NCERT Solutions for Class 9th Maths Chapter 3 .4 Filling the Spaces: Fractions and Rational Numbers — Think and Reflect

Book page 493 Updated on2026-09-08

Q1.

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

Answer

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

a/b and c/d, with m = LCM(b, d)
a/b = (a × m/b)/m and c/d = (c × m/d)/m
then a/b + c/d = (a·m/b + c·m/d)/m

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

12 = 2² × 3, 8 = 2³ → LCM = 2³ × 3 = 24
7/12 = 14/24 and 5/8 = 15/24
sum = (14 + 15)/24 = 29/24

Why it happens: multiplying numerator and denominator by the same non-zero number does not change the value, because (ka)/(kb) = a/b whenever k ≠ 0. Once both fractions count parts of the same size, the counts can simply be added — which is exactly the rule a/b + c/b = (a + c)/b that the chapter states.

Tip: the product b × d always works as a common denominator, but the LCM keeps the numbers small and often saves the final reduction step.

Q2.

Verify the distributive law for rational numbers.

Answer

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

Left side
q + r = 1/4 + 5/6 = 3/12 + 10/12 = 13/12
p(q + r) = 2/3 × 13/12 = 26/36 = 13/18

Right side
pq = 2/3 × 1/4 = 2/12 = 1/6
pr = 2/3 × 5/6 = 10/18 = 5/9
pq + pr = 1/6 + 5/9 = 3/18 + 10/18 = 13/18

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

A single example only illustrates. Here is the general proof, using p = a/b, q = c/d, r = e/d with a common denominator d:

p(q + r) = (a/b) × (c + e)/d = a(c + e)/(bd)
= (ac + ae)/(bd) (integers distribute)
= ac/(bd) + ae/(bd)
= (a/b)(c/d) + (a/b)(e/d) = pq + pr

Why it happens: the whole argument rests on one line — a(c + e) = ac + ae for integers. Rational arithmetic is built out of integer arithmetic, so every law the integers obey is inherited by ℚ.

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