NCERT Solutions for Class 9th Maths Chapter 3 .4.2 The Density of Rational Numbers — Exercise Set 3.4

Book page 523 Updated on2026-09-08

Q1.
Represent the rational numbers 2/3, –5/4 and 1 1/2 on a single number line.
Answer

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

NumberLies betweenConstruction
2/30 and 1cut 0–1 into 3 parts, take 2 to the right
−5/4 = −1 1/4−2 and −1cut −1 to −2 into 4 parts, take 1 to the left of −1
1 1/2 = 3/21 and 2halfway between 1 and 2
−2−1012 −5/4 2/3 quarters thirds half
All three rational numbers marked on one line between −2 and 2.
Tip: to draw all three on a single ruler, use twelfths — 2/3 = 8/12, −5/4 = −15/12 and 3/2 = 18/12, since 12 = LCM(3, 4, 2).
Q2.
Find three distinct rational numbers that lie strictly between – 1/2 and 1/4.
Answer

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

LCM(2, 4) = 4, so −1/2 = −2/4 and 1/4 = 1/4
only 0/4 lies strictly between — one number, not three
so enlarge the denominator: use eighths
−1/2 = −4/8 and 1/4 = 2/8

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

−3/8, −1/8, 1/8
check: −4/8 < −3/8 < −1/8 < 1/8 < 2/8 ✓
Why it happens: with a common denominator d, the numbers between the two endpoints are exactly the fractions k/d whose numerators k lie between the two numerators. So the number of candidates is controlled entirely by how far apart the numerators are — and multiplying the denominator by 2 doubles that gap without moving the endpoints.
Check it yourself: repeated averaging works too. (−1/2 + 1/4)/2 = −1/8; then average −1/8 with each endpoint to get −5/16 and 1/16.
Q3.
Simplify the expression: (−1/4) + (5/12).
Answer
LCM(4, 12) = 12
−1/4 = −3/12
−3/12 + 5/12 = 2/12
divide numerator and denominator by 2
= 1/6
Why it happens: 5/12 is larger than 1/4 = 3/12, so the sum lands on the positive side of 0. Reducing 2/12 to 1/6 is allowed because 2/12 and 1/6 are equivalent rational numbers, and the chapter asks us to name a rational number by its co-prime form.
Q4.
A tailor has 15 3/4 metres of fine silk. If making one kurta requires 2 1/4 metres of silk, exactly how many kurtas can he make?
Answer

Convert both mixed numbers to improper fractions and divide.

15 3/4 = (15 × 4 + 3)/4 = 63/4
2 1/4 = (2 × 4 + 1)/4 = 9/4
63/4 ÷ 9/4 = 63/4 × 4/9 = 63/9 = 7

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

Why it happens: both lengths are whole numbers of quarter-metres — 63 quarters and 9 quarters. Dividing them is just 63 ÷ 9, and since 9 divides 63 exactly, nothing remains. The common denominator has done all the work.
Check it yourself: 7 × 2 1/4 = 7 × 9/4 = 63/4 = 15 3/4 metres. ✓
Q5.
Find three rational numbers between 3.1415 and 3.1416.
Answer

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

3.1415 = 31415/10000 = 314150/100000
3.1416 = 31416/10000 = 314160/100000

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

3.14152 = 314152/100000 = 39269/12500
3.14155 = 314155/100000 = 62831/20000
3.14158 = 314158/100000 = 157079/50000
Why it happens: every terminating decimal is a rational number, and inserting a further decimal place multiplies the denominator by 10 while multiplying both numerators by 10 as well. The gap between the numerators widens from 1 to 10, creating nine new rationals — and the trick can be repeated for ever. This is the density of ℚ seen in decimal clothing.
Did you know? π = 3.14159265… lies inside this very interval, but π is not one of the rationals you can list this way — its decimal never terminates and never repeats.
Q6.
Can you think of other way(s) to find a rational number between any two rational numbers?
Answer

Yes — the average is only one recipe. Here are three more, each with its own proof.

  • 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 mediator 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.

Proof of the third, the least obvious one:

p/q < r/s means ps < qr
(p + r)/(q + s) − p/q = (q(p + r) − p(q + s))/(q(q + s))
= (qr − ps)/(q(q + s)) > 0
similarly r/s − (p + r)/(q + s) = (qr − ps)/(s(q + s)) > 0

So p/q < (p + r)/(q + s) < r/s.

Why it happens: in every method the answer is built from a, b and integers using only +, −, × and ÷ by non-zero numbers. Since ℚ is closed under all four, the result is bound to be rational; the inequalities then pin it inside the interval. Try the mediator on 1/2 and 2/3: it gives (1 + 2)/(2 + 3) = 3/5, and indeed 1/2 < 3/5 < 2/3.
Tip: the mediator is not the same as adding fractions. (p + r)/(q + s) is a well-known construction called the mediant, and it is the reason the fractions on a Farey sequence line up so neatly.
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