NCERT Solutions Class 9 Maths Chapter 3 End-of-Chapter Exercises — The World of Numbers

NCERT Class 9 Mathematics — Ganita Manjari Part I, The World of Numbers, End-of-Chapter Exercises. All 16 questions solved.

These sixteen questions run through the whole journey of the chapter — natural numbers to integers to rationals to irrationals. Four techniques cover almost all of them:

• Fraction to decimal: divide, and predict in advance whether it will stop by factorising the denominator.
• Decimal to fraction: for a repeating decimal, let $x$ be the number, multiply by a power of $10$ that shifts one whole block, and subtract.
• Proof by contradiction: assume a number is rational, write it in lowest terms, and derive an impossibility.
• Finding numbers in between: raise both to a common denominator large enough to leave gaps, or take averages.

Questions $13$, $14$ and $15$ carry an asterisk, marking them as enrichment that is not assessed. All three are solved in full.

Key insight. The denominator decides everything about the decimal. Whether it stops (Questions 1, 11, 12), how many places it runs to (Questions 10, 11, 12), and whether a repeating block can be cyclic — all are read off the prime factorisation of the denominator in lowest terms. If a question asks about decimal places, factorise before doing anything else.

Question 1

Convert the following rational numbers in the form of a terminating decimal or non-terminating and repeating decimal, whichever the case may be, by the process of long division.

Solution. Predict first from the denominator, then divide.

(i) $\dfrac{3}{50}$

$50 = 2 \times 5^2$ — only $2$s and $5$s, so this terminates. Multiply top and bottom by $2$ to reach a power of ten:

$$\frac{3}{50} = \frac{6}{100} = 0.06$$

(ii) $\dfrac{2}{9}$

$9 = 3^2$ — the factor $3$ means it repeats. Long division of $2.000\ldots$ by $9$ gives $2$ remainder $2$ at every step, so the digit $2$ repeats for ever:

$$\frac{2}{9} = 0.2222\ldots = 0.\overline{2}$$

Question 2

Prove that $\sqrt5$ is an irrational number.

Solution. By contradiction, following the chapter’s proof for $\sqrt2$.

Step 1 — assume the opposite. Suppose $\sqrt5$ is rational. Then it can be written as

$$\sqrt5 = \frac{p}{q}, \qquad q \neq 0$$

with $p$ and $q$ integers sharing no common factor — that is, the fraction is in its lowest terms. Every rational number can be written this way, so this costs nothing.

Step 2 — square both sides.

$$5 = \frac{p^2}{q^2} \quad\Longrightarrow\quad p^2 = 5q^2$$

Step 3 — deduce a factor of $p$. The right-hand side is $5$ times an integer, so $p^2$ is divisible by $5$. Now, $5$ is prime, and a prime dividing a square must divide the number itself. So $5$ divides $p$, and we may write

$$p = 5k \quad \text{for some integer } k$$

Step 4 — substitute back.

$$(5k)^2 = 5q^2 \quad\Longrightarrow\quad 25k^2 = 5q^2 \quad\Longrightarrow\quad q^2 = 5k^2$$

Step 5 — deduce a factor of $q$. By exactly the same reasoning, $q^2$ is divisible by $5$, so $q$ is divisible by $5$.

Step 6 — the contradiction. Both $p$ and $q$ are divisible by $5$, so they share the common factor $5$. But Step 1 said the fraction was in lowest terms, with no common factor at all. The assumption has produced an impossibility.

Therefore $\sqrt5$ cannot be written as a ratio of integers: it is irrational.

The step that makes this work is “$5$ is prime”. The same argument fails for $\sqrt4$, as it must, because $4$ is not prime: $p^2$ divisible by $4$ does not force $p$ to be divisible by $4$ — take $p = 2$.

Question 3

Convert the following decimal numbers in the form of $\dfrac{p}{q}$.

Solution. Two cases. A terminating decimal is written over the appropriate power of $10$ and reduced. A repeating decimal needs the shift-and-subtract method: multiply $x$ by whatever power of $10$ moves the decimal point past one full repeating block, subtract the original, and solve.

(i) $12.6$

Terminating:

$$12.6 = \frac{126}{10} = \frac{63}{5}$$

(ii) $0.0120$

Terminating; the trailing zero adds nothing:

$$0.0120 = \frac{120}{10000} = \frac{3}{250}$$

(iii) $3.0\overline{52}$

One non-repeating digit after the point, then a two-digit block. Let $x = 3.0525252\ldots$

$$1000x = 3052.525252\ldots, \qquad 10x = 30.525252\ldots$$ $$990x = 3022 \quad\Longrightarrow\quad x = \frac{3022}{990} = \frac{1511}{495}$$

(iv) $1.2\overline{35}$

Same shape: one non-repeating digit, two repeating.

$$1000x = 1235.353535\ldots, \qquad 10x = 12.353535\ldots$$ $$990x = 1223 \quad\Longrightarrow\quad x = \frac{1223}{990}$$

($1223$ is prime, so this will not reduce.)

(v) $0.\overline{23}$

A pure repeating decimal with a two-digit block:

$$100x = 23.2323\ldots, \qquad x = 0.2323\ldots$$ $$99x = 23 \quad\Longrightarrow\quad x = \frac{23}{99}$$

(vi) $2.0\overline{5}$

One non-repeating digit, one repeating:

$$100x = 205.555\ldots, \qquad 10x = 20.555\ldots$$ $$90x = 185 \quad\Longrightarrow\quad x = \frac{185}{90} = \frac{37}{18}$$

(vii) $2.12\overline{5}$

Two non-repeating digits, one repeating:

$$1000x = 2125.555\ldots, \qquad 100x = 212.555\ldots$$ $$900x = 1913 \quad\Longrightarrow\quad x = \frac{1913}{900}$$

(viii) $3.12\overline{5}$

The same with a different whole part:

$$1000x = 3125.555\ldots, \qquad 100x = 312.555\ldots$$ $$900x = 2813 \quad\Longrightarrow\quad x = \frac{2813}{900}$$

(ix) $2.\overline{1625}$

A four-digit block, starting straight after the point:

$$10000x = 21625.16251625\ldots, \qquad x = 2.16251625\ldots$$ $$9999x = 21623 \quad\Longrightarrow\quad x = \frac{21623}{9999}$$

Question 4

Locate the following rational numbers on the number line.

Solution. The technique is successive magnification: zoom in on a shorter and shorter stretch of the line, one decimal place at a time.

(i) $0.532$

• It lies between $0$ and $1$. Divide that into ten parts and it falls between $0.5$ and $0.6$.
• Divide $0.5$ to $0.6$ into ten parts; it falls between $0.53$ and $0.54$.
• Divide $0.53$ to $0.54$ into ten parts and take the second mark: $0.532$.

(ii) $1.1\overline{5}$

First work out what the number is: $1.1555\ldots$, which by the method of Question 3 is

$$\frac{115 – 11}{90} = \frac{104}{90} = \frac{52}{45}$$

• It lies between $1$ and $2$, and then between $1.1$ and $1.2$.
• Between $1.15$ and $1.16$.
• Between $1.155$ and $1.156$ — and so on for ever, since the decimal never terminates.

Each magnification traps the number in an interval ten times shorter. The point exists exactly, even though no finite number of steps reaches it, and it sits just above $1.1\overline{5} \approx 1.1556$.

Question 5

Find $6$ rational numbers between $3$ and $4$.

Solution. Give both endpoints a common denominator large enough to leave at least six gaps. To fit six numbers, use a denominator of $7$:

$$3 = \frac{21}{7}, \qquad 4 = \frac{28}{7}$$

The numerators strictly between $21$ and $28$ are $22, 23, 24, 25, 26, 27$ — exactly six.

$$\frac{22}{7}, \quad \frac{23}{7}, \quad \frac{24}{7}, \quad \frac{25}{7}, \quad \frac{26}{7}, \quad \frac{27}{7}$$

Question 6

Find $5$ rational numbers between $\dfrac{2}{5}$ and $\dfrac{3}{5}$.

Solution. The denominators already match, but the numerators $2$ and $3$ are consecutive, so there is no room. Multiply top and bottom of both by $6$ to create it:

$$\frac{2}{5} = \frac{12}{30}, \qquad \frac{3}{5} = \frac{18}{30}$$

Now the numerators $13, 14, 15, 16, 17$ all lie strictly between:

$$\frac{13}{30}, \quad \frac{14}{30}, \quad \frac{15}{30}, \quad \frac{16}{30}, \quad \frac{17}{30}$$

Question 7

Find $5$ rational numbers between $\dfrac{1}{6}$ and $\dfrac{2}{5}$.

Solution. Different denominators this time, so start with a common one. The LCM of $6$ and $5$ is $30$:

$$\frac{1}{6} = \frac{5}{30}, \qquad \frac{2}{5} = \frac{12}{30}$$

The numerators strictly between $5$ and $12$ are $6, 7, 8, 9, 10, 11$ — six of them, so there is already room for five. Taking the first five:

$$\frac{6}{30}, \quad \frac{7}{30}, \quad \frac{8}{30}, \quad \frac{9}{30}, \quad \frac{10}{30}$$

Check the outer ones: $\frac{5}{30} < \frac{6}{30}$ and $\frac{10}{30} < \frac{12}{30}$ ✓

Question 8

If $\dfrac{x}{3} + \dfrac{x}{5} = \dfrac{16}{15}$, find the rational number $x$.

Solution. Combine the left-hand side over the common denominator $15$:

$$\frac{x}{3} = \frac{5x}{15}, \qquad \frac{x}{5} = \frac{3x}{15}$$ $$\frac{5x + 3x}{15} = \frac{8x}{15}$$

So the equation is

$$\frac{8x}{15} = \frac{16}{15}$$

The denominators are equal, so the numerators must be:

$$8x = 16 \quad\Longrightarrow\quad x = 2$$

Check: $\frac{2}{3} + \frac{2}{5} = \frac{10}{15} + \frac{6}{15} = \frac{16}{15}$ ✓

Question 9

Let $a$ and $b$ be two non-zero rational numbers such that $a + \dfrac{1}{b} = 0$. Without assigning any numerical values, determine whether $ab$ is positive or negative. Justify your answer.

Solution. Rearrange the condition to express $a$ in terms of $b$:

$$a + \frac{1}{b} = 0 \quad\Longrightarrow\quad a = -\frac{1}{b}$$

Now form the product $ab$ directly:

$$ab = \left( -\frac{1}{b} \right) \times b = -\frac{b}{b} = -1$$

So $ab$ is not merely negative — it is exactly $-1$, whatever non-zero rational $b$ may be. Being $-1$, it is certainly negative.

The condition says that $a$ and $\frac{1}{b}$ are additive inverses, so they have opposite signs. And $\frac{1}{b}$ always has the same sign as $b$. So $a$ and $b$ have opposite signs, and their product must be negative — which is the sign argument the question is inviting, with the algebra above pinning down the exact value as a bonus.

Check with any values: $b = 4$ gives $a = -\frac{1}{4}$ and $ab = -1$; $b = -\frac{2}{3}$ gives $a = \frac{3}{2}$ and $ab = -1$.

Question 10

A rational number has a terminating decimal expansion whose last non-zero digit occurs in the $4^{th}$ decimal place. Show that such a number can be written in the form $\dfrac{p}{10^4}$, where $p$ is an integer not divisible by $10$. Is it necessary that the denominator of this rational number, when written in the lowest form, is divisible by $2^4$ or $5^4$? Give reasons.

Solution. First part. A decimal that stops at the $4^{\text{th}}$ place can be written by putting all four decimal digits over $10^4$: a number $0.abcd$ is $\frac{abcd}{10000}$, and the same works with any whole-number part. So the number is $\frac{p}{10^4}$ for some integer $p$.

Why is $p$ not divisible by $10$? Because the last non-zero digit is in the $4^{\text{th}}$ place, which means the $4^{\text{th}}$ decimal digit is not zero. That digit is the units digit of $p$. A number whose units digit is non-zero is not divisible by $10$.

Second part. Now write $\frac{p}{10^4} = \frac{p}{2^4 \times 5^4}$ and reduce it.

Since $p$ is not divisible by $10$, it cannot have both $2$ and $5$ as factors. So one of two things happens:

• $5 \nmid p$. Then no factor of $5$ cancels, and the reduced denominator still contains $5^4$ — so it is divisible by $5^4$.
• $2 \nmid p$. Then no factor of $2$ cancels, and the reduced denominator still contains $2^4$ — so it is divisible by $2^4$.

At least one of these must hold, so yes — the lowest-form denominator is always divisible by $2^4$ or by $5^4$.

But not necessarily by both, and an example makes that concrete. Take $0.0002$:

$$0.0002 = \frac{2}{10^4} = \frac{2}{10000} = \frac{1}{5000} = \frac{1}{2^3 \times 5^4}$$

Here $p = 2$, which is not divisible by $10$, and the last non-zero digit is indeed in the fourth place. The reduced denominator $5000$ is divisible by $5^4 = 625$ but not by $2^4 = 16$, since $5000 \div 16 = 312.5$.

Question 11

Without performing division, determine whether the decimal expansion of $\dfrac{18}{125}$ is terminating or non-terminating. If it terminates, state the number of decimal places.

Solution. Check lowest terms first: $18 = 2 \times 3^2$ and $125 = 5^3$ share no factor, so the fraction is already reduced.

$$125 = 5^3$$

Only the prime $5$ appears, so the expansion terminates.

How many places? Turn the denominator into a power of ten. It needs three $5$s matched by three $2$s, so multiply top and bottom by $2^3 = 8$:

$$\frac{18}{125} = \frac{18 \times 8}{125 \times 8} = \frac{144}{1000} = 0.144$$

Three decimal places. In general, for a denominator $2^m \times 5^n$ in lowest terms, the number of decimal places is the larger of $m$ and $n$ — here $m = 0$ and $n = 3$, giving $3$.

Question 12

A rational number in its lowest form has denominator $2^3 \times 5$. How many decimal places will its decimal expansion have? Explain your answer.

Solution. The denominator contains only $2$s and $5$s, so the expansion terminates. To find the number of places, convert the denominator to a power of ten.

$$2^3 \times 5 = 8 \times 5 = 40$$

There are three $2$s but only one $5$, so two more $5$s are needed to pair them off. Multiply top and bottom by $5^2 = 25$:

$$2^3 \times 5 \times 5^2 = 2^3 \times 5^3 = 10^3 = 1000$$

The denominator becomes $10^3$, so the expansion has $3$ decimal places.

The rule: for $2^m \times 5^n$ in lowest terms, the answer is $\max(m, n)$ — here $\max(3, 1) = 3$. The larger exponent sets the pace, because the smaller one has to be topped up to match it.

Check with an example: $\frac{1}{40} = \frac{25}{1000} = 0.025$, three places ✓. And $\frac{7}{40} = \frac{175}{1000} = 0.175$ ✓. “In lowest form” matters — $\frac{5}{40}$ would reduce to $\frac{1}{8}$, which is a different denominator.

Question 13

*Let $a = \dfrac{7}{12}$ and $b = \dfrac{5}{6}$. Express both $a$ and $b$ in the form $\dfrac{k_1}{m}$ and $\dfrac{k_2}{m}$ where $k_1$, $k_2$ and $m$ are integers and $k_2 – k_1 > 6$. Using the same denominator $m$, write exactly five distinct rational numbers lying between $a$ and $b$ keeping an integer numerator. Explain why the condition $k_2 – k_1 > n + 1$ is necessary to find $n$ such rational numbers between $a$ and $b$ using this method.

Solution. Step 1 — a first common denominator. The LCM of $12$ and $6$ is $12$:

$$a = \frac{7}{12}, \qquad b = \frac{10}{12}$$

Here $k_2 – k_1 = 10 – 7 = 3$, which is not greater than $6$. Only $\frac{8}{12}$ and $\frac{9}{12}$ lie between them — two numbers, not five.

Step 2 — enlarge it. Multiply top and bottom of both by $3$, giving $m = 36$:

$$a = \frac{21}{36}, \qquad b = \frac{30}{36}$$

Now $k_2 – k_1 = 30 – 21 = 9 > 6$ ✓

Step 3 — write five numbers. Any five of the numerators $22$ to $29$ will do:

$$\frac{22}{36}, \quad \frac{24}{36}, \quad \frac{25}{36}, \quad \frac{27}{36}, \quad \frac{28}{36}$$

Check the outermost: $\frac{21}{36} < \frac{22}{36}$ and $\frac{28}{36} < \frac{30}{36}$ ✓

Step 4 — why the gap condition. The method only produces numbers of the form $\frac{k}{m}$ with $k$ a whole number strictly between $k_1$ and $k_2$. How many such whole numbers are there? The integers from $k_1 + 1$ up to $k_2 – 1$, and counting them gives

$$k_2 – k_1 – 1$$

To find $n$ of them you therefore need

$$k_2 – k_1 – 1 \;\geq\; n, \qquad \text{that is} \qquad k_2 – k_1 \;\geq\; n + 1$$

The book’s condition $k_2 – k_1 > n + 1$ guarantees this comfortably, with one slot to spare. For $n = 5$ it asks for a gap of at least $7$, while a gap of exactly $6$ would already give the five numbers needed — as $m = 24$ would, with $a = \frac{14}{24}$ and $b = \frac{20}{24}$.

The essential point stands either way: enlarging $m$ enlarges the gap, because scaling both numerators by $k$ multiplies the gap by $k$ too. So however many numbers are wanted, a large enough $m$ supplies them — which is the density of the rationals, seen through numerators.

Question 14

*Three rational numbers $x$, $y$, $z$ satisfy $x + y + z = 0$ and $xy + yz + zx = 0$. Show that all the rational numbers $x$, $y$, $z$ must be simultaneously zero.

Solution. The two conditions look unrelated until you square the first one — the expansion of a squared sum contains exactly the expression in the second condition.

$$(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)$$

Substitute both given facts. The left-hand side is $0^2 = 0$, and the bracket on the right is $0$:

$$0 = x^2 + y^2 + z^2 + 2(0)$$ $$x^2 + y^2 + z^2 = 0$$

Now use the fact that a square is never negative. Each of $x^2$, $y^2$, $z^2$ is greater than or equal to $0$, and three non-negative numbers can only add to $0$ if every one of them is $0$:

$$x^2 = 0, \qquad y^2 = 0, \qquad z^2 = 0$$

and the only number whose square is zero is zero itself. Hence

$$x = y = z = 0$$

Check: $0 + 0 + 0 = 0$ ✓ and $0 + 0 + 0 = 0$ ✓

The argument works for rational numbers — and indeed for any real numbers — because it relies only on squares being non-negative.

Question 15

*Show that the rational number $\dfrac{(a + b)}{2}$ lies between the rational numbers $a$ and $b$.

Solution. Suppose without loss of generality that $a < b$ — if they were the other way round the argument is identical with the letters swapped, and if $a = b$ there is nothing between them to speak of.

Two things must be shown: that the average exceeds $a$, and that it falls short of $b$.

It is greater than $a$. Compare by subtracting:

$$\frac{a+b}{2} – a = \frac{a + b – 2a}{2} = \frac{b – a}{2}$$

Since $a < b$, the difference $b – a$ is positive, so $\frac{b-a}{2}$ is positive. A positive difference means

$$\frac{a+b}{2} > a$$

It is less than $b$. Subtract the other way:

$$b – \frac{a+b}{2} = \frac{2b – a – b}{2} = \frac{b – a}{2}$$

Positive again, so

$$\frac{a+b}{2} < b$$

Putting the two together:

$$a \;<\; \frac{a+b}{2} \;<\; b$$

Notice that both gaps came out as the same quantity $\frac{b-a}{2}$ — the average sits exactly halfway, which is why it is called the midpoint.

It is also rational: $a$ and $b$ are rational, and sums and quotients of rationals are rational (the denominator $2$ being non-zero). So the number found is of the right kind, which is what makes this the proof that between any two rationals lies another.

Question 16

Find the lengths of the hypotenuses of all the right triangles in Fig. 3.14, which is referred to as the square root spiral.

Solution. The spiral is built one triangle at a time, and each new hypotenuse becomes a leg of the next triangle. Every short leg added is of length $1$.

First triangle. Both legs are $1$, so by the Baudhāyana–Pythagoras theorem

$$h_1 = \sqrt{1^2 + 1^2} = \sqrt2$$

Second triangle. Its legs are the previous hypotenuse $\sqrt2$ and a new unit:

$$h_2 = \sqrt{(\sqrt2)^2 + 1^2} = \sqrt{2 + 1} = \sqrt3$$

Third triangle.

$$h_3 = \sqrt{(\sqrt3)^2 + 1^2} = \sqrt4 = 2$$

The pattern is now plain. At each stage the square of the hypotenuse gains exactly $1$:

$$h_n = \sqrt{n + 1}$$

Fig. 3.14 shows ten triangles — ten coloured wedges, with eleven segments marked $1$ (two for the first triangle and one for each of the other nine). So the hypotenuses are

$$\sqrt2, \; \sqrt3, \; \sqrt4, \; \sqrt5, \; \sqrt6, \; \sqrt7, \; \sqrt8, \; \sqrt9, \; \sqrt{10}, \; \sqrt{11}$$

Most of these are irrational, which is the point of the figure: it is a ruler-and-compass recipe for marking $\sqrt2$, $\sqrt3$, $\sqrt5$ and the rest on the number line, showing that irrational lengths are entirely constructible even though no fraction names them. Two are not irrational — $\sqrt4 = 2$ and $\sqrt9 = 3$ — a reminder from Exercise Set 3.5 Question 3(i) that a square root sign does not by itself mean irrational.

Common mistakes

• Question 2, forgetting to say “in lowest terms”. Without that assumption there is no contradiction at the end — $p$ and $q$ sharing a factor is only impossible because the fraction was assumed reduced.
• Question 2, using the argument for $\sqrt4$. It breaks at the step “a prime dividing $p^2$ divides $p$”, which needs $5$ to be prime. $4$ is not, and $\sqrt4 = 2$ is rational — so any version of the proof that seems to work for $4$ has an error in it.
• Question 3(iii) to (ix), using the wrong powers of ten. The two multipliers must differ by exactly one full block: for one non-repeating digit and a two-digit block, use $1000$ and $10$. Using $100$ and $10$ leaves a mismatched tail and the subtraction does not cancel.
• Question 3(vii) and (viii), treating $2.12\overline{5}$ as $2.\overline{125}$. The overline covers only the final $5$, so the number is $2.125555\ldots$, not $2.125125125\ldots$. These give quite different fractions.
• Question 5, answering $3.5$ alone. Six numbers are asked for. Using a denominator of $7$ produces exactly six at once.
• Question 6, saying there is nothing between $\frac{2}{5}$ and $\frac{3}{5}$. Consecutive numerators leave no room, but the rationals are dense: scaling both to thirtieths opens up five gaps immediately.
• Question 9, testing numbers and generalising. The question says “without assigning any numerical values”. Rearranging to $a = -\frac{1}{b}$ settles every case at once, and shows the product is not just negative but exactly $-1$.
• Question 10, claiming the denominator must be divisible by both $2^4$ and $5^4$. $0.0002 = \frac{1}{5000}$ is the counterexample: only one of the two survives cancellation, which is why the question says “or”.
• Questions 11 and 12, adding the exponents. For $2^m \times 5^n$ the number of decimal places is $\max(m, n)$, not $m + n$. In Question 12 that is $3$, not $4$.
• Question 12, skipping “in lowest form”. The rule applies only to a reduced fraction. If the numerator shared a factor with the denominator, the denominator would shrink and so would the number of places.
• Question 14, dividing by $x + y + z$. It equals zero, and dividing by zero is not allowed. Squaring the first condition is what connects it to the second.
• Question 15, checking a single pair of numbers. Showing $\frac{2+6}{2} = 4$ lies between $2$ and $6$ proves nothing in general. The two subtractions, done with letters, cover every pair at once.
• Question 16, expecting every hypotenuse to be irrational. $\sqrt4$ and $\sqrt9$ are $2$ and $3$. The spiral produces $\sqrt{n+1}$ for each $n$, and perfect squares come up along the way.

Practise next

• Chapter 4, Exploring Algebraic Identities — the next chapter, which returns to algebra now that the number system underneath it is complete.
• Exercise Set 3.5 — worth revisiting alongside Questions 11 and 12, since predicting whether a decimal terminates and predicting how many places it runs to are the same factorisation done twice.
• Exercise Set 3.4, Question 6 — lists several ways of finding a rational number between two others, which is the method Questions 5, 6, 7 and 13 all rely on.