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PostPosted: Fri Jun 22, 2012 11:00 am 
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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
Solution:
(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.
Solution:
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?
Solution:
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.
Solution:
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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