Exercise 3.5 Practice
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Overview
Step-by-step NCERT solutions and practice problems for CBSE Class 9 Maths Coordinate Geometry (Exercise 3.5 Practice), providing clear algebraic methods and exam-oriented solutions.
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 $p - 1$.