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## Chapter 1: Real Number

TEXTBOOK’S EXERCISE – 1.1

1. Use Euclid’s division algorithm to find the HCF of :

(i) 135 and 225

(ii) 196 and 38220

(iii) 867 and 255

Sol. (i) In 135 and 225, 225 is a larger integer.

Using Euclid’s division algorithm,

225 = 135 × 1 + 90

[where 135 is the divisor, 90 is remainder]

Since, remainder 90 ≠ 0, we apply Euclid’s division algorithm to 135 and 90

∴ 135 = 90 × 1 + 45

Again since, remainder 45 ≠ 0, we apply Euclid’s division algorithm to 90 and 45

∴ 90 = 45 × 2 + 0

Now, the remainder is zero so, our procedure stops.

Hence, HCF of 135 and 225 is 45.

(ii) In 196 and 38220, 38220 is a larger integer.

Using Euclid’s division algorithm,

38220 = 196 × 195 + 0

The remainder is zero, so our procedure stops.

Hence, HCF of 196 and 38220 is 196.

(iii) In 867 and 255, 867 is a larger integer.

Using Euclid’s division algorithm

867 = 255 × 3 + 102

[where 255 is the divisor, 102 is remainder]

Since, remainder 102 ≠ 0, we apply Euclid’s division algorithm to 255 and 102

∴ 255 = 102 × 2 + 51

Since, remainder 51 ≠ 0, we apply Euclid’s division algorithm to 102 and 51

So, 102 = 51 × 2 + 0

Remainder = 0

The remainder is zero, so our procedure stops.

Hence, HCF of 867 and 255 and 51.

2. Show that any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5, where q is some integer.

Sol. Let a be any positive odd integer. We apply Euclid’s division algorithm with a and b = 6.

a = 6q + r, 0 ≤ r < 6

Since 0 ≤ r < 6, the possible remainders are 0, 1, 2, 3, 4 and 5.

i.e., a can be 6q or 6q + 1 or 6q + 2, or 6q + 3, or 6q + 4, or 6q + 5, where q is the quotient. But, a is odd, so a cannot be equal to 6q, 6q + 2, 6q + 4, which are even numbers.

⇒ Any odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5. Proved.

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?

Sol. Total members in the army are 616 and 32.

616 is larger integer i.e., 616 > 32

Using Euclid’s division algorithm,

616 = 32 × 19 + 8 [where 32 is divisor, 8 is remainder]

Since, remainder 8 ≠ 0, by applying Euclid’s division algorithm, we get

32 = 8 × 4 + 0

The remainder is zero,

4. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Sol. Let x be any positive integer, then it is of the form 3q, 3q + 1 or 3q + 2.

If x = 3q

⇒ (x)2 = (3q)2 [Squaring both sides]

⇒ x2 = 9q2 = 3(3q2) = 3m, where m = 3q2

Hence, x2 = 3m …(i)

If x = 3q + 1

⇒ x2 = (3q + 1)2 [Squaring both sides]

⇒ x2 = 9q2 + 1 + 2 × 3q × 1

⇒ x2 = 3(3q2 + 2q) + 1 = 3m + 1,

where m = 3q2 + 2q

⇒ x2 = 3m + 1 …(ii)

If x = 3q + 2

⇒ x2 = (3q + 2)2 = 9q2 + 4 + 12q

= 3(3q2 + 4q +1) + 1 = 3m + 1; where m = 3q2 + 4q + 1

⇒ x2 = 3m + 1 …(iii)

From (i) , (ii), and (iii), we get

x2 = 3m, 3m + 1

∴ Square of any positive integer is either of the form 3m or 3m + 1 for some integer m. Proved.

5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Sol. Let a be any positive integer, then it is of the form 3q, 3q + 1 or 3q + 2.

If a = 3q,

Then a3 = 27q3 [Cubing both sides]

⇒ a3 = 9 × 3q3 = 9m …(i)

where, m = 3q3

If a = (3q + 1)

⇒ a3 = (3q + 1)3 [Cubing both sides]

 (a + b)3 = a3 + b3 + 3a2b + 3b2a

⇒ a3 = (3q)3 + (1)3 + 3 × (3q)2 × 1 + 3 × 1 × 3q

⇒ a3 = 27q3 + 1 + 27q2 + 9q

⇒ a3 = 27q3 + 27q2 + 9q + 1

⇒ a3 = 9(3q3 + 3q2 + q) + 1

⇒ a3 = 9m + 1 …(ii)

where, m = 3q3 + 3q2 + q

If a = 3q + 2

⇒ a3 = (3q + 2)3 [Cubing both sides]

⇒ a3 = (3q)3 + (2)3 + 3 × (3q)2 × 2 + 3 × (2)2 × 3q

⇒ a3 = 27q3 + 8 + 54q2 + 36q

⇒ a3 = 27q3 + 54q2 + 36q + 8

⇒ a3 = 9(3q3 + 6q2 + 4q) + 8

⇒ a3 = 9m + 8 …(iii)

where, m = 3q3 + 6q2 + 4q

Hence, from (i), (ii) and (iii) we can say that the cube of any integer is of the form 9m, 9m + 1 or 9m + 8. Proved.