Ganita Manjari Class 9 Maths Chapter 3: The World of Numbers — NCERT Solutions

Explanation: 22/7 is a ratio of two integers, so it is rational, even though it is often used to approximate π. √25 = 5 and 3.272727... is a repeating decimal, so both are rational. The decimal 0.101001000100001... never terminates and never repeats because the number of zeros keeps increasing, so it cannot be written as p/q and is irrational.

Explanation: By definition, a rational number is any number that can be written as p/q where p and q are integers and q ≠ 0. The condition q ≠ 0 is needed because division by zero is not defined. Restricting to natural numbers would wrongly exclude negative rationals like −3/4, and demanding p ≠ 0 would wrongly exclude 0, which is rational since 0 = 0/1.

Explanation: A rational number always has a decimal expansion that either terminates or repeats a block of digits. An irrational number cannot be written as p/q, so its decimal expansion goes on forever without any repeating block. For example, √2 = 1.41421356... never terminates and never repeats.

Explanation: From √2 = p/q, squaring gives p² = 2q², so p² is even, which forces p to be even, say p = 2m. Substituting gives 4m² = 2q², so q² = 2m², which forces q to be even as well. Then p and q share the common factor 2, contradicting the assumption that they are coprime, so √2 cannot be written as p/q.

Explanation: √2 is the length of the diagonal of the unit square in Fig. 3.10, since 1² + 1² = 2. Suppose, to the contrary, that √2 = p/q where p and q are integers with no common factor and q ≠ 0. Squaring gives p² = 2q², so p² is even, which forces p to be even; write p = 2m. Then 4m² = 2q², so q² = 2m², which makes q even as well. Now p and q share the common factor 2, contradicting our assumption that the fraction was in lowest terms. Hence no such fraction exists and √2 is irrational.

Explanation: In Fig. 3.11, OB = √2 is obtained as the hypotenuse of the right triangle with legs OA = 1 and AB = 1, and an arc of radius OB centred at O locates √2 at P on the number line. For √3, draw a segment BD of length 1 perpendicular to OB at B; by the Pythagoras theorem OD = √((√2)² + 1²) = √3, and swinging an arc of radius OD onto the number line marks √3. For √5, take a right triangle with legs 2 and 1, giving hypotenuse √(2² + 1²) = √5 (equivalently, repeat the unit-perpendicular step on √3 to get √4 = 2, and once more to get √5). In general, once a segment of length √(n-1) has been constructed, erect a perpendicular of unit length at its endpoint; the hypotenuse of the resulting right triangle is √((n-1) + 1) = √n. Starting from 1 and repeating this step, a segment of length √n can be constructed for every positive integer n and transferred to the number line with a compass.