# Class 10 -Chapter 1- Real Numbers Exercise -1.1-( All Sums Q/A)

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### 10th-Maths- SCERT- Real Numbers- Chapter 1- Exercise 1.1

#### Class 10 Chapter 1 Real Numbers Exercise 1.1

1. If a and b are two odd positive integers such that a > b, then prove that one of the two numbers (a+b)/2 and (a-b)/2 is odd and the other is even.

Solution:

We know that any odd positive integer is of the form 4q+1 or, 4q+3 for some whole number q.

Now that it’s given a > b

So, we can choose a= 4q+3 and b= 4q+1.

∴ (a+b)/2 = [(4q+3) + (4q+1)]/2

⇒ (a+b)/2 = (8q+4)/2

⇒ (a+b)/2 = 4q+2 = 2(2q+1) which is clearly an even number.

Now, doing (a-b)/2

⇒ (a-b)/2 = [(4q+3)-(4q+1)]/2

⇒ (a-b)/2 = (4q+3-4q-1)/2

⇒ (a-b)/2 = (2)/2

⇒ (a-b)/2 = 1 which is an odd number.

Hence, one of the two numbers (a+b)/2 and (a-b)/2 is odd and the other is even.

2. Prove that the product of two consecutive positive integers is divisible by 2.

Solution:

Let’s consider two consecutive positive integers as (n-1) and n.

∴ Their product = (n-1) n

= n2 – n

And then we know that any positive integer is of the form 2q or 2q+1. (From Euclid’s division lemma for b= 2)

So, when n= 2q

We have,

⇒ n2 – n = (2q)2 – 2q

⇒ n2 – n = 4q2 -2q

⇒ n2 – n = 2(2q2 -q)

Thus, n2 – n is divisible by 2.

Now, when n= 2q+1

We have,

⇒ n2 – n = (2q+1)2 – (2q-1)

⇒ n2 – n = (4q2+4q+1 – 2q+1)

⇒ n2 – n = (4q2+2q+2)

⇒ n2 – n = 2(2q2+q+1)

Thus, n2 – n is divisible by 2 again.

Hence, the product of two consecutive positive integers is divisible by 2.

3. Prove that the product of three consecutive positive integers is divisible by 6.

Solution:

Let n be any positive integer.

Thus, the three consecutive positive integers are n, n+1 and n+2.

We know that any positive integer can be of the form 6q, or 6q+1, or 6q+2, or 6q+3, or 6q+4, or 6q+5. (From Euclid’s division lemma for b= 6)

So,

For n= 6q,

⇒ n(n+1)(n+2)= 6q(6q+1)(6q+2)

⇒ n(n+1)(n+2)= 6[q(6q+1)(6q+2)]

⇒ n(n+1)(n+2)= 6m, which is divisible by 6. [m= q(6q+1)(6q+2)]

For n= 6q+1,

⇒ n(n+1)(n+2)= (6q+1)(6q+2)(6q+3)

⇒ n(n+1)(n+2)= 6[(6q+1)(3q+1)(2q+1)]

⇒ n(n+1)(n+2)= 6m, which is divisible by 6. [m= (6q+1)(3q+1)(2q+1)]

For n= 6q+2,

⇒ n(n+1)(n+2)= (6q+2)(6q+3)(6q+4)

⇒ n(n+1)(n+2)= 6[(3q+1)(2q+1)(6q+4)]

⇒ n(n+1)(n+2)= 6m, which is divisible by 6. [m= (3q+1)(2q+1)(6q+4)]

For n= 6q+3,

⇒ n(n+1)(n+2)= (6q+3)(6q+4)(6q+5)

⇒ n(n+1)(n+2)= 6[(2q+1)(3q+2)(6q+5)]

⇒ n(n+1)(n+2)= 6m, which is divisible by 6. [m= (2q+1)(3q+2)(6q+5)]

For n= 6q+4,

⇒ n(n+1)(n+2)= (6q+4)(6q+5)(6q+6)

⇒ n(n+1)(n+2)= 6[(3q+2)(3q+1)(2q+2)]

⇒ n(n+1)(n+2)= 6m, which is divisible by 6. [m= (3q+2)(3q+1)(2q+2)]

For n= 6q+5,

⇒ n(n+1)(n+2)= (6q+5)(6q+6)(6q+7)

⇒ n(n+1)(n+2)= 6[(6q+5)(q+1)(6q+7)]

⇒ n(n+1)(n+2)= 6m, which is divisible by 6. [m= (6q+5)(q+1)(6q+7)]

Hence, the product of three consecutive positive integers is divisible by 6.

4. For any positive integer n, prove that n3 – n divisible by 6.

Solution:

Let, n be any positive integer. And since any positive integer can be of the form 6q, or 6q+1, or 6q+2, or 6q+3, or 6q+4, or 6q+5. (From Euclid’s division lemma for b= 6)

We have n3 – n = n(n2-1)= (n-1)n(n+1),

For n= 6q,

⇒ (n-1)n(n+1)= (6q-1)(6q)(6q+1)

⇒ (n-1)n(n+1)= 6[(6q-1)q(6q+1)]

⇒ (n-1)n(n+1)= 6m, which is divisible by 6. [m= (6q-1)q(6q+1)]

For n= 6q+1,

⇒ (n-1)n(n+1)= (6q)(6q+1)(6q+2)

⇒ (n-1)n(n+1)= 6[q(6q+1)(6q+2)]

⇒ (n-1)n(n+1)= 6m, which is divisible by 6. [m= q(6q+1)(6q+2)]

For n= 6q+2,

⇒ (n-1)n(n+1)= (6q+1)(6q+2)(6q+3)

⇒ (n-1)n(n+1)= 6[(6q+1)(3q+1)(2q+1)]

⇒ (n-1)n(n+1)= 6m, which is divisible by 6. [m= (6q+1)(3q+1)(2q+1)]

For n= 6q+3,

⇒ (n-1)n(n+1)= (6q+2)(6q+3)(6q+4)

⇒ (n-1)n(n+1)= 6[(3q+1)(2q+1)(6q+4)]

⇒ (n-1)n(n+1)= 6m, which is divisible by 6. [m= (3q+1)(2q+1)(6q+4)]

For n= 6q+4,

⇒ (n-1)n(n+1)= (6q+3)(6q+4)(6q+5)

⇒ (n-1)n(n+1)= 6[(2q+1)(3q+2)(6q+5)]

⇒ (n-1)n(n+1)= 6m, which is divisible by 6. [m= (2q+1)(3q+2)(6q+5)]

For n= 6q+5,

⇒ (n-1)n(n+1)= (6q+4)(6q+5)(6q+6)

⇒ (n-1)n(n+1)= 6[(6q+4)(6q+5)(q+1)]

⇒ (n-1)n(n+1)= 6m, which is divisible by 6. [m= (6q+4)(6q+5)(q+1)]

Hence, for any positive integer n, n3 – n is divisible by 6.

5.  Prove that if a positive integer is of form 6q + 5, then it is of the form 3q + 2 for some integer q, but not conversely.

Solution:

Let n= 6q+5 be a positive integer for some integer q.

We know that any positive integer can be of the form 3k, or 3k+1, or 3k+2.

∴ q can be 3k or, 3k+1 or, 3k+2.

If q= 3k, then

⇒ n= 6q+5

⇒ n= 6(3k)+5

⇒ n= 18k+5 = (18k+3)+ 2

⇒ n= 3(6k+1)+2

⇒ n= 3m+2, where m is some integer

If q= 3k+1, then

⇒ n= 6q+5

⇒ n= 6(3k+1)+5

⇒ n= 18k+6+5 = (18k+9)+ 2

⇒ n= 3(6k+3)+2

⇒ n= 3m+2, where m is some integer

If q= 3k+2, then

⇒ n= 6q+5

⇒ n= 6(3k+2)+5

⇒ n= 18k+12+5 = (18k+15)+ 2

⇒ n= 3(6k+5)+2

⇒ n= 3m+2, where m is some integer

Hence, if a positive integer is of form 6q + 5, then it is of the form 3q + 2 for some integer q.

Conversely,

Let n= 3q+2

And we know that a positive integer can be of the form 6k, or 6k+1, or 6k+2, or 6k+3, or 6k+4, or 6k+5.

So, now if q=6k+1 then

⇒ n= 3q+2

⇒ n= 3(6k+1)+2

⇒ n= 18k + 5

⇒ n= 6m+5, where m is some integer

So, now if q=6k+2 then

⇒ n= 3q+2

⇒ n= 3(6k+2)+2

⇒ n= 18k + 6 +2 = 18k+8

⇒ n= 6 (3k + 1) + 2

⇒ n= 6m+2, where m is some integer

Now, this is not of the form 6q + 5.

Therefore, if n is of the form 3q + 2, then is necessary won’t be of the form 6q + 5.

6.  Prove that square of any positive integer of the form 5q + 1 is of the same form.

Solution:

Here, the integer ‘n’ is of the form 5q+1.

⇒ n= 5q+1

On squaring it,

⇒ n2= (5q+1)2

⇒ n2= (25q2+10q+1)

⇒ n2= 5(5q2+2q)+1

⇒ n2= 5m+1, where m is some integer. [For m = 5q2+2q]

Therefore, the square of any positive integer of the form 5q + 1 is of the same form.

7. Prove that the square of any positive integer is of the form 3m or 3m + 1 but not of the form 3m + 2.

Solution:

Let any positive integer ‘n’ be of the form 3q or, 3q+1 or 3q+2. (From Euclid’s division lemma for b= 3)

If n= 3q,

Then, on squaring

⇒ n2= (3q)2 = 9q2

⇒ n2= 3(3q2)

⇒ n2= 3m, where m is some integer [m = 3q2]

If n= 3q+1,

Then, on squaring

⇒ n2= (3q+1)2 = 9q2 + 6q + 1

⇒ n2= 3(3q2 +2q) + 1

⇒ n2= 3m + 1, where m is some integer [m = 3q2 +2q]

If n= 3q+2,

Then, on squaring

⇒ n2= (3q+2)2 = 9q2 + 12q + 4

⇒ n2= 3(3q2 + 4q + 1) + 1

⇒ n2= 3m + 1, where m is some integer [m = 3q2 + 4q + 1]

Thus, it is observed that the square of any positive integer is of the form 3m or 3m + 1 but not of the form 3m + 2.

8.  Prove that the square of any positive integer is of the form 4q or 4q + 1 for some integer q.

Solution:

Let ‘a’ be any positive integer.

Then,

According to Euclid’s division lemma,

a=bq+r

According to the question, when b = 4.

a = 4k + r, n < r < 4

When r = 0, we get, a = 4k

a2 = 16k2 = 4(4k2) = 4q, where q = 4k2

When r = 1, we get, a = 4k + 1

a2 = (4k + 1)2 = 16k2 + 1 + 8k = 4(4k + 2) + 1 = 4q + 1, where q = k(4k + 2)

When r = 2, we get, a = 4k + 2

a2 = (4k + 2)2 = 16k2 + 4 + 16k = 4(4k2 + 4k + 1) = 4q, where q = 4k2 + 4k + 1

When r = 3, we get, a = 4k + 3

a2 = (4k + 3)2 = 16k2 + 9 + 24k = 4(4k2 + 6k + 2) + 1

= 4q + 1, where q = 4k2 + 6k + 2

Therefore, the square of any positive integer is either of the form 4q or 4q + 1 for some integer q.

9. Prove that the square of any positive integer is of the form 5q or 5q + 1, 5q + 4 for some integer q.

Solution:

Let ‘a’ be any positive integer.

Then,

According to Euclid’s division lemma,

a = bq+r

According to the question, when b = 5.

a = 5k + r, n < r < 5

When r = 0, we get, a = 5k

a2 = 25k2 = 5(5k2) = 5q, where q = 5k2

When r = 1, we get, a = 5k + 1

a2 = (5k + 1)2 = 25k2 + 1 + 10k = 5k(5k + 2) + 1 = 5q + 1, where q = k(5k + 2)

When r = 2, we get, a = 5k + 2

a2 = (5k + 2)2 = 25k2 + 4 + 20k = 5(5k2 + 4k) + 4 = 4q + 4, where q = 5k2 + 4k

When r = 3, we get, a = 5k + 3

a2 = (5k + 3)2 = 25k2 + 9 + 30k = 5(5k2 + 6k + 1) + 4

= 5q + 4, where q = 5k2 + 6k + 1

When r = 4, we get, a = 5k + 4

a2 = (5k + 4)2 = 25k2 + 16 + 40k = 5(5k2 + 8k + 3) + 1

= 5q + 1, where q = 5k2 + 8k + 3

Therefore, the square of any positive integer is of the form 5q or, 5q + 1 or 5q + 4 for some integer q.

10. Show that the square of odd integer is of the form 8q + 1, for some integer q.

Solution:

From Euclid’s division lemma,

a = bq+r ; where 0 < r < b

Putting b=4 for the question,

⇒ a = 4q + r, 0 < r < 4

For r = 0, we get a = 4q, which is an even number.

For r = 1, we get a = 4q + 1, which is an odd number.

On squaring,

⇒ a2 = (4q + 1)2 = 16q2 + 1 + 8q = 8(2q2 + q) + 1 = 8m + 1, where m = 2q2 + q

For r = 2, we get a = 4q + 2 = 2(2q + 1), which is an even number.

For r = 3, we get a = 4q + 3, which is an odd number.

On squaring,

⇒ a2 = (4q + 3)2 = 16q2 + 9 + 24q = 8(2q2 + 3q + 1) + 1

= 8m + 1, where m = 2q2 + 3q + 1

Thus, the square of an odd integer is of the form 8q + 1, for some integer q.

11. 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.

Then from Euclid’s division lemma,

a = bq+r ; where 0 < r < b

Putting b=6 we get,

⇒ a = 6q + r, 0 < r < 6

For r = 0, we get a = 6q = 2(3q) = 2m, which is an even number. [m = 3q]

For r = 1, we get a = 6q + 1 = 2(3q) + 1 = 2m + 1, which is an odd number. [m = 3q]

For r = 2, we get a = 6q + 2 = 2(3q + 1) = 2m, which is an even number. [m = 3q + 1]

For r = 3, we get a = 6q + 3 = 2(3q + 1) + 1 = 2m + 1, which is an odd number. [m = 3q + 1]

For r = 4, we get a = 6q + 4 = 2(3q + 2) + 1 = 2m + 1, which is an even number. [m = 3q + 2]

For r = 5, we get a = 6q + 5 = 2(3q + 2) + 1 = 2m + 1, which is an odd number. [m = 3q + 2]

Thus, from the above it can be seen that any positive odd integer can be of the form 6q +1 or 6q + 3 or 6q + 5, where q is some integer.

12. Show that the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m.

Solution:

Let the positive integer = a

According to Euclid’s division algorithm,

a = 6q + r, where 0 ≤ r < 6

a2 = (6q + r)2 = 36q2 + r2 + 12qr [∵(a+b)2 = a2 + 2ab + b2]

a2 = 6(6q2 + 2qr) + r2   …(i), where,0 ≤ r < 6

When r = 0, substituting r = 0 in Eq.(i), we get

a2 = 6 (6q2) = 6m, where, m = 6q2 is an integer.

When r = 1, substituting r = 1 in Eq.(i), we get

a2 = 6 (6q2 + 2q) + 1 = 6m + 1, where, m = (6q2 + 2q) is an integer.

When r = 2, substituting r = 2 in Eq(i), we get

a2 = 6(6q2 + 4q) + 4 = 6m + 4, where, m = (6q2 + 4q) is an integer.

When r = 3, substituting r = 3 in Eq.(i), we get

a2 = 6(6q2 + 6q) + 9 = 6(6q2 + 6q) + 6 + 3

a2 = 6(6q2 + 6q + 1) + 3 = 6m + 3, where, m = (6q + 6q + 1) is integer.

When r = 4, substituting r = 4 in Eq.(i) we get

a2 = 6(6q2 + 8q) + 16

= 6(6q2 + 8q) + 12 + 4

⇒ a2 = 6(6q2 + 8q + 2) + 4 = 6m + 4, where, m = (6q2 + 8q + 2) is integer.

When r = 5, substituting r = 5 in Eq.(i), we get

a2 = 6 (6q2 + 10q) + 25 = 6(6q2 + 10q) + 24 + 1

a2 = 6(6q2 + 10q + 4) + 1 = 6m + 1, where, m = (6q2 + 10q + 4) is integer.

Hence, the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m.

Hence Proved.

13. Show that the cube of a positive integer of the form 6q + r, q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the form 6m + r.

Solution:

Given, 6q + r is a positive integer, where q is an integer and r = 0, 1, 2, 3, 4, 5

Then, the positive integers are of the form 6q, 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5.

Taking cube on L.H.S and R.H.S,

For 6q,

(6q)3 = 216 q3 = 6(36q)3 + 0

= 6m + 0, (where m is an integer = (36q)3)

For 6q+1,

(6q+1)3 = 216q3 + 108q2 + 18q + 1

= 6(36q3 + 18q2 + 3q) + 1

= 6m + 1, (where m is an integer = 36q3 + 18q2 + 3q)

For 6q+2,

(6q+2)3 = 216q3 + 216q2 + 72q + 8

= 6(36q3 + 36q2 + 12q + 1) +2

= 6m + 2, (where m is an integer = 36q3 + 36q2 + 12q + 1)

For 6q+3,

(6q+3)3 = 216q3 + 324q2 + 162q + 27

= 6(36q3 + 54q2 + 27q + 4) + 3

= 6m + 3, (where m is an integer = 36q3 + 54q2 + 27q + 4)

For 6q+4,

(6q+4)3 = 216q3 + 432q2 + 288q + 64

= 6(36q3 + 72q2 + 48q + 10) + 4

= 6m + 4, (where m is an integer = 36q3 + 72q2 + 48q + 10)

For 6q+5,

(6q+5)3 = 216q3 + 540q2 + 450q + 125

= 6(36q3 + 90q2 + 75q + 20) + 5

= 6m + 5, (where m is an integer = 36q3 + 90q2 + 75q + 20)

Hence, the cube of a positive integer of the form 6q + r, q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the form 6m + r.

14. Show that one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5, where n is any positive integer.

Solution:

According to Euclid’s division Lemma,

Let the positive integer = n

And, b=5

n = 5q+r, where q is the quotient and r is the remainder

0 < r < 5 implies remainders may be 0, 1, 2, 3, 4 and 5

Therefore, n may be in the form of 5q, 5q+1, 5q+2, 5q+3, 5q+4

So, this gives us the following cases:

CASE 1:

When, n = 5q

n+4 = 5q+4

n+8 = 5q+8

n+12 = 5q+12

n+16 = 5q+16

Here, n is only divisible by 5

CASE 2:

When, n = 5q+1

n+4 = 5q+5 = 5(q+1)

n+8 = 5q+9

n+12 = 5q+13

n+16 = 5q+17

Here, n + 4 is only divisible by 5

CASE 3:

When, n = 5q+2

n+4 = 5q+6

n+8 = 5q+10 = 5(q+2)

n+12 = 5q+14

n+16 = 5q+18

Here, n + 8 is only divisible by 5

CASE 4:

When, n = 5q+3

n+4 = 5q+7

n+8 = 5q+11

n+12 = 5q+15 = 5(q+3)

n+16 = 5q+19

Here, n + 12 is only divisible by 5

CASE 5:

When, n = 5q+4

n+4 = 5q+8

n+8 = 5q+12

n+12 = 5q+16

n+16 = 5q+20 = 5(q+4)

Here, n + 16 is only divisible by 5

So, we can conclude that one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5.

Hence Proved

15. Show that the square of an odd integer can be of the form 6q + 1 or 6q + 3, for some integer q.

Solution:

Let ‘a’ be an odd integer and b = 6.

According to Euclid’s algorithm,

a = 6m + r for some integer m ≥ 0

And r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 6.

So, we have that,

a = 6m or, 6m + 1 or, 6m + 2 or, 6m + 3 or, 6m + 4 or 6m + 5

Thus, we are choosing for a = 6m + 1 or, 6m + 3 or 6m + 5 for it to be an odd integer.

For a = 6m + 1,

(6m + 1)2 = 36m2 + 12m + 1

= 6(6m2 + 2m) + 1

= 6q + 1, where q is some integer and q = 6m2 + 2m.

For a = 6m + 3

(6m + 3)2 = 36m2 + 36m + 9

= 6(6m2 + 6m + 1) + 3

= 6q + 3, where q is some integer and q = 6m2 + 6m + 1

For a = 6m + 5,

(6m + 5)2 = 36m2 + 60m + 25

= 6(6m2 + 10m + 4) + 1

= 6q + 1, where q is some integer and q = 6m2 + 10m + 4.

Therefore, the square of an odd integer is of the form 6q + 1 or 6q + 3, for some integer q.

Hence Proved.

16. A positive integer is of the form 3q + 1, q being a natural number. Can you write its square in any form other than 3m + 1, 3m or 3m + 2 for some integer m? Justify your answer.

Solution:

No.

Justification:

By Euclid’s Division Lemma,

a = bq + r, 0 ≤ r < b

Here, a is any positive integer and b = 3,

⇒ a = 3q + r

So, a can be of the form 3q, 3q + 1 or 3q + 2.

Now, for a = 3q

(3q)2 = 3(3q2) = 3m [where m = 3q2]

for a = 3q + 1

(3q + 1)2 = 9q2 + 6q + 1 = 3(3q2 + 2q) + 1 = 3m + 1 [where m = 3q2 + 2q]

for a = 3q + 2

(3q + 2)2 = 9q2 + 12q + 4 = 9q2 + 12q + 3 + 1 = 3(3q2 + 4q + 1) + 1

= 3m + 1 [where m = 3q2 + 4q + 1]

Thus, square of a positive integer of the form 3q + 1 is always of the form 3m + 1 or 3m for some integer m.

17. Show that the square of any positive integer cannot be of the form 3m + 2, where m is a natural number.

Solution:

Let the positive integer be ‘a’

According to Euclid’s division lemma,

a = bm + r

According to the question, we take b = 3

a = 3m + r

So, r = 0, 1, 2.

When r = 0, a = 3m.

When r = 1, a = 3m + 1.

When r = 2, a = 3m + 2.

Now,

When a = 3m

a2 = (3m)2 = 9m2

a2 = 3(3m2) = 3q, where q = 3m2

When a = 3m + 1

a2 = (3m + 1)2 = 9m2 + 6m + 1

a2 = 3(3m2 + 2m) + 1 = 3q + 1, where q = 3m2 + 2m

When a = 3m + 2

a2 = (3m + 2)2