RD Sharma Solutions Chapter 1 Real Numbers Exercise 1.1 Class 10 Maths
Chapter Name  RD Sharma Chapter 1 Real Numbers 
Book Name  RD Sharma Mathematics for Class 10 
Other Exercises 

Related Study  NCERT Solutions for Class 10 Maths 
Exercise 1.1 Solutions
1. Prove that the product of two consecutive positive integers is divisible by 2.
Solution
Let, (n1) and n be two consecutive positive integers
∴ Their product = n(n1)
= n^{2} − n
We know that any positive integer is of the form 2q or 2q+1, for some integer q
When n = 2q, we have
n^{2} − n = (2q)^{2} − 2q
= (4q)^{2} − 2q
2q(2q − 1)
Then n^{2} − n is divisible by 2.
When n = 2q + 1, we have
n^{2} − n = (2q+1)^{2} − (2q+1)
= (4q)^{2} +4q +1 − 2q −1)
= 4q^{2} + 2q
= 2q(2q+1)
Then n^{2} − n is divisible by 2.
Hence the product of two consecutive positive integers is divisible by 2.
2. 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 (ab)/2 is odd and the other is even.
Solution
Let a = 2q + 3 and b = 2q+1 be two positive odd integers such that a > b
Hence one of the two numbers (a+b)/2 and (ab)/2 is odd and the other is even for any two positive odd integer.
3. Show that the square of an odd positive integer is of the form 8q+1, for some integer q.
Solution
By Euclid's division algorithm
a = bq + r, where 0 ≤ r ≤ b
Put b = 4
a = 4q + r, where 0 ≤ r ≤ 4
If r = 0, then a = 4q even
If r = 1, then a = 4q + 1 odd
If r = 2, then a = 4q + 2 even
If r = 3, then a = 4q + 3 odd
Now, (4q+1)^{2} = (4q)^{2} + 2(4q)(1) + (1)^{2}
= 16q^{2} + 8q + 1
= 8(2q^{2} + 2q) + 1
= 8m + 1 where m is some integer
Hence the square of an odd integer is of the form 8q + 1, for some integer q.
4. Show that any positive odd integer is of the form 6q + 1 or, 6q + 5, where q is some integer.
Solution
Let a be any odd positive integer we need to prove that a is of the form 6q+ 1, or 6q+ 3, 6q+5, where q is some integer
Since a is an integer consider b = 6 another integer applying Euclid's division lemma we get
a = 6q + r for some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 since 0 ≤ r < 6.
Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q+ 5
However since a is odd so a can not take the values 6q, 6q + 2 and 6q + 4
(since all these air divisible by 2)
Also, 6q + 1 = 2× 3q + 1 = 2k_{1} + 1, where k_{1} is a positive integer
6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k_{2} + 1, where k_{2} is an integer
6q + 5 = (6q+ 4) + 1 = 2(3q + 2) + 1 = 2k_{3} + 1, where k_{3} is an integer
Clearly, 6q +1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer
Therefore, 6q + 1, 6q + 3, 6q + 5 are odd numbers.
Therefore, any odd integer can be expressed is of the form
6q +1, or 6q +3, 6q + 5 where q is some integer
Concept insight: In order to solve such problems Euclid's division lemma is applied to two integers a and b the integer b must be taken in accordance with what is to be proved, for example here the integer b was taken 6 because a must be of the form 6q + 1, 6q + 3, 6q + 5
Basic definition of even (divisible by 2) and odd numbers (not divisible by 2) and the fact that addiction and multiplication of integers is always an integer are applicable here.
5. Prove that the square of any positive integer is of the form 3m or, 3m + 1 but not of the form 3m + 2.
Solution
By Euclid's division algorithm
a = bq + r, where 0 ≤ r ≤ b
Put b = 3
a = 3q + r, where 0 ≤ r ≤ 3
If r = 0, then a = 3q
If r = 1, then a = 3q + 1
If r = 2, then a = 3q + 2
Now, (3q)^{2} = 9q^{2}
= 3 × 3q^{2}
= 3m, where m is some integer
(3q + 1)^{2} = (3q)^{2} + 2(3q)(1) + (1)^{2}
= 9q^{2} + 6q+ 1
= 3(3q^{2} + 2q) + 1
= 3 m + 1, where m is some integer
(3q + 2)^{2} = (3q)^{2} + 2(3q)(2) + (2)^{2}
= 9q^{2} + 12q + 4
= 9q^{2} + 12q + 4
= 3(3q^{2} + 4q + 1) + 1
= 3m + 1 , where m is some integer
Hence the square of any positive integer is of the form 3m, or 3m + 1
But not of the form 3m + 2
6. Prove that the square of any positive integer is of the form 4q or 4q + 1 for some integer q.
Solution
By Euclid's division algorithm
a = bm + r, where 0 ≤ r ≤ b
Put b = 4
a = 4m + r, where 0 ≤ r ≤ 4
If r = 0 , then a = 4m
If r = 1, then a = 4m + 1
If r = 2, then a = 4m + 2
If r = 3, then a = 4m + 3
Now, (4m)^{2} = 16m^{2}
= 4 × 4m^{2}
= 4q where q is some integer
(4m + 1)^{2} = (4m)^{2} + 2(4m)(1) + (1)^{2}
= 16m^{2} + 8m + 1
= 4(4m^{2} + 2m) + 1
= 4q + 1 where q is some integer
(4m + 2)^{2} = (4m)^{2} + 2(4m)(2) + (2)^{2}
= 16m^{2} + 24m + 9
= 16m^{2} + 24 m + 8 + 1
4(4m^{2} + 6m + 2) + 1
= 4q + 1, where q is some integer
Hence, the square of any positive integer is of the form 4q or 4q + 1
7. Prove that the square of any positive integer is of the form 5q, 5q + 1, 5q + 4 for some integer q.
Solution
By Euclid's division algorithm
a = bm + r, where 0 ≤ r ≤ b
Put b= 5
a = 5m + r, where 0 ≤ r ≤ b
If r = 0, then a = 5m
If r = 1, then a = 5m + 1
If r = 2, then a = 5m + 2
If r = 3, then a = 5m + 3
If r = 4, then a = 5m + 4
Now, (5m)^{2} = 25m^{2}
= 5(5m^{2})
= 5q where q is some integer
(5m + 1)^{2} = (5m)^{2} + 2(5m)(1) + (1)^{2}
= 25m^{2} + 10m + 1
= 5(5m^{2} + 2m) + 1
= 5q + 1where q is some integer
(5m + 1)^{2} = (5m)^{2} + 2(5m)(1)(1)^{2}
= 25m^{2} + 10m + 1
= 5(5m^{2} + 2m) + 1
= 5q + 1 where q is some integer
= (5m + 2)^{2} = (5m)^{2} + 2(5m)(2) + (2)^{2}
= 25m^{2} + 20m + 4
= 5(5m^{2} + 4m) + 4
= 5q + 4, where q is some integer
= (5m + 3)^{2} = (5m)^{2} + 2(5m)(3) + (3)^{2}
= 25m^{2} + 30m + 9
= 25m^{2} + 30m + 5+ 4
= 5(5m^{2} + 6m + 1) + 4
= 5q + 1, where q is some integer
= (5m + 4)^{2} = (5m)^{2} + 2(5m)(4) + (4)^{2}
= 25m^{2} + 40m + 16
= 25m^{2} + 40m + 15 + 1
= 5(5m)^{2} + 2(5m)(4) + (4)^{2}
= 5q + 1, where q is some integer
Hence the square of any positive integer is of the form 5q or 5q + 1, 5q + 4 for some integer q.
8. Prove that if a positive integer is of the form 6q + 5, then it is of the form 3q + 2 for some integer q, but not conversely.
Solution
Let, n = 6q + 5, when q is a positive integer
We know that any positive integer is of the form 3k, or 3k + 1, or 3k + 2
∴ q = 3k or 3k + 1, or 3k + 2
If q = 3k, then
n = 6q + 5
= 6(3k) + 5
= 18k+ 5
= 18k + 3 + 2
= 3(6k + 1)+ 2
= 3m + 2, where m is some integer
If q = 3k + 1, then
n = 6q + 5
= 6(3k + 1) + 5
= 18k + 6 + 5
= 18k +11
= 3(6k + 3) + 2
= 3m +2, where m is some integer
Hence, if a positive integer is of the form 6q + 5, then it is of the form 3q + 2 for some integer q.
Conversely
Let n = 3q + 2
We know that a positive integer can be of the form 6k + 1, 6k + 2, 6k + 3, 6k + 4 or 6k + 5
So, now if q = 6k + 1 then
n = 3(6k + 1) + 2
= 18k + 5
= 6(3k) + 5
= 6m + 5, where m is some integer
So, now if q = 6k + 2 then
n = 3(6k + 2) + 2
= 18k + 8
= 6(3k + 1) + 2
= 6m + 2, where m is some integer
Now, this is not of the form 6m + 5
Hence, if n is of the form 3q + 2, then it necessarily won't be of the form 6q + 5 always.
9. Prove that the square of any positive integer of the form 5q + 1 is of the same form.
Solution
Let n = 5q + 1 where q is a positive integer
∴ n^{2} = (5q + 1)^{2}
= 25q^{2} + 10q + 1
= 5(5q^{2} + 2q) + 1
= 5m + 1, where m is some integer
Hence, the square of any positive integer of the form 5q + 1 is of the same form.
10. Prove that the product of three consecutive positive integer is divisible by 6.
Solution
Let, n be any positive integer. Since any positive integer is of the form 6q or 6q + 1 or 6q +2 or 6q + 3 or 6q + 4 or 6q + 5.
If n = 6q, then
n(n+ 1)(n+2) = (6q + 1)(6q +2)(6q + 3)
= 6[(6q + 1)(3q + 1)(2q + 1)]
= 6m, which is divisible by 6 ?
If n = 6q + 1, then
n(n+1)(n+2) = (6q + 2)(6q +3)(6q +4)
= 6[(6q + 1)(3q+ 1)(2q + 1)]
= 6m, which is divisible by 6
If n = 6q + 2, then
n(n+1)(n+2) = (6q + 2)(6q +3)(6q +4)
= 6[(3q + 1)(2q + 1)(6q+4)]
= 6m, which is divisible by 6.
If n = 6q + 3, then
n(n+1)(n+2) = (6q + 3)(6q +4)(6q +5)
= 6[(6q + 1)(3q + 2)(2q + 5)]
= 6m, which is divisible by 6.
If n = 6q + 4, then
n(n+1)(n+2) = (6q + 4)(6q + 5)(6q + 6)
= 6[(6q + 4)(3q + 5)(2q + 1)]
= 6m, which is divisible by 6.
If n = 6q + 5, then
n(n+1)(n+2) = (6q + 5)(6q + 6)(6q + 7)
= 6[(6q + 5 )(q + 1)(6q + 7)
= 6m, which is divisible by 6.
Hero, the product of three consecutive positive integer is divisible by 6.
11. For any positive integer n, prove that n^{3}  n divisible by 6.
Solution
We have n^{3}  n = n(n^{2}  1) = (n1)(n)(n+1)
Let, n be any positive integer. Since any positive integer is of the form 6q or 6q + 1 or, 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5.
If n = 6q, then
(n1)(n)(n+1) = (6q1)(6q)(6q + 1)
= 6[(6q  1)(q)(6q + 1)]
= 6m, which is divisible by 6
If n = 6q + 1, then
(n1)(n+1) = (6q)(6q + 1)(6q+2)
= 6[(q)(6q+1)(6q+2)]
= 6m, which is divisible by 6
If n = 6q + 2, then
(n1)(n)(n+1) = (6q + 1)(6q +2)(6q + 3)
= 6[(6q + 1)(3q + 1)(2q + 1)]
= 6m, which is divisible by 6
If n = 6q + 4, then
(n1)(n)(n+1) = (6q + 3)(6q +4)(6q + 5)
= 6[(2q + 1)(3q+2)(6q+5)]
= 6m, which is divisible by 6
If n = 6q + 5, then
(n1)(n)(n+1) = (6q + 4)(6q +5)(6q + 6)
= 6[(6q + 4)(6q + 5)(q + 1)]
= 6m, which is divisible by 6
Hence, for any positive integer n, n^{3}  n is divisible by 6.