Ch 1: Real Numbers (Euclid's Lemma & Decimals)
This page covers the remaining exercises from the traditional NCERT textbook: Euclid's Division Algorithm and Terminating/Non-terminating Decimals.
Euclid's Division Lemma Solutions
Q1. Use Euclid's division algorithm to find the HCF of: (i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255
Solution for (i) 135 and 225:
Since 225 > 135, we apply Euclid’s division lemma (a = bq + r) to 225 and 135:
225 = 135 × 1 + 90
Since the remainder 90 ≠ 0, we apply the lemma to 135 and 90:
135 = 90 × 1 + 45
Since the remainder 45 ≠ 0, we apply the lemma to 90 and 45:
90 = 45 × 2 + 0
The remainder is now 0. The divisor at this stage is 45.
Therefore, HCF(135, 225) = 45.
Solution for (iii) 867 and 255:
Since 867 > 255, we apply the lemma:
867 = 255 × 3 + 102
Since remainder 102 ≠ 0, apply lemma to 255 and 102:
255 = 102 × 2 + 51
Since remainder 51 ≠ 0, apply lemma to 102 and 51:
102 = 51 × 2 + 0
The remainder is 0. The divisor is 51.
Therefore, HCF(867, 255) = 51.
Q2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Solution:
Let a be any positive integer and b = 6.
By Euclid's division algorithm, a = 6q + r, for some integer q ≥ 0, and the possible remainders are 0, 1, 2, 3, 4, 5 (because 0 ≤ r < 6).
Therefore, a can be: 6q, 6q+1, 6q+2, 6q+3, 6q+4, or 6q+5.
- If a = 6q, it is divisible by 2 (Even).
- If a = 6q + 2 = 2(3q + 1), it is divisible by 2 (Even).
- If a = 6q + 4 = 2(3q + 2), it is divisible by 2 (Even).
Since a is an odd integer, it cannot be 6q, 6q+2, or 6q+4.
Hence, any positive odd integer must be of the form 6q + 1, 6q + 3, or 6q + 5.
Rational Numbers & Decimal Expansions
💡 Trick: A rational number p/q has a terminating decimal if the prime factorization of 'q' is strictly of the form 2ⁿ × 5ᵐ.
Q1. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:
(i) 13 / 3125
Denominator (q) = 3125
Prime factorization of 3125 = 5 × 5 × 5 × 5 × 5 = 5⁵
Since the denominator is of the form 2⁰ × 5⁵ (only contains prime factor 5), the rational number has a terminating decimal expansion.
(iii) 64 / 455
Denominator (q) = 455
Prime factorization of 455 = 5 × 7 × 13
Since the denominator contains prime factors 7 and 13 (other than 2 and 5), the rational number has a non-terminating repeating decimal expansion.
(viii) 6 / 15
Careful! Always simplify the fraction first.
6 / 15 = 2 / 5 (Dividing numerator and denominator by 3)
Now, Denominator (q) = 5 = 5¹
Since the denominator only has the prime factor 5, it is a terminating decimal expansion.