Class Xth Maths - Real Numbers Solutions - Download PDF Files and Solved your real number quaries
Q.1: Use Euclid’s division algorithm to find the HCF of:
(i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255
(i) We have,
a = bq + r
Applying division lemma to 225 and 135 we obtain,
225 = 135 x 1 + 90
and 135 = 90 x 1 + 45
and 90 = 45 x 2 + 0
Therefore, HCF of 225, 135 = 45
(ii) We have,
a = bq + r
Applying division lemma to 196 and 38220 we obtain,
38220 = 196 x 195 + 0
Therefore, HCF of 196 and 38220 = 196
(iii) We have,
a = bq + r
Applying division lemma to 867 and 255 we obtain,
867 = 255 x 3 + 102
255 = 102 x 2 + 51
102 = 51 x 2 + 0
Therefore, HCF of 867 and 255 is 51.
Q.2: Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or6q + 5, where q is some integer.
Let a be any positive number and a = 6. Then, by Euclid’s algorithm,
a = 6q + r (0 ≤ r < 6)
say, r = 0, 1, 2, 3, 4, 5
or, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5
Also, 6q + 1 = 2 × 3q + 1 = 2k1 + 1, where k1 is a positive integer,
Similarly, 6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k2 + 1, where k2 is a positive integer,
and, 6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k3 + 1, where k3 is a positive integer.
From these we observe that 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1. So, these numbers are not divisible by 2 and hence, are odd positive integers.
Q.3: An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
To find the maximum number of columns, we have to find HCF of 616 and 32.
Applying Euclid’s algorithm to find the HCF we get,
616 = 32 × 19 + 8
32 = 8 × 4 + 0
Or, the HCF (616, 32) = 8.
Therefore, maximum number of column is 8.
Q.4: Use Euclid’s division lemma to show that the square of any positive integer is either of form 3m or 3m + 1 for some integer m.
a = bq + r;
Let a be any positive integer, b = 3 and r = 0, 1, 2 because 0 ≤ r < 3
Then a = 3q + r for some integer q ≥ 0
Therefore, a = 3q + 0 or 3q + 1 or 3q + 2
From the above we can say that the square of any positive integer is either of the form 3m or 3m + 1.
Other Questions -
1. Euclid’s division lemma :
Given positive integers a and b, there exist whole numbers q and r satisfying a = bq + r, 0 ≤ r < b.
2. Euclid’s division algorithm : This is based on Euclid’s division lemma. According to this, the HCF of any two positive integers a and b, with a > b, is obtained as follows:
Step 1 : Apply the division lemma to find q and r where a = bq + r, 0 ≤ r < b.
Step 2 : If r = 0, the HCF is b. If r ≠ 0, apply Euclid’s lemma to b and r.
Step 3 : Continue the process till the remainder is zero. The divisor at this stage will be
HCF (a, b). Also, HCF(a, b) = HCF(b, r).
3. The Fundamental Theorem of Arithmetic :
Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.
4. If p is a prime and p divides a2, then p divides q, where a is a positive integer.
5. To prove that 2, 3 are irrationals.
6. Let x be a rational number whose decimal expansion terminates. Then we can express x in the form p
q , where p and q are coprime, and the prime factorisation of q is of the form 2n5m, where n, m are non-negative integers.
7. Let x = p/q be a rational number, such that the prime factorisation of q is of the form 2n5m,where n, m are non-negative integers. Then x has a decimal expansion which terminates.
8. Let x = p/q be a rational number, such that the prime factorisation of q is not of the form 2n 5m, where n, m are non-negative integers. Then x has a decimal expansion which isnon-terminating repeating (recurring).
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