NCERT Solutions Class 9 Maths Exercise 3.1 — The World of Numbers

NCERT Class 9 Mathematics — Ganita Manjari Part I, The World of Numbers, Exercise Set 3.1. All 4 questions solved.

This first exercise sits at the start of the chapter’s history of counting, and its questions are about the natural numbers $\mathbb{N} = \{1, 2, 3, \ldots\}$ — where they came from and what they cannot do.

One word does a lot of work here. A set is closed under an operation if performing that operation on members of the set always lands you back inside the set. The naturals are closed under addition and multiplication, and the whole of the rest of the chapter is driven by the operations they are not closed under.

Key insight. To show a set is closed you must argue about every possible case; to show it is not closed, a single counterexample is enough and no more is needed. That asymmetry is why Question 3 asks only for “a couple of examples” — finding one pair whose difference escapes the naturals settles the matter completely.

Question 1

A merchant in the port city of Lothal is exchanging bags of spices for copper ingots. He receives $15$ ingots for every $2$ bags of spices. If he brings $12$ bags of spices to the market, how many copper ingots will he leave with?

Solution. The rate is $15$ ingots per $2$ bags, so work out how many lots of $2$ bags there are and multiply.

$$12 \div 2 = 6 \text{ lots of two bags}$$ $$6 \times 15 = 90 \text{ ingots}$$

Equivalently, set it out as a proportion and scale up. Twelve bags is six times two bags, so the ingots must also be six times as many:

$$\frac{15}{2} = \frac{n}{12} \quad\Longrightarrow\quad n = \frac{15 \times 12}{2} = 90$$

Question 2

Look at the sequence of numbers on one column of the Ishango bone: $11$, $13$, $17$, $19$. What do these numbers have in common? List the next three numbers that fit this pattern.

Solution. Two answers can be defended, and it is worth seeing both.

They are all odd. True, but it does not explain the gaps — $15$ is odd and is missing from the list, and the differences run $2, 4, 2$ rather than a steady $2$.

They are all prime. Each has no factor other than $1$ and itself:

$$11, \quad 13, \quad 17, \quad 19$$

and the numbers skipped over — $12, 14, 15, 16, 18$ — are all composite. In fact these four are every prime between $10$ and $20$, which is what makes the column remarkable: a carving from some twenty thousand years ago listing exactly the primes of a decade.

Continuing with the next primes after $19$:

$$23, \quad 29, \quad 31$$

($20$, $21$, $22$ are composite, as are $24$ to $28$ and $30$.)

Question 3

We know that natural numbers are closed under addition (the sum of any two natural numbers is always a natural number). Are they closed under subtraction? Provide a couple of examples to justify your answer.

Solution. No. To show that a set is not closed, one counterexample is enough — but the question asks for a couple, so here are two of different kinds.

Example 1 — a negative result.

$$3 – 5 = -2$$

Both $3$ and $5$ are natural numbers, but $-2$ is not: the naturals are $\{1, 2, 3, \ldots\}$ and contain nothing below $1$.

Example 2 — a result of zero.

$$7 – 7 = 0$$

Again both inputs are natural, and $0$ is not among $\{1, 2, 3, \ldots\}$ either.

So subtraction can take you out of the set in two different ways, and the naturals are not closed under it.

This failure is exactly what the rest of the chapter repairs. Zero is added to make differences like $7 – 7$ meaningful, and the negatives are added to make $3 – 5$ meaningful — producing the integers, which are closed under subtraction.

Question 4

*Ancient Indians used the joints of their fingers to count, a practice still seen today. Each finger has $3$ joints, and the thumb is used to count them. How many can you count on one hand? How does this relate to the ancient base-$12$ counting systems?

Solution. The thumb is the pointer, not one of the things counted, so only the other four fingers are available:

$$4 \text{ fingers} \times 3 \text{ joints each} = 12$$

Touch the thumb to each joint in turn — three on the index finger, three on the middle finger, and so on — and one hand takes you up to $12$.

Why that leads to base 12. A counting system’s base is usually whatever one hand or one body naturally counts up to. Counting fingers gives ten and hence base $10$; counting finger joints gives twelve and hence base $12$, the duodecimal system.

Twelve is a convenient base for trade because of its factors: $12$ is divisible by $1, 2, 3, 4, 6$ and $12$, while $10$ is divisible only by $1, 2, 5$ and $10$. A quantity can be split into halves, thirds and quarters without fractions.

Traces of it survive all around: $12$ inches in a foot, $12$ months, $24$ hours in two sets of $12$, $12$ items in a dozen and $144$ in a gross — which is $12^2$. And using the second hand to count completed twelves gives $12 \times 5 = 60$, which is where the sixty minutes in an hour and sixty seconds in a minute come from.

Common mistakes

• Question 1, multiplying by $15$ without halving. $12 \times 15 = 180$ treats the rate as $15$ ingots per bag. The rate is per two bags, so the number of bags must be halved first.
• Question 2, answering “they go up by $2$”. The steps are $2$, $4$, $2$ — not constant. Any rule offered has to account for $15$ being absent, and “odd” does not.
• Question 3, offering $5 – 3 = 2$ as evidence of closure. One example that stays inside the set proves nothing; closure is a claim about every pair. The logic runs the other way: a single escape disproves it.
• Question 3, forgetting $0$. Whether $7 – 7 = 0$ is a counterexample depends on where the naturals start, and this book defines them as $\{1, 2, 3, \ldots\}$ — so it is. Saying which definition you are using is part of a complete answer.
• Question 4, counting $15$ joints. Including the thumb’s own joints double-counts the hand: the thumb is doing the pointing. Four fingers, three joints each, is $12$.
• Question 4, stopping at the number. The question has two halves, and the second — why $12$ became a base, and where it survives — is the part that is actually being taught.

Practise next

• Exercise Set 3.2 — the next repair to the number system: zero and the negative numbers, which is precisely what Question 3 shows the naturals to be missing.
• The End-of-Chapter Exercises, Question 2 — proving $\sqrt5$ irrational, the last and largest gap in the number line that this chapter fills.