NCERT Solutions Class 9 Maths Exercise 3.4 — The World of Numbers

NCERT Class 9 Mathematics — Ganita Manjari Part I, The World of Numbers, Exercise Set 3.4. All 6 questions solved.

This exercise moves the rational numbers onto the number line, and the property it is really about is density: between any two rational numbers, however close, there is always another one — and therefore infinitely many.

The tool for finding one is the average:

$$\text{between } a \text{ and } b \text{ lies } \frac{a + b}{2}$$

To place $\frac{p}{q}$ on the line, cut each unit interval into $q$ equal parts and count $p$ of them — to the right of $0$ if positive, to the left if negative.

Key insight. Density is what separates the rationals from the integers. Between $3$ and $4$ there is no integer at all, but there are infinitely many rationals. That abundance makes it feel as though the rationals must fill the line completely — and the next section shows they do not, which is the surprise the chapter is building towards.

Question 1

Represent the rational numbers $\dfrac{2}{3}$, $-\dfrac{5}{4}$ and $1\dfrac{1}{2}$ on a single number line.

Solution. All three lie between $-2$ and $2$, so one line from $-2$ to $2$ holds them all. Take each in turn.

$\dfrac{2}{3}$. The denominator is $3$, so divide the interval from $0$ to $1$ into three equal parts and take two of them. The point sits two-thirds of the way from $0$ to $1$ — just past the middle.

$-\dfrac{5}{4}$. The denominator is $4$, so divide each unit into four equal parts. Five of those parts is one whole unit and one more quarter, and the number is negative, so count to the left: past $-1$ and a quarter further, landing a quarter of the way from $-1$ to $-2$.

$$-\frac{5}{4} = -1\frac{1}{4} = -1.25$$

$1\dfrac{1}{2}$. A mixed number: one whole and a half, so exactly halfway between $1$ and $2$.

$$1\frac{1}{2} = \frac{3}{2} = 1.5$$

In order along the line, from left to right:

$$-\frac{5}{4} \;<\; \frac{2}{3} \;<\; 1\frac{1}{2}$$

$$-1.25 \;<\; 0.67 \;<\; 1.5$$

A practical tip for drawing all three on one line: the denominators are $3$, $4$ and $2$, whose LCM is $12$, so marking each unit in twelfths lets all three points fall on marks — $\frac{2}{3} = \frac{8}{12}$, $-\frac{5}{4} = -\frac{15}{12}$, $\frac{3}{2} = \frac{18}{12}$.

Question 2

Find three distinct rational numbers that lie strictly between $-\dfrac{1}{2}$ and $\dfrac{1}{4}$.

Solution. Two methods work, and both are worth knowing.

Method 1 — a common denominator. Write both endpoints over the same denominator and then look for numerators in between. The LCM of $2$ and $4$ is $4$:

$$-\frac{1}{2} = -\frac{2}{4}, \qquad \frac{1}{4} = \frac{1}{4}$$

The numerators available strictly between $-2$ and $1$ are $-1$ and $0$ — only two, and three are needed. So enlarge the denominator. Multiplying top and bottom by $4$:

$$-\frac{1}{2} = -\frac{8}{16}, \qquad \frac{1}{4} = \frac{4}{16}$$

Now there is plenty of room: any of $-\frac{7}{16}, -\frac{6}{16}, \ldots, \frac{3}{16}$ will do. Taking three:

$$-\frac{1}{4}, \qquad 0, \qquad \frac{1}{8}$$

Method 2 — repeated averaging. The midpoint of the two endpoints is

$$\frac{-\frac{1}{2} + \frac{1}{4}}{2} = \frac{-\frac{1}{4}}{2} = -\frac{1}{8}$$

then average that with an endpoint again, and so on. This produces as many as you like, one at a time.

Check that all three answers are strictly inside. In decimals the interval runs from $-0.5$ to $0.25$, and $-0.25$, $0$ and $0.125$ all lie within it, none touching an endpoint. ✓

Question 3

Simplify the expression: $\left( -\dfrac{1}{4} \right) + \left( \dfrac{5}{12} \right)$.

Solution. A common denominator is needed. The LCM of $4$ and $12$ is $12$:

$$-\frac{1}{4} = -\frac{3}{12}$$

$$-\frac{3}{12} + \frac{5}{12} = \frac{-3 + 5}{12} = \frac{2}{12}$$

Now reduce — $2$ and $12$ share a factor of $2$:

$$\frac{2}{12} = \frac{1}{6}$$

The answer is positive because $\frac{5}{12}$ is larger in size than $\frac{1}{4} = \frac{3}{12}$.

Question 4

A tailor has $15\dfrac{3}{4}$ metres of fine silk. If making one kurta requires $2\dfrac{1}{4}$ metres of silk, exactly how many kurtas can he make?

Solution. Convert both mixed numbers into improper fractions before dividing — mixed numbers cannot be divided as they stand.

$$15\frac{3}{4} = \frac{15 \times 4 + 3}{4} = \frac{63}{4}, \qquad 2\frac{1}{4} = \frac{2 \times 4 + 1}{4} = \frac{9}{4}$$

Now divide, which means multiplying by the reciprocal:

$$\frac{63}{4} \div \frac{9}{4} = \frac{63}{4} \times \frac{4}{9} = \frac{63}{9} = 7$$

The fours cancel outright, which is what makes this so clean: the question is really just $63 \div 9$.

The answer is exactly $7$ with nothing left over — which is why the question says “exactly how many”. Check: $7 \times 2\frac{1}{4} = 7 \times \frac{9}{4} = \frac{63}{4} = 15\frac{3}{4}$ metres, the whole piece used up. ✓

Question 5

Find three rational numbers between $3.1415$ and $3.1416$.

Solution. The two numbers differ only in the fourth decimal place, so go to the fifth decimal place and there is room to spare:

$$3.14151, \qquad 3.14155, \qquad 3.14159$$

Each is strictly larger than $3.14150$ and strictly smaller than $3.14160$, so all three lie inside.

These are rational numbers because every terminating decimal is: for instance

$$3.14155 = \frac{314155}{100000} = \frac{62831}{20000}$$

a ratio of two integers.

This interval is worth a second look. It contains $\pi = 3.14159265\ldots$, which is irrational — so between these two rationals sit both infinitely many rationals and infinitely many irrationals. That the two kinds are interleaved everywhere is the point the next section makes.

Question 6

*Can you think of other way(s) to find a rational number between any two rational numbers?

Solution. The book’s method is the average, $\frac{a+b}{2}$. Here are three genuine alternatives.

1. Raise the common denominator. Write $a$ and $b$ over a shared denominator and then multiply top and bottom of both by any whole number $k$ large enough to leave gaps between the numerators. Between $\frac{1}{3} = \frac{10}{30}$ and $\frac{2}{5} = \frac{12}{30}$ there is only $\frac{11}{30}$; scaling by $10$ gives $\frac{100}{300}$ and $\frac{120}{300}$, with nineteen numbers between them. This is Method 1 of Question 2, and it produces many numbers at once rather than one.

2. Use the decimal expansions. Write both numbers as decimals, find the first place where they differ, and choose any value in between at the next place along — the method used in Question 5. It is the quickest by hand.

3. Take a weighted average. Instead of the midpoint, use any point a fraction of the way across:

$$a + t(b – a) \quad \text{for any rational } t \text{ with } 0 < t < 1$$

$t = \frac{1}{2}$ gives the ordinary average; $t = \frac{1}{3}$ gives a point a third of the way from $a$ to $b$. Since $t$ can be any rational strictly between $0$ and $1$, this one formula generates infinitely many answers directly.

A fourth, for positive numbers only, is the mediant: between $\frac{a}{b}$ and $\frac{c}{d}$ lies $\frac{a+c}{b+d}$ — so between $\frac{1}{3}$ and $\frac{2}{5}$ lies $\frac{3}{8}$. Check: $0.333 < 0.375 < 0.4$ ✓. This one is surprising, because adding numerators and denominators looks like exactly the mistake warned against in Exercise Set 3.3 — it is not a way of adding fractions, but it does reliably land between them.

Common mistakes

• Question 1, reading $1\frac{1}{2}$ as $\frac{11}{2}$. A mixed number is a sum, $1 + \frac{1}{2} = \frac{3}{2}$, not the digits written side by side. $\frac{11}{2} = 5.5$ would not even fit on the same short number line as the other two.
• Question 1, marking $-\frac{5}{4}$ between $0$ and $-1$. Five quarters is more than one whole, so the point lies beyond $-1$, at $-1.25$.
• Question 2, choosing $-\frac{1}{2}$ or $\frac{1}{4}$ themselves. “Strictly between” excludes the endpoints. Converting to decimals is the quickest way to confirm each answer is genuinely inside.
• Question 2, giving up when the common denominator is too small. Over $4$ there are only two numerators available. Enlarging the denominator creates as much room as needed — the supply of rationals in any interval is infinite.
• Question 4, dividing the mixed numbers as they stand. Writing $15\frac{3}{4} \div 2\frac{1}{4}$ as “$15 \div 2$ and $\frac{3}{4} \div \frac{1}{4}$” gives nonsense. Convert to $\frac{63}{4}$ and $\frac{9}{4}$ first.
• Question 4, multiplying instead of dividing. The question asks how many lots of $2\frac{1}{4}$ fit into $15\frac{3}{4}$, which is a division. Multiplying gives $\frac{567}{16}$, about $35.4$, which is plainly more kurtas than the silk allows.
• Question 5, offering $3.1416$ or $3.1415$. These are the endpoints. Also, “between” needs the numbers strictly inside, and $3.14160$ is the same number as $3.1416$.
• Question 5, claiming there is no room. Decimals do not stop at four places. Going one place further gives nine choices immediately, and each further place multiplies that by ten.

Practise next

• Exercise Set 3.5 — asks which rationals have terminating decimals and which repeat, which is the natural sequel to writing rationals as decimals in Questions 5 and 6 here.
• The End-of-Chapter Exercises, Questions 5, 6, 7 and 15 — four more “find rationals in between” questions, with Question 15 asking you to prove that the average always lies between the two numbers rather than just using it.