## RD Sharma Solutions for Class 10 Maths Chapter 1 Exercise 1.1

**RD Sharma Solutions for Class 10 Chapter 1 Exercise 1.1** Real Number are available in this article. The answers have been created from RD Sharma’s latest edition books. Practicing and answering questions from Study Path will improve your understanding of the topic and increase your confidence for the examination. In the end, our solutions will help you to get high marks on the test. Answers for RD Sharma Maths Chapter 1 Real Numbers Exercise 1.1 can be found in **Knowledge Glow**.

**Question 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.

**Answer****: **

a and b are two odd numbers such that a > b

Let a = 2n + 1, then b = 2n + 3

**Question 2.**

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

**Answer : ** Let n and n + 1 are two consecutive positive integer

– We know that n is of the form n = 2q and n + 1 = 2q + 1

n (n + 1) = 2q (2q + 1) = 2 (2q^{2} + q)

Which is divisible by 2

If n = 2q + 1, then n (n + 1) = (2q + 1) (2q + 2)

= (2q + 1) x 2(q + 1)

= 2(2q + 1)(q + 1)

Which is also divisible by 2

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

**Question 3.**

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

**Answer : ** Let n be the positive any integer Then

n(n + 1) (n + 2) = (n^{2} + n) (n + 2)

= n³ + 2n² + n² + 2n

= n³ + 3n² + 2n Now if n

= 6q, then n³ + 3n² + 2n

= (6q)3 + 3(6g)² + 2 (69)

=216q3+ 108q2 + 12q

= 6g (36g2+18q+2q) **Which is divisible by 6**

If n6q+1, then n³ + 3n² + 2n

= (6q + 1)3 + 3 (6g+ 1)²+2 (6q + 1)

=216q3+ 108q2+18q+1+3 (36q² + 12q + 1) + 2 (6g + 1)

=216g+108q2+18q+1+108q2 +36g+ 3+12q+2

=216q3+216q2 + 669 +6

=6 (36q2 + 36q2+11q+1) **Which is divisible by 6**

If n = 6q+2, then n³ + 3n² + 2n

= (6q + 2)3 + 3 (6g + 2)² + 2 (6q + 2)

=216q3+216q2+72g+8+3 (36q² + 24q +4)+2 (6q + 2)

=216q³ +216q2 + 72q+8+ 108q² + 72g+ 12 + 12q + 4 -216q3+324g2+ 156q+24

=6 (36q³ + 54q² + 26g+4) **Which is also divisible by 6**

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

**Question 4.**

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

**Answer : ** Let n = 6q or 6q + 1, 6q +2, 6q + 3 … 6q +5

If n = 6q,

Then n³n (69)³ – 6q=216g’-6q

= 6 (36q³ – q) Which is divisible by 6

If n = 6q+1, then n³n = (6q + 1)³ – (6g+1)

=216g3+ 108g2+18q+1-6q-1

=216q+108q² + 12q

=6(36q3+ 18q2+2q) **Which is also divisible by 6**

If n = 6q+2, then n³n = (6q+2)³ – (6q+2)

=216g3 +216q2 + 72g+8-6q-2

=216g3+216q2 + 669 +6

=6 (36q³ +36q² + 11g + 1) **Which is divisible by 6****Hence **we can similarly, **prove that n ^{2} – n is divisible by 6** for any positive integer n.

Hence proved.

**Question 5.**

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.

** Answer : **Let n = 6q + 5, where q is a positive integer

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

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 = 6k + 1

If q = 3k+ 1, then n = 6q+5=6 (3k+ 1) + 5

= 18k + 6 + 5 = 18k + 11

= 18k+9+2=3 (6k+3)+2 =

= 3m + 2 where m = 6k + 3

If q = 3k+2, then n = 6q+5=6 (3k+2)+5

18k+ 12 + 5 = 18k + 17

= 18k + 15 + 2 = 3 (6k+5)+2

3m+2 where m = 6k + 5 =3m+2 6k+

**Hence proved.**

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

** Answer : **Let a be any positive integer

Then a = 5m + 1

a2 = (5m + 1 )2 = 25m2 + 10m + 1

= 5 (5m2 + 2m) + 1

= 5q + 1 where q = 5m2 + 2m

Which is of the same form as given

**Hence proved.**

**Question 7.**

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

** Answer : **Let a be any positive integer

In the form of 3m or 3m + 1

Lets a = 3q, then

a²=9q2=3(3q2) = 3m

Where m = 3q2

**Which is odd**

Let a = 39+ 1, then a²= (39 + 1)² = 9q²+6q + 1

= 3 (3q² + 2q) + 1

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

Which is odd

If a 3q+2, then a²= (3q + 2)²

=9q² + 12q +4 = 9q² + 12q +3 +1

= 3 (3q2+4q+1)+1

=3m+1 Which is odd

**Hence proved.**

**Question 8.**

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

** Answer :** Let a be the positive integer and

Let a = 4m

:. a² (4m)2= 16m² = 4(4m²)

= 4g where q = 4m² and let a = 4m + 1

:. a² = (4m + 1)2

16m²+8m+1

= 4 (4m² + 2m) + 1

= 4q + 1 where q = 4m² + 2m

**Hence proved.**

**Question 9.**

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

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

Let a be the positive integer, and

Let a = 5m, then

a² = (5m)² = 25m²

= 5 (5m²) = 5q Where q = 5m²

and a = (5m+ 1) then

a² = (5m + 1)²

=25m² + 10m+1

=5 (5m² + 2m) +1

= 5q+1 where q = 5m² + 2m

and let a = 5m + 1, then

a²= (5m+4)2=25m² + 40m + 16

=25m² + 40m + 15 + 1

=5 (5m² + 8m+3)+1

= 5q+1 where q = 5m² + 8m +3

and a = 5m + 2, then

a² = (5m + 2)²

=25m² + 20m + 4

= 5 (5m² + 4m) + 4

=5q+4 where q = 5m² + 4m =

and a= 5m+ 3,

then a² =(5m+3)²

=25m² + 30m +9

=25m² + 30m +5+4

=5 (5m² + 6m+1)+4

= 5q+4 where q = 5m² + 6m + 1

**Hence proved.**

**Question 10.**

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

** Answer :** Let n is any positive odd integer

Let n = 4p + 1, then

(4p + 1)

^{2}= 16p

^{2}+ 8p + 1

n

^{2}= 8p (2p + 1) + 1

= 8q + 1 where q = p(2p + 1)

Hence proved.

**Question 11.**

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

**Answer : **Let n be any positive odd integer and

let n = 6q + r

=> 6q + r, b = 6, and 0 ≤ r < 6

or r = 0, 1, 2, 3, 4, 5

If n = 6q = 2 x 3q

But it is not odd

When n = 6q + 1 which is odd

When n = 6q + 2 which is not odd = 2 (3q+ 1)

When n = 6q + 3 which is odd

When n = 6q + 4 = 2 (3q + 2) which is not odd

When n = 6q + 5, which is odd

Hence 6q + 1 or 6q + 3 or 6q + 5 are odd numbers.

**Question 12.**

Show that the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m. [NCERT Exemplar]

**Answer : ** Let a be an arbitrary positive integer, then by Euclid’s division algorithm, corresponding to the positive integers a and 6, there exist non-negative integers q and r such that.

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

**Question 13.**

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

**Answer : ** Let a be an arbitrary positive integer. Then, by Euclid’s division algorithm, corresponding to the positive integers ‘a’ and 6, there exist non-negative integers q and r such that

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 forms 6m, 6m + 1, 6m + 3, 6m + 3, 6m + 4 and 6m + 5 i.e., 6m + r.

**Question 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. (NCERT)

**Answer : ** Given numbers are n, (n + 4), (n + 8), (n + 12) and (n + 16), where n is any positive integer.

Then, let n = 5q, 5q + 1, 5q + 2, 5q + 3, 5q + 4 for q ∈N [By Euclid’s algorithm]

Then, in each case if we put the different values of n in the given numbers. We definitely get one and only one of given numbers is divisible by 5.

Hence, one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5.

*Alternate Method*On dividing on n by 5, let q be the quotient and r be the remainder.

Then n = 5q + r, where 0 ≤ r < 5. n = 5q + r, where r = 0, 1, 2, 3, 4

=> n = 5q or 5q + 1 or 5q + 2 or 5q + 3 or 5q + 4

Case I: If n = 5q, then n is only divisible by 5. .

Case II: If n = 5q + 1, then n + 4 = 5q + 1 + 4 = 5q + 5 = 5(q + 1), which is only divisible by 5.

So, in this case, (n + 4) is divisible by 5.

Case III : If n = 5q + 3, then n + 2 = 5q + 3 + 12 = 5q + 15 = 5(q + 3), which is divisible by 5.

So, in this case (n + 12) is only divisible by 5.

Case IV : If n = 5q + 4, then n + 16 = 5q + 4 + 16 = 5q + 20 = 5(q + 4), which is divisible by 5.

So, in this case, (n + 16) is only divisible by 5.

Hence, 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.

**Question 15.**

Show that the square of an odd positive integer can be of the form 6q + 1 or 6q + 3 for some integer ?

**Answer : ** We know that any positive integer can be of the form 6m, 6m + 1, 6m + 2, 6m + 3, 6m + 4 or 6m + 5, for some integer m.

Thus, an odd positive integer can be of the form 6m + 1, 6m + 3, or 6m + 5 Thus we have:

(6m+1)²= 36m² + 12m+ 1 = 6(6m² + 2m) +1=6q+1, q is an integer

(6m+3)2=36m² +36m+9=6(6m² + 6m+ 1)+3=6q +3, q is an integer

(6m+5)2=36m² + 60m+25=6(6m² + 10m +4)+1=6g+1, q is an integer.

Thus, the **square of an odd positive integer can be of the form 6q + 1 or 6q + 3.**

**Question 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.

**Answer : ** No, by Euclid’s Lemma, b = aq + r, 0 ≤ r < a

Here, b is any positive integer.

a=3, b=3q+r for 0 <r<3

So, this must be in the form 3q, 3q + 1 or 3q+2**Now,** (3q)2=9q2=3m ** [here, m = 3q2]**

and (3q + 1)² = 9q²+6g+1

= 3(3q2+2q) + 1 = 3m + 1 **[where, m = 3q2 + 2q]**

Also, (3q + 2)²=9q² + 12g + 4

=9q2+12g+3+1

= 3(3g² + 4q + 1) +1

= 3m + 1 **[here, m = 3q+4q + 1]**

Hence, square of a positive integer is of the form 39 + 1 is always in the form 3m + 1 for some integer m.

**Question 17.**

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

**Answer : ** By Euclid’s lemma, b = aq + r, 0 ≤ r ≤ a

Here, b is any positive integer,

a = 3, b = 3q + r for 0 ≤ r ≤ 2

So, any positive integer is of the form 3k, 3k + 1 or 3k + 2

Now, (3k)2=9k2= 3m

**[where, m = 3k²]**

and (3k+ 1)29k² + 6k+ 1

= 3(3k² + 2k) + 1 = 3m+1

**[where, m = 3k² + 2k]**

Also, (3k+ 2)²= 9k² + 12k + 4

**[: (a + b)²= a²+2ab+ b²]**

= 9k2+12k+3+1

= 3(3k2+4k+ 1) + 1

=3m+1

**[where, m = 3k² + 4k + 1]**.

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