This means the decimal does not repeat any fixed block — it is non-terminating and non-repeating.
→ Irrational
Look carefully at the pattern: the number of zeros between each "1" increases by one each time (1 zero, 2 zeros, 3 zeros, 4 zeros, …).
This means the decimal does not repeat any fixed block — it is non-terminating and non-repeating.
→ Irrational
(vi) $23.560185612239874790120$:
This is a finite (terminating) decimal — it ends after a fixed number of digits. Every terminating decimal is rational. $$23.560185612239874790120 = \frac{23560185612239874790120}{1000000000000000000000}$$ This fraction can be simplified but is already clearly a ratio of two integers.
→ Rational
This is a finite (terminating) decimal — it ends after a fixed number of digits. Every terminating decimal is rational. $$23.560185612239874790120 = \frac{23560185612239874790120}{1000000000000000000000}$$ This fraction can be simplified but is already clearly a ratio of two integers.
→ Rational
Rational: (i) $\sqrt{81} = \dfrac{9}{1}$, (iii) $0.\overline{3} = \dfrac{1}{3}$, (iv) $0.\overline{12345} = \dfrac{4115}{33333}$, (vi) terminating decimal.
Irrational: (ii) $\sqrt{12}$, (v) $1.01001000100001\ldots$
Irrational: (ii) $\sqrt{12}$, (v) $1.01001000100001\ldots$
Q4: Why 0.9̄ = 1 Exactly
The number $0.\overline{9}$ (which means $0.99999\ldots$) is a rational number. Using algebra (let $x = 0.\overline{9}$, multiply by 10, and subtract), explain why $0.\overline{9}$ is exactly equal to 1.
Step 1 – Set up the algebra:
Let $x = 0.\overline{9} = 0.99999\ldots$
Let $x = 0.\overline{9} = 0.99999\ldots$
Step 2 – Multiply both sides by 10:
$$10x = 9.99999\ldots = 9.\overline{9}$$
Step 3 – Subtract the original equation:
$$10x - x = 9.\overline{9} - 0.\overline{9}$$
$$9x = 9$$
$$x = 1$$
Why does this work?
The key insight is that $9.\overline{9} - 0.\overline{9}$ eliminates the infinite repeating tail exactly, leaving just 9. This is valid because the infinite decimal $0.\overline{9}$ is defined as the limit of the sequence $0.9, 0.99, 0.999, \ldots$, which converges to exactly 1.
Another way to see it: $0.\overline{9} = \dfrac{9}{9} = 1$ (using the repeating decimal formula).
The key insight is that $9.\overline{9} - 0.\overline{9}$ eliminates the infinite repeating tail exactly, leaving just 9. This is valid because the infinite decimal $0.\overline{9}$ is defined as the limit of the sequence $0.9, 0.99, 0.999, \ldots$, which converges to exactly 1.
Another way to see it: $0.\overline{9} = \dfrac{9}{9} = 1$ (using the repeating decimal formula).
$0.\overline{9} = 1$ exactly. The algebra gives $9x = 9 \Rightarrow x = 1$. There is no "gap" between $0.\overline{9}$ and 1 — they represent the same real number.
*Q5: Finding More Cyclic Numbers
We have seen that the repeating block of $\dfrac{1}{7}$ is a cyclic number. Try to find more numbers $n$ whose reciprocals $\left(\dfrac{1}{n}\right)$ produce decimals with repeating blocks that are cyclic.
Recall: $\dfrac{1}{7}$
$$\frac{1}{7} = 0.\overline{142857}$$
The repeating block "142857" is a cyclic number: multiplying it by 1, 2, 3, 4, 5, or 6 gives cyclic rotations of the same 6 digits.
What makes a cyclic number?
A prime $p$ generates a cyclic number if the period of $\dfrac{1}{p}$ (the length of the repeating block) equals $p - 1$. These primes are called full-reptend primes in base 10.
A prime $p$ generates a cyclic number if the period of $\dfrac{1}{p}$ (the length of the repeating block) equals $p - 1$. These primes are called full-reptend primes in base 10.
Known full-reptend primes in base 10:
$n = 7$: period 6 ($= 7 - 1$). Cyclic number: 142857.
$n = 17$: period 16 ($= 17 - 1$). $\dfrac{1}{17} = 0.\overline{0588235294117647}$. Cyclic number: 0588235294117647.
$n = 19$: period 18 ($= 19 - 1$). $\dfrac{1}{19} = 0.\overline{052631578947368421}$. Cyclic: 052631578947368421.
$n = 23$: period 22 ($= 23 - 1$). $\dfrac{1}{23} = 0.\overline{0434782608695652173913}$.
$n = 7$: period 6 ($= 7 - 1$). Cyclic number: 142857.
$n = 17$: period 16 ($= 17 - 1$). $\dfrac{1}{17} = 0.\overline{0588235294117647}$. Cyclic number: 0588235294117647.
$n = 19$: period 18 ($= 19 - 1$). $\dfrac{1}{19} = 0.\overline{052631578947368421}$. Cyclic: 052631578947368421.
$n = 23$: period 22 ($= 23 - 1$). $\dfrac{1}{23} = 0.\overline{0434782608695652173913}$.
Verification for $n = 17$:
Multiply the cyclic number 0588235294117647 by 1, 2, 3, … , 16 — each product is a cyclic rotation of the same 16 digits. For example: $$2 \times 0588235294117647 = 1176470588235294 \quad \checkmark$$ This is indeed a rotation of the original block.
Multiply the cyclic number 0588235294117647 by 1, 2, 3, … , 16 — each product is a cyclic rotation of the same 16 digits. For example: $$2 \times 0588235294117647 = 1176470588235294 \quad \checkmark$$ This is indeed a rotation of the original block.
Non-full-reptend primes (for contrast):
$n = 13$: period 6 (not $13-1 = 12$), so it has two separate cyclic groups — not a single cyclic number.
$n = 11$: period 2 (repeating block "09"), not a cyclic number.
$n = 13$: period 6 (not $13-1 = 12$), so it has two separate cyclic groups — not a single cyclic number.
$n = 11$: period 2 (repeating block "09"), not a cyclic number.
Full-reptend primes (generators of cyclic numbers in base 10): 7, 17, 19, 23, 29, 47, 59, 61, 97, …
Their reciprocals each produce a cyclic number of length $p - 1$.
Their reciprocals each produce a cyclic number of length $p - 1$.