NCERT Solutions for Class 9 Maths Chapter 3 The World of Numbers (PDF) – 2027

He leaves with 90 copper ingots.

All four are prime numbers — in fact they are exactly the primes between 10 and 20.

Continuing past 19, we test each number for a factor other than 1 and itself:

The next three are 23, 29 and 31.

No — the natural numbers are not closed under subtraction.

On one hand you can count up to 12.

The thumb touches the joints one by one — three on each of the four fingers — so one full hand counts one dozen. This is the link with base 12: counting is done in groups of 12, and a fresh count starts after every 12, just as base 10 starts again after every 10.

Because removing a debt makes you richer — and “removing” is what the second negative sign does.

The midnight temperature is −11 °C.

Writing a debt as a negative integer and a fortune as a positive one:

His final standing is −₹100, that is, he is ₹100 in debt.

Suppose you have ₹10 in hand and your account also carries a debt of ₹5 written as −5. If a relative takes that debt away — subtracts the −5 — nothing is removed from your pocket, yet you are ₹5 better off.

So taking away a debt of ₹5 has the same effect as adding ₹5: 10 − (−5) = 10 + 5.

Because division by zero has no answer — and the symbol p/q means “the number which, multiplied by q, gives p”.

Now two separate disasters appear:

• If p ≠ 0, say 5/0, we would need x × 0 = 5. No number does that, so no value exists.
• If p = 0, that is 0/0, then x × 0 = 0 is true for every x. So every number would qualify, and the symbol names nothing in particular.

Replace each rational number by an equivalent one whose denominator is a common multiple of the two denominators — most conveniently their LCM.

For example, to add 7/12 and 5/8:

The law claims p(q + r) = pq + pr for all rational p, q, r. Take p = 2/3, q = 1/4, r = 5/6.

Both sides give 13/18, so the law holds for this triple.

The chapter’s test is: a/b = c/d exactly when ad = bc. Apply it to each pair.

The rule is a/b × c/d = ac/bd. Cancelling common factors before multiplying keeps the numbers small.

Dividing by c/d means multiplying by its reciprocal d/c, which needs c ≠ 0.

Both sides equal 10/3, so the statement is proved.

Distribute 7/9 across the bracket:

Expand the left side using the distributive law:

The equation reduces to a true statement with no x left in it. So it is an identity: every rational number x satisfies it.

Follow the chapter’s rule: to mark p/q, cut the unit interval into q equal parts and count p of them from 0 — right if the number is positive, left if it is negative.

Two things have to be shown: that the average is rational, and that it lies between a and b.

1. It is rational. ℚ is closed under addition, and closed under division except by zero.

Here ps + qr and 2qs are integers and 2qs ≠ 0, so (a + b)/2 is rational.

2. It lies between. Suppose a < b. Then

Both gaps are positive and, in fact, equal — so a < (a + b)/2 < b, with the average exactly halfway.

Each number is placed by cutting its own unit interval into the number of parts named by the denominator.

Rewrite both endpoints over a common denominator so that the gap contains enough whole-numbered steps.

Now the integers strictly between −4 and 2 are −3, −2, −1, 0, 1 — five choices. Take any three:

Convert both mixed numbers to improper fractions and divide.

He can make exactly 7 kurtas, with no silk left over.

Add one more decimal place. Between the fourth-place numbers 3.1415 and 3.1416 there is room for nine fifth-place numbers.

Any numerator strictly between 314150 and 314160 works, for example:

Yes — the average is only one recipe. Here are three more:

• Common denominator, bigger denominator. Write a = p/m and b = r/m with r − p ≥ 2 (enlarge m if needed). Then (p + 1)/m lies strictly between them.
• Weighted average. For any rational t with 0 < t < 1, the number ta + (1 − t)b lies between a and b. Taking t = 1/2 gives the ordinary average; t = 1/3 gives a point one-third of the way along.
• The mediant of two fractions. For positive a = p/q and b = r/s with p/q < r/s, the fraction (p + r)/(q + s) always lies strictly between them.

No. There is no pair of integers p, q with q ≠ 0 for which √2 = p/q.

A first look at why the answer must be no, before the formal proof of Section 3.5.1:

Claim: √3 is irrational. Follow the eight steps of Section 3.5.1, replacing “even” by “divisible by 3”.

The assumption is the only thing that can be wrong, so √3 is irrational.

Does it work for √5, √7, √10? Yes, all three.

By using the Baudhāyana–Pythagoras Theorem instead of division. Cutting a unit into q equal parts can only ever produce rational lengths p/q; a right triangle produces lengths that are square roots.

The construction of Fig. 3.11 then transfers that length onto the number line:

• Step 1. Mark OA = 1 unit on the number line and erect a perpendicular at A.
• Step 2. On that perpendicular mark B with AB = 1 unit and join OB. Then OB = √2.
• Step 3. With centre O and radius OB draw an arc cutting the number line at P. Then OP = √2, so P is the point √2.

Length √3. Start from the √2 already built at P.

Length √5. Repeat twice more, or take a shortcut with legs 1 and 2:

The general rule. Having constructed √n, erect a perpendicular of length 1 at its far end. The new hypotenuse satisfies

Exactly those p/q which, in lowest terms, have a denominator whose only prime factors are 2 and 5.

What we observe. Both are non-terminating and repeating, but the repetition begins in different places:

• In 10/3 the repeating block starts immediately after the decimal point — a pure repeating decimal.
• In 11/12 two digits, 9 and 1, come first and only then does 6 repeat — a general repeating decimal.

Because then the denominator can be made a power of 10, by multiplying the numerator and the denominator by a suitable number. (Throughout, p/q is in lowest terms.)

If q = 2^m × 5^n, the decimal terminates. Let k be the larger of m and n. Multiply top and bottom by whatever is missing:

A whole number divided by 10^k is a decimal that stops after k places. The chapter’s own example is 3/20 with 20 = 2² × 5, k = 2:

Prediction from the denominators. Each fraction is already in lowest terms, so factorise the denominator and look for primes other than 2 and 5.

Now the long divisions.

All three predictions are confirmed.

The repeating block is 076923, of length 6. Now the other thirteenths:

What we notice. The twelve blocks are not twelve different sets of digits — they fall into two families, each a cyclic rotation of one block:

• Family A (rotations of 076923): 1/13, 3/13, 4/13, 9/13, 10/13, 12/13
• Family B (rotations of 153846): 2/13, 5/13, 6/13, 7/13, 8/13, 11/13

So 2/13 does not give a rotation of 1/13’s block — it starts a second cycle. This is the difference from 1/7.

The working for the rational ones.

So 0.9 and 1 are two decimal names for the same number.

The reciprocal 1/n gives a fully cyclic block exactly when the repeating block has the greatest possible length, n − 1. That forces n to be prime.

So n = 7, 17, 19, 23, 29 all work, and the list continues with 47, 59, 61, 97, …

There is no real number whose square is −1. The reason is Brahmagupta’s own sign rule.

Every real number falls into one of these three cases, so x² ≥ 0 for every real x. Since −1 < 0, the equation x² = −1 has no solution anywhere on the number line.

Proof by contradiction, following the pattern of Section 3.5.1.

Every step after the assumption is forced, so the assumption itself must be false. Hence √5 cannot be written as p/q — it is irrational.

(i) 0.532 is located by successive magnification: each decimal place cuts the interval into ten equal parts.

(ii) 1.15 is a repeating decimal, so first turn it into a fraction and then place it exactly.

So 1.15 = 52/45 = 1 7/45. To mark it exactly, divide the interval from 1 to 2 into 45 equal parts and count 7 of them from 1. By magnification it lies between 1.15 and 1.16, just past one half of that gap.

Write both endpoints over a denominator large enough to leave six whole numerators in between. Since we want 6 numbers, use 6 + 1 = 7.

So six rational numbers between 3 and 4 are

Five rational numbers between 2/5 and 3/5:

Choosing five of them:

Check the order in decimals: 0.16 < 0.2 < 0.23 < 0.26 < 0.3 < 0.3 < 0.4 ✓

ab is negative — in fact ab = −1 exactly.

Since −1 < 0, the product ab is negative for every such pair a, b.

Part 1: the form p/10⁴. A decimal that stops at the 4th place has the shape

Here d₄, the 4th decimal digit, is the last non-zero digit, and d₄ is exactly the units digit of p. Since d₄ ≠ 0, the integer p does not end in 0, so 10 does not divide p. ∎

Part 2: yes, it is necessary. N = p/10⁴ = p/(2⁴ × 5⁴). Since 10 does not divide p, p is not divisible by both 2 and 5:

• if 2 does not divide p, no factor 2 cancels, so the lowest-form denominator keeps the whole 2⁴;
• if 5 does not divide p, no factor 5 cancels, so it keeps the whole 5⁴.

So the denominator in lowest form is always divisible by 2⁴ or by 5⁴ (possibly both).

The denominator has the form 2⁰ × 5³, so the decimal terminates. The number of places is the larger of the two exponents:

Confirming without long division:

The expansion terminates after 3 decimal places.

A whole number over 1000 stops after three places. For example, 3/40:

Step 1: a common denominator that is large enough.

Step 2: five rational numbers between them. Any five of 22/36, 23/36, …, 29/36 will do, for example:

Step 3: why the size condition is needed.

Square the first condition and use the second.

Now x², y², z² are each squares of rational numbers, so each is ≥ 0. Three non-negative numbers add to 0 only if every one of them is 0:

Take a ≠ b and, without loss of generality, let a < b (otherwise swap the names).

Combining, a < (a + b)/2 < b. ∎

It is also a rational number: a + b is rational because ℚ is closed under addition, and dividing by 2 ≠ 0 keeps it rational.

Fig. 3.14 is built by the rule of Section 3.5.2: each new right triangle takes the previous hypotenuse as one leg and a fresh segment of length 1 as the other.

The figure prints ten right triangles, so their hypotenuses are:

So the ten hypotenuses are √2, √3, 2, √5, √6, √7, 2√2, 3, √10 and √11.