NCERT Solutions Class 9 Maths Exercise 3.5 — The World of Numbers

NCERT Class 9 Mathematics — Ganita Manjari Part I, The World of Numbers, Exercise Set 3.5. All 5 questions solved.

This exercise is about the decimal face of a rational number, and one theorem governs nearly all of it:

A fraction in its lowest terms has a terminating decimal expansion exactly when its denominator’s only prime factors are $2$ and $5$. Otherwise the expansion repeats.

The reason is that our decimals are built on powers of ten, and $10 = 2 \times 5$. A denominator made only of $2$s and $5$s can be turned into a power of $10$ by multiplying; any other prime factor never can be.

The converse also holds, and matters for Question 3: every terminating or repeating decimal is a rational number, and any decimal that neither terminates nor repeats is irrational.

Key insight. “Non-terminating” and “irrational” are not the same thing. $\frac{1}{3} = 0.333\ldots$ goes on for ever and is perfectly rational, because it repeats. The test for irrationality is that the digits never end and never fall into a repeating block — which is why Question 3 asks you to look for the block rather than at the length.

Question 1

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.

Solution. Factorise each denominator and look for primes other than $2$ and $5$. Each fraction is already in lowest terms, which the theorem requires.

$\dfrac{7}{20}$:   $20 = 2^2 \times 5$ — only $2$s and $5$s, so it terminates.

$\dfrac{4}{15}$:   $15 = 3 \times 5$ — the factor $3$ is fatal, so it repeats.

$\dfrac{13}{250}$:   $250 = 2 \times 5^3$ — only $2$s and $5$s, so it terminates.

Now the check by division.

$\frac{7}{20}$: multiply top and bottom by $5$ to reach a power of ten —

$$\frac{7}{20} = \frac{35}{100} = 0.35$$

$\frac{4}{15}$: long division gives $4.000\ldots \div 15$ —

$$\frac{4}{15} = 0.2666\ldots = 0.2\overline{6}$$

The $6$ repeats for ever, as predicted. The remainders in the division cycle back to a value already seen, which is what forces the repetition.

$\frac{13}{250}$: multiply top and bottom by $4$ —

$$\frac{13}{250} = \frac{52}{1000} = 0.052$$

All three predictions confirmed.

Question 2

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?

Solution. Since $13$ is neither $2$ nor $5$, the expansion must repeat.

$$\frac{1}{13} = 0.076923076923\ldots = 0.\overline{076923}$$

The repeating block is $076923$, six digits long.

Now the multiples:

$$\frac{2}{13} = 0.\overline{153846} \qquad \frac{3}{13} = 0.\overline{230769} \qquad \frac{4}{13} = 0.\overline{307692}$$ $$\frac{5}{13} = 0.\overline{384615} \qquad \frac{6}{13} = 0.\overline{461538} \qquad \frac{9}{13} = 0.\overline{692307}$$

What to notice. The blocks fall into two families, each a cyclic rotation of a single sequence of six digits:

• rotations of $076923$, arising from the numerators $1, 3, 4, 9, 10, 12$ — $076923$, $230769$, $307692$, $692307$, $769230$, $923076$;
• rotations of $153846$, arising from the numerators $2, 5, 6, 7, 8, 11$ — $153846$, $384615$, $461538$, $538461$, $615384$, $846153$.

So $\frac{1}{13}$ is cyclic, but only within its own family of six numerators — unlike $\frac{1}{7}$, whose single block $142857$ covers all six of $\frac{1}{7}$ to $\frac{6}{7}$.

Why the difference. Every block has length $6$, and there are $12$ possible numerators $1$ to $12$. Six rotations cannot account for twelve fractions, so the twelve split into two groups of six. For $\frac{1}{7}$ the block is also six long but there are only six numerators, so one family suffices. The length of the block is what decides this, and it is always a divisor of $n – 1$.

Question 3

Classify the following numbers as rational or irrational. Find the explicit fractions in case they are rational.

Solution. Apply the test: terminating or repeating means rational; never-ending and never-repeating means irrational.

(i) $\sqrt{81}$

Work out the root before judging it:

$$\sqrt{81} = 9$$

A square root is not automatically irrational — this one is a whole number. Rational, equal to $\frac{9}{1}$.

(ii) $\sqrt{12}$

$$\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$$

$12$ is not a perfect square, and $\sqrt3$ is irrational (by the same argument the chapter gives for $\sqrt2$), so twice it is too. Irrational — no fraction exists.

(iii) $0.33333\ldots$

Non-terminating but repeating, with block $3$. Rational. Let $x = 0.333\ldots$; then $10x = 3.333\ldots$, and subtracting gives $9x = 3$, so

$$x = \frac{3}{9} = \frac{1}{3}$$

(iv) $0.123451234512345\ldots$

The block $12345$ repeats. Rational. With a block of five digits, multiply by $10^5$:

$$x = 0.\overline{12345}, \qquad 100000x = 12345.\overline{12345}$$ $$99999x = 12345 \quad\Longrightarrow\quad x = \frac{12345}{99999}$$

Both divide by $3$:

$$x = \frac{4115}{33333}$$

(v) $1.01001000100001\ldots$

Look closely: the gaps between the $1$s grow — one zero, then two, then three, then four. No block ever repeats, because each group is longer than the last. Non-terminating and non-repeating, so irrational.

This is the part of the question doing the teaching. The digits follow a clear pattern, yet the number is irrational — a pattern and a repeating block are not the same thing.

(vi) $23.560185612239874790120$

This one stops. It is written out in full, ending in $0$, with no dots afterwards. Every terminating decimal is rational. Rational, and its fraction is immediate:

$$23.560185612239874790120 = \frac{23560185612239874790120}{10^{21}}$$

which reduces, but the unreduced form already proves rationality.

Question 4

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 $1$.

Solution. Follow the recipe the question gives.

$$x = 0.99999\ldots$$

Multiply both sides by $10$. This shifts the decimal point one place, and because the $9$s go on for ever the tail after the point is unchanged:

$$10x = 9.99999\ldots$$

Subtract the first equation from the second. The infinite tails are identical and cancel exactly:

$$10x – x = 9.99999\ldots – 0.99999\ldots$$ $$9x = 9$$ $$x = 1$$

So $0.\overline{9} = 1$ — not approximately, but exactly.

Why this is hard to believe, and why it is true. The instinct is that $0.999\ldots$ must fall short of $1$ by some tiny amount. But name that amount: whatever positive number you propose, the $9$s go on far enough to get closer than it. A difference smaller than every positive number is $0$, so the two are the same number.

Another route to the same place: $\frac{1}{3} = 0.333\ldots$, and multiplying both sides by $3$ gives $1 = 0.999\ldots$.

What this really shows is that a number can have two decimal representations. Just as $1 = \frac{10}{10} = \frac{100}{100}$, so $1 = 1.000\ldots = 0.999\ldots$, and likewise $2.47 = 2.46999\ldots$. Every terminating decimal has a second form ending in repeated $9$s.

Question 5

*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.

Solution. The block of $\frac{1}{7}$ is $142857$, six digits for a denominator of $7$ — that is, of length $n – 1$. That is exactly the condition to look for: a full-length block leaves no room for a second family, so all the rotations must be the multiples of the same block.

Testing small primes:

So $17$, $19$ and $23$ all work, and after them come $29$, $47$, $59$, $61$ and $97$. These are called full-reptend primes.

Try it for $17$: $\frac{1}{17} = 0.\overline{0588235294117647}$, and doubling that block gives $1176470588235294$ — the same sixteen digits, rotated, which is $\frac{2}{17}$.

Two things to notice. Only primes can do this, since the block length must reach $n-1$. And not all primes do: $11$ and $13$ fail, so it depends on the particular prime, not merely on being prime.

Common mistakes

• Question 1, factorising before reducing. The theorem applies to the fraction in lowest terms. $\frac{6}{15}$ looks as though the $3$ makes it repeat, but it reduces to $\frac{2}{5} = 0.4$, which terminates. Always cancel first.
• Question 1, applying the test to the numerator. Only the denominator matters. $\frac{4}{15}$ and $\frac{7}{15}$ both repeat, whatever the top.
• Question 3(i), assuming every square root is irrational. $\sqrt{81} = 9$. Evaluate the root before classifying it; only roots of non-perfect-squares are irrational.
• Question 3(iii) and (iv), calling a repeating decimal irrational. Non-terminating is not the test. $0.333\ldots$ repeats, so it is rational — and the algebra produces its fraction.
• Question 3(v), calling it rational because there is a pattern. The pattern is real but the blocks grow, so nothing ever repeats. A pattern is not a repeating block.
• Question 3(vi), calling it irrational because it is long and messy. It is written out in full and ends. Length is not the test; whether it terminates is.
• Question 4, answering “they are nearly equal”. The algebra gives $9x = 9$ exactly. Treating $0.\overline{9}$ as slightly less than $1$ requires a positive gap, and no positive number is small enough to be one.
• Question 5, assuming every prime is cyclic. $11$ and $13$ are primes whose blocks are too short. The condition is that the block reaches the full length $n – 1$.

Practise next

• The End-of-Chapter Exercises, Questions 1, 3, 11 and 12 — the same conversions in both directions, plus two questions on predicting the number of decimal places from the denominator.
• Exercise Set 3.4 — worth revisiting after Question 3 here, since the interval $3.1415$ to $3.1416$ in its Question 5 contains both rationals and the irrational $\pi$.