NCERT Solutions Class 9 Maths Chapter 3 Think and Reflect — The World of Numbers

NCERT Class 9 Mathematics — Ganita Manjari Part I, The World of Numbers. All 12 Think and Reflect prompts answered.

Also here. Every exercise in this chapter is solved, one page per exercise.

Chapter 3 tells the story of the number line being filled in — naturals, then zero and the negatives, then the rationals, then the irrationals. Its eleven Think and Reflect boxes are the places where the book stops and asks why a step was necessary, rather than how to carry it out.

Several of them ask for a justification of something you already know how to do, which is a harder and more valuable task than the computation. The two about constructing irrational lengths are the ones most worth attempting before reading the answers.

Key insight. Nearly every prompt here is about a gap in the number system and the repair that fills it. Subtraction breaks the naturals; division by zero cannot be repaired at all; the diagonal of a square breaks the rationals; and the square root of $-1$ breaks the reals. Recognising which prompts are repairs and which are genuine impossibilities is what the chapter is teaching.

Prompt 1 comes from the box on page 46, alongside Brahmagupta’s sign rules.

Question 1

Why does a negative times a negative equal a positive? Think of it in terms of action and debt. If a negative number represents a debt, then multiplying by a negative number represents the removal of that debt.

Solution. Take the hint literally and count the money.

Suppose you hold four IOUs, each recording a debt of ₹$3$. Each is worth $-3$ to you, so your position is

$$4 \times (-3) = -12$$

Now someone takes away those four debts — tears up all four notes. Taking away is the negative of adding, so this is $-4$ lots of $-3$:

$$(-4) \times (-3) = ?$$

Nothing physical has entered your pocket, yet you are now ₹$12$ better off than you were, because you no longer owe it. So the answer must be $+12$.

A second argument, from the pattern. Watch the products as the multiplier falls:

$$3 \times (-3) = -9, \quad 2 \times (-3) = -6, \quad 1 \times (-3) = -3, \quad 0 \times (-3) = 0$$

Each step up the list adds $3$. Continuing the same steady pattern past zero:

$$(-1) \times (-3) = 3, \qquad (-2) \times (-3) = 6, \qquad (-3) \times (-3) = 9$$

The rule is not an arbitrary convention — it is the only choice that keeps arithmetic consistent.

Prompt 2 comes from the box on page 47, just after the definition of a rational number as $\frac{p}{q}$ with $q \neq 0$.

Question 2

Can you explain why we need $q \neq 0$ in the definition of a rational number?

Solution. Division is defined as the reverse of multiplication: $\frac{a}{b} = c$ means exactly that $c \times b = a$. Apply that to a zero denominator and two separate things go wrong.

Case 1: $\frac{5}{0}$. This would be a number $c$ with $c \times 0 = 5$. But anything multiplied by $0$ is $0$, never $5$. No such number exists.

Case 2: $\frac{0}{0}$. This would be a number $c$ with $c \times 0 = 0$ — and every number satisfies that. Every number would qualify, so the symbol names nothing in particular.

So a zero denominator gives either no answer or all answers, and neither is a number. Excluding $q = 0$ is not a restriction imposed for tidiness; it is the recognition that the expression would otherwise be meaningless.

Note the asymmetry: $\frac{0}{5} = 0$ is perfectly fine. Zero on top is a number; zero underneath is not.

Prompts 3 and 4 come from the box on page 49, before Exercise Set 3.3.

Question 3

While adding or subtracting two rational numbers having different denominators, how will you make the denominators equal?

Solution. Use the fact that multiplying the numerator and denominator by the same non-zero number leaves a fraction unchanged — it renames the number without altering it.

The method.

1. Find the LCM of the two denominators. This is the smallest number both divide into.
2. For each fraction, work out what the denominator must be multiplied by to reach the LCM, and multiply the numerator by the same amount.
3. Add or subtract the numerators, keeping the common denominator.
4. Reduce the result if possible.

Example. $\frac{7}{12} + \frac{5}{8}$. The LCM of $12$ and $8$ is $24$:

$$\frac{7}{12} = \frac{7 \times 2}{12 \times 2} = \frac{14}{24}, \qquad \frac{5}{8} = \frac{5 \times 3}{8 \times 3} = \frac{15}{24}$$ $$\frac{14}{24} + \frac{15}{24} = \frac{29}{24}$$

Multiplying the two denominators together also works and needs no LCM, but gives larger numbers: over $96$ the same sum is $\frac{116}{96}$, which must then be reduced back to $\frac{29}{24}$.

Why it is needed at all. A fraction counts pieces of a particular size. Thirds and quarters are different-sized pieces, and counting them together means nothing until both are re-cut into twelfths. Multiplication needs none of this, because there the two fractions are not being counted together but scaled.

Question 4

Verify the distributive law for rational numbers.

Solution. The law states that for any rational $a$, $b$, $c$:

$$a \times (b + c) = a \times b + a \times c$$

“Verify” means test it with actual numbers, working out each side independently.

Take $a = \frac{2}{3}$, $b = \frac{1}{4}$, $c = \frac{5}{6}$.

Left-hand side — bracket first:

$$b + c = \frac{1}{4} + \frac{5}{6} = \frac{3}{12} + \frac{10}{12} = \frac{13}{12}$$ $$a \times \frac{13}{12} = \frac{2}{3} \times \frac{13}{12} = \frac{26}{36} = \frac{13}{18}$$

Right-hand side — two products, then add:

$$\frac{2}{3} \times \frac{1}{4} = \frac{2}{12} = \frac{1}{6}, \qquad \frac{2}{3} \times \frac{5}{6} = \frac{10}{18} = \frac{5}{9}$$ $$\frac{1}{6} + \frac{5}{9} = \frac{3}{18} + \frac{10}{18} = \frac{13}{18}$$

Both sides give $\frac{13}{18}$ ✓

A second test with negatives, since those are where laws are most likely to fail. Take $a = -\frac{1}{2}$, $b = \frac{3}{4}$, $c = -\frac{1}{4}$:

$$\text{LHS} = -\frac{1}{2}\left( \frac{3}{4} – \frac{1}{4} \right) = -\frac{1}{2} \times \frac{1}{2} = -\frac{1}{4}$$ $$\text{RHS} = -\frac{3}{8} + \frac{1}{8} = -\frac{2}{8} = -\frac{1}{4} \; \checkmark$$

Verification is not proof — two examples cannot establish a law for every rational. But it does confirm the rule behaves as claimed, and Question 6 of Exercise Set 3.3 asks for exactly this kind of check on a specific case.

Prompt 5 comes from the box on page 51, after Fig. 3.6 has placed $\frac{9}{4}$ on the number line.

Question 5

Try and represent $\dfrac{8}{5}$ and $-\dfrac{7}{4}$ on a number line.

Solution. For each, the denominator says how finely to cut each unit and the numerator says how many pieces to count.

$\dfrac{8}{5}$. Cut each unit into five equal parts. Eight fifths is one whole unit and three fifths more, so count to the right of $0$: past $1$, then three of the five sub-marks onwards.

$$\frac{8}{5} = 1\frac{3}{5} = 1.6$$

It sits between $1$ and $2$, closer to $2$.

$-\dfrac{7}{4}$. Cut each unit into four equal parts. Seven quarters is one whole unit and three quarters more, and the number is negative, so count to the left of $0$: past $-1$, then three of the four sub-marks further.

$$-\frac{7}{4} = -1\frac{3}{4} = -1.75$$

It sits between $-1$ and $-2$, closer to $-2$.

To draw both on one line, use a common denominator: the LCM of $5$ and $4$ is $20$, so marking each unit in twentieths puts both on marks — $\frac{8}{5} = \frac{32}{20}$ and $-\frac{7}{4} = -\frac{35}{20}$.

Prompt 6 comes from the box on page 53, after the diagonal of a unit square is found to be $\sqrt2$.

Question 6

Can $\sqrt2$ be written as a rational number $\dfrac{p}{q}$?

Solution. No — and this is the discovery that forces the number line to be enlarged again.

The chapter gives Hippasus’s proof by contradiction. Suppose $\sqrt2 = \frac{p}{q}$ in lowest terms. Squaring gives $p^2 = 2q^2$, so $p^2$ is even and therefore $p$ is even; writing $p = 2k$ gives $q^2 = 2k^2$, so $q$ is even too. But then $p$ and $q$ share the factor $2$, contradicting “lowest terms”.

What makes this so striking is that $\sqrt2$ is not an exotic quantity. It is the diagonal of a square with sides of one unit — a length you can draw with a ruler in a second. The rationals, dense as they are, miss it completely.

So there are points on the number line that no fraction names. The rationals do not fill the line, and the numbers filling the gaps are the irrationals.

Prompts 7 and 8 come from the two boxes on page 55, during the proof of the irrationality of $\sqrt2$ and the construction that follows it.

Question 7

Try to prove the irrationality of $\sqrt3$ using the approach of proof by contradiction. Will the same approach work for $\sqrt5$, $\sqrt7$, or $\sqrt{10}$?

Solution. The proof for $\sqrt3$. Suppose $\sqrt3 = \frac{p}{q}$ with $p$, $q$ integers sharing no common factor. Squaring:

$$3 = \frac{p^2}{q^2} \quad\Longrightarrow\quad p^2 = 3q^2$$

So $3$ divides $p^2$. Since $3$ is prime, a prime dividing a square must divide the number, so $3$ divides $p$. Write $p = 3k$:

$$9k^2 = 3q^2 \quad\Longrightarrow\quad q^2 = 3k^2$$

By the same step, $3$ divides $q$. So $p$ and $q$ share the factor $3$ — contradicting “lowest terms”. Hence $\sqrt3$ is irrational.

Does it work for the others?

$\sqrt5$ and $\sqrt7$: yes, unchanged. Both $5$ and $7$ are prime, so the step “a prime dividing $p^2$ divides $p$” applies word for word. Replace every $3$ by $5$ or $7$.

$\sqrt{10}$: yes, but with one extra step. $10$ is not prime, so “$10$ divides $p^2$ implies $10$ divides $p$” cannot be quoted directly. Handle the prime factors separately: from $p^2 = 10q^2$, both $2$ and $5$ divide $p^2$, so each divides $p$, and therefore $10$ divides $p$. From there the argument runs as before. It works because $10 = 2 \times 5$ has no repeated prime factor.

Where it must fail — and does. Try it on $\sqrt4$. From $p^2 = 4q^2$, it is not true that $4$ dividing $p^2$ forces $4$ to divide $p$ — take $p = 2$. The argument collapses, as it has to, because $\sqrt4 = 2$ is rational. Any version of the proof that appears to work for a perfect square has a mistake in it.

Question 8

We have seen how to obtain a line whose length is a rational number. How do we obtain lines whose lengths are irrational?

Solution. A rational length is easy: to get $\frac{p}{q}$, divide a unit into $q$ equal parts and lay out $p$ of them. Every step is counting, so nothing but rationals can come out.

Irrational lengths need a different tool — and the Baudhāyana–Pythagoras theorem is it, because squaring and then taking the square root leaves the rationals.

Build a right-angled triangle with both legs of length $1$. Its hypotenuse is

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

Now transfer that length to the number line: put a compass point at $0$, open it to the length of the hypotenuse, and draw an arc cutting the line. The point where it cuts is exactly $\sqrt2$.

The construction is entirely straight-edge and compass — no measurement, no approximation — and yet it produces a length that no fraction can name. This is how irrational numbers are located rather than merely asserted to exist.

Prompt 9 comes from the box on page 56, after $\sqrt2$ has been marked on the line.

Question 9

Try to extend this method for constructing line segments of lengths $\sqrt3$ and $\sqrt5$ using a ruler and a compass. Generalise this method to construct a line segment of any length of the form $\sqrt{n}$, where $n$ is a positive integer.

Solution. The trick is to feed each hypotenuse back in as a leg of the next triangle.

Constructing $\sqrt3$. Start from the segment of length $\sqrt2$ just built. At its far end, erect a perpendicular of length $1$ and join back to the origin. The new hypotenuse is

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

Constructing $\sqrt5$. Repeat twice more:

$$\sqrt{\left( \sqrt3 \right)^2 + 1^2} = \sqrt4 = 2, \qquad \sqrt{2^2 + 1^2} = \sqrt5$$

Or take a shortcut, since $5 = 4 + 1$: a right-angled triangle with legs $2$ and $1$ has hypotenuse $\sqrt5$ directly.

The general method. To construct $\sqrt{n}$, build a chain of right-angled triangles:

• Triangle $1$: legs $1$ and $1$, hypotenuse $\sqrt2$.
• Triangle $k$: legs $\sqrt{k}$ (the previous hypotenuse) and $1$, hypotenuse $\sqrt{k+1}$.

After $n-1$ triangles the hypotenuse is $\sqrt{n}$. Then set a compass to that length from $0$ and mark the number line.

Drawing all of these from a common point produces the square root spiral of Fig. 3.14, which is Question 16 of the End-of-Chapter Exercises. Every irrational $\sqrt{n}$ can be reached this way — with a ruler, a compass, and patience.

Prompt 10 comes from the box on page 57, after terminating and repeating decimals are defined.

Question 10

Try to find the decimal expansions of $\dfrac{10}{3}$ and $\dfrac{11}{12}$. What do you observe about the repetition of the digits after the decimal point?

Solution. Divide each out.

$\dfrac{10}{3}$. Three goes into ten three times with remainder $1$, and the remainder $1$ then repeats for ever:

$$\frac{10}{3} = 3.3333\ldots = 3.\overline{3}$$

The repetition starts immediately after the decimal point.

$\dfrac{11}{12}$. Long division gives

$$\frac{11}{12} = 0.91666\ldots = 0.91\overline{6}$$

Here two digits, $9$ and $1$, do not repeat and only then does the $6$ begin looping.

What to observe. Both repeat, but they start repeating at different places — and the denominators explain why. Factorise each:

$$3 = 3, \qquad 12 = 2^2 \times 3$$

Both contain the prime $3$, which is why neither terminates. But $12$ also contains $2^2$, and it is exactly those two factors of $2$ that produce the two non-repeating digits before the loop begins. The count of leading non-repeating digits is the larger of the powers of $2$ and $5$ in the denominator — here $\max(2, 0) = 2$.

$\frac{10}{3}$ has no factors of $2$ or $5$ at all, so nothing delays the repetition.

Prompt 11 comes from the box on page 58, stating the terminating-decimal rule.

Question 11

The decimal expansion of $\dfrac{p}{q}$ will be terminating precisely when the prime factors of $q$ are only $2$, only $5$, or both $2$ and $5$. Can you explain why?

Solution. The reason is that our decimals are written in base ten, and

$$10 = 2 \times 5$$

A terminating decimal is a fraction over a power of ten. The number $0.375$ means $\frac{375}{1000}$, and in general a decimal stopping after $k$ places is $\frac{\text{integer}}{10^k}$.

So the question becomes: for which $q$ can $\frac{p}{q}$ be rewritten with a denominator of $10^k$?

If $q$ has only $2$s and $5$s, say $q = 2^m \times 5^n$, then take $k = \max(m, n)$ and multiply top and bottom by whatever is missing. For instance $\frac{3}{8}$:

$$\frac{3}{2^3} = \frac{3 \times 5^3}{2^3 \times 5^3} = \frac{375}{1000} = 0.375$$

The extra $5$s supply the partners the $2$s need to make tens. So it terminates.

If $q$ has any other prime factor — a $3$, a $7$, an $11$ — that factor can never be removed. It is not a factor of $10^k$ for any $k$, and since $\frac{p}{q}$ is in lowest terms it cannot be cancelled against $p$ either. No power of ten is ever reachable, so the division never comes out exactly and the decimal cannot terminate.

Why it must then repeat. When dividing by $q$, each remainder is one of $1, 2, \ldots, q-1$ — only finitely many. Carry on long enough and a remainder must recur; from that moment the digits repeat the same cycle for ever. So a non-terminating decimal expansion of a fraction always repeats, and the block is at most $q – 1$ digits long.

Prompt 12 comes from the box on page 64, at the very end of the chapter.

Question 12

Consider this puzzle: What is the square root of $-1$? We know that $1 \times 1 = 1$. We also know that $(-1) \times (-1) = 1$. There is no real number that, when multiplied by itself, results in a negative number. Thus $\sqrt{-1}$ cannot exist on the number line.

Solution. The reasoning in the prompt is correct, and it is worth seeing exactly why it closes off every possibility.

Take any real number $x$ and ask what $x^2$ can be:

• if $x > 0$, then $x^2 > 0$ (positive times positive);
• if $x < 0$, then $x^2 > 0$ (negative times negative — Question 1);
• if $x = 0$, then $x^2 = 0$.

Those three cases exhaust the real numbers, and in none of them is $x^2$ negative. So no real number squares to $-1$, and $\sqrt{-1}$ is not on the number line at all.

Notice how different this is from $\sqrt2$. That was on the line — it just had no fractional name, so the rationals had to be enlarged to the reals. Here the number is missing from the line entirely, and no amount of filling gaps will produce it.

What mathematicians did about it. Rather than fill a gap, they added a new direction. The symbol $i$ is defined by

$$i^2 = -1$$

and numbers of the form $a + bi$ are the complex numbers. They live in a plane, not on a line — which is why the number line could never have contained them.

This is the chapter’s last move in a pattern it has repeated throughout: subtraction broke the naturals and gave the integers; division broke the integers and gave the rationals; the diagonal of a square broke the rationals and gave the reals; and the square root of a negative breaks the reals and gives the complex numbers. Each time, an operation with no answer was the reason to invent one. Far from being fictional, these numbers are essential to electrical engineering and quantum mechanics — as the book notes, to the technology in a mobile phone.

Common mistakes

• Question 1, answering “because that’s the rule”. The prompt asks why. The debt picture and the pattern argument both give a reason, and the pattern one shows the rule is forced rather than chosen.
• Question 2, saying “division by zero is infinity”. It is not. $\frac{5}{0}$ has no answer at all, and $\frac{0}{0}$ has every answer — two different failures, neither of which is a number.
• Question 3, describing the method for multiplication. Only addition and subtraction need a common denominator. Answering with “multiply the numerators and denominators” answers a different question.
• Question 4, rewriting one side into the other. Verification means evaluating both sides separately and comparing. Using the law to transform the left into the right assumes what is being checked.
• Question 5, marking $\frac{8}{5}$ between $0$ and $1$. Eight fifths is more than one whole, so it lies past $1$, at $1.6$. Converting to a mixed number first prevents this.
• Question 7, using the argument on $\sqrt4$. It appears to work and reaches a false conclusion, because the step “$n$ divides $p^2$ implies $n$ divides $p$” needs $n$ prime. Testing a proof on a case where you know the answer is the way to find this out.
• Question 7, claiming the method fails for $\sqrt{10}$. It succeeds, but the prime step must be applied to $2$ and $5$ separately before concluding $10 \mid p$.
• Question 9, expecting a different construction for each $\sqrt{n}$. One construction repeated is enough: each hypotenuse becomes the next leg, and the chain reaches every $\sqrt{n}$ in turn.
• Question 10, saying only one of the two repeats. Both repeat. What differs is where the repetition starts, and the factors of $2$ in the denominator are the reason.
• Question 11, answering “because $2$ and $5$ are small”. The reason is that $10 = 2 \times 5$, so only those two primes can be built up into a power of ten. Nothing about their size matters.
• Question 12, concluding that $\sqrt{-1}$ is meaningless. It has no meaning among the real numbers, which is a different claim. Enlarging the number system gives it one — the same move that produced the integers, the rationals and the reals in this very chapter.

Practise next

• The End-of-Chapter Exercises, Question 2 — proving $\sqrt5$ irrational, which is Question 7 above carried out in full.
• The End-of-Chapter Exercises, Questions 11 and 12 — the terminating-decimal rule of Question 11 above applied to predict the exact number of decimal places.
• Exercise Set 3.5 — the decimal expansions of Question 10 explored systematically, including the cyclic behaviour of $\frac{1}{7}$ and $\frac{1}{13}$.