NCERT Exemplar Solutions for Class 10 Maths Chapter 1 Real number Exercise 1.3

Chapter 1 Real Number NCERT Exemplar Solutions Exercise 1.3 Class 10 Maths

Chapter Name	NCERT Maths Exemplar Solutions for Chapter 1 Real Numbers Exercise 1.3
Book Name	NCERT Exemplar for Class 10 Maths
Other Exercises	Exercise 1.1 Exercise 1.2 Exercise 1.4
Related Study	NCERT Solutions for Class 10 Maths

Exercise 1.3 Solutions

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

Solution

By Euclid’s division lemma,

a = 4m + r ...(i)

Where r = 0 ≤ r < 4

or

r = 0, 1, 2, 3

When r = 0, a = 4m  [From (i)]

When r = 0, we get,

a = 4k
a² = 16m² 
= 4(4m² )
= 4q, where q = 4m²
When r = 1, we get,
a = 4m + 1
a² = (4m + 1) ²
= 16m² + 1 + 8m
= 4(4m + 2)+ 1
= 4q + 1, where q = m(4m + 2)
When r = 2, we get, a = 4m + 2
a² = (4m + 2) ²
= 16m² + 4 + 16m
= 4(4m² + 4m + 1)
= 4q, where q = 4m² + 4m + 1
When r = 3, we get,
a = 4m + 3
a² = (4m + 3)²
= 16m² + 9 + 24m
= 4(4m² + 6m +2) + 1
= 4q + 1, where q = 4k² + 6k + 2
Hence, the square of any positive integer is either of the form 4q or 4q + 1 for some integer q.

2. Show that the cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3 for some integer m. 

Solution 

Let x be any positive integer and b = 4.

Using Euclid Algorithm,

x = bq + r

⇒ x = 4q + r

The possible values of r are,

r = 0,1,2,3

If r = 0,

x = 4q + 0

⇒ x = 4q

Taking cubes on LHS and RHS,

x³ = (4q)³ 
⇒ x³ = 4(16q³)
x³ = 4m  [where m = 16q³]
If r = 1,
x = 4q + 1 
Taking cubes on LHS and RHS,

3. Show that the square of any positive integer cannot be of the form 5q + 2 or 5q + 3 for any integer q.

Solution

Let the positive integer = x 

Using Euclid algorithm, 

x = bm + r 
According to the question, b = 5 
x = 5m + r
So, r = 0, 1, 2, 3, 4 
For r = 0, c = 5 m.

For r = 1, x = 5 m + 1
For r = 2, x = 5m + 2.
For r = 3, x = 5m + 3.
For r = 4, x = 5m + 4.
Now, When x = 5m 
x² = (5m)² 
= 25m² 
x² = 5(5m² )
= 5q, where q = 5m² 

So, square of any positive integer cannot be of the form 5q + 2 or 5q + 3.

4. 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 = x

Using Euclid’s algorithm,

x = 6q + r, where 0 ≤ r < 6
x² = (6q + r)² 
= 36q² + r² + 12qr
x² = 6(6q² + 2qr) + r² 
Here, 0 < r < 6

If r = 0, we get

x² = 6(6q² )

= 6m,

(Here, m = 6q² ).

When r = 1, we get

x² = 6(6q²  + 2q) + 1

= 6m + 1,

Here, m = (6q2 + 2q)

If r = 2, we get

x² = 6(6q²  + 4q) + 4

= 6m + 4,

Here, m = (6q²  + 4q).

If r = 3, we get

x² = 6(6q² + 6q) + 9

= 6(6q²  + 6a) + 6 + 3

x² = 6(6q² + 6q + 1) + 3

= 6m + 3,

Here, m = (6q + 6q + 1)

If r = 4, we get

x² = 6(6q²  + 8q) + 16

= 6(6q² + 8q) + 12 + 4

x²  = 6(6q² + 8q + 2) + 4

= 6m + 4,

Here, m = (6q²  + 8q + 2)

If r = 5, we get

x² = 6(6q²  + 10q) + 25

= 6(6q²  + 10q) + 24 + 1

x²  = 6(6q²  + 10q + 4) + 1

= 6m + 1,

Here, m = (6q²  + 10q + 1).

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

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

Solution

Let c be any odd integer and d = 4.

Using Euclid’s algorithm,

c = 4m + r

and

r = 0,1,2,3 as, 0 ≤ r < 4.

Therefore,

c = 4m, 4m + 1, 4m + 2, 4m + 3

So, c can be 4m + 1 or 4m + 3

c cannot be 4m or 4m + 2, as they are even integer

Now,

(4m + 1)²  = 16m²  + 8m + 1

= 4(4m²  + 2m) + 1

= 4q + 1, where q = 4m²  + 2m.

(4m + 3)²  = 16m²  + 24m + 9

= 4(4m²  + 6m + 2) + 1

= 4q + 1, where q = 4m²  + 6m + 2

Thus we can conclude that square of any odd integer is of the form 4q + 1, for some integer q.

6. If n is an odd integer, then show that n²  – 1 is divisible by 8.

Solution

Let 

x = n² - 1
As n is odd number so, 

n = 1, 3, 5, 7
If n = 1 
x = 1² - 1
= 0, which is divisible by 8. 
If n = 3 

x = 3² - 1 
= 9 -1 
= 8, which is also divisible by 8. 
If n = 5, 

x = 5² - 1 
= 25 - 1
= 24 
= 8 × 3, which is divisible by 8.

Therefore, n² - 1 is divisible by 8 when n is odd. 

7. Prove that, if x and y, both are odd positive integers, then (x² + y² ) is even but not divisible by 4.

Solution

Let the two odd positive numbers x and y be 2m + 1 and 2n + 1 respectively. 

So, 

x² + y² = (2m + 1)²  + (2n + 1)² 
= 4m²  + 4m + 1 + 4n² + 4n + 1 
= 4m² + 4n² + 4m + 4n + 2
= 4(m² + n² + m + n) + 2 

In the above expression we see that the sum of square is even, but the number is not divisible

by 4. Therefore, we can say that if x and y are odd positive integer, then x² + y²  is even but not divisible by four. 

8. Use Euclid’s division algorithm to find HCF of 441, 567 and 693.

Solution

Let p = 693 and q = 567 
Using Euclid's division lemma, 
p = qm + r 

693 = 567 × 1 + 126

567 = 126 × 4 + 63

126 = 63 × 2 + 0

So, HCF (693 and 567) = 63.

Further,

Taking 441 and HCF = 63

Using Euclid’s algorithm,

c = dm + r

⇒ c = 441 and d = 63

⇒ 441 = 63 × 7 + 0

So, HCF of (693, 567, 441) = 63.

9. Using Euclid’s division algorithm, find the largest number that divides 1251, 9377 and 1562 8 leaving remainders, 1, 2, and 3 respectively.

Solution

According to question 1, 2, and 3 are the remainders when the largest number divides 1251,

9377 and 15628 respectively.

So, we have to find HCF of (1251 – 1), (9377 – 2) and (15628 – 3)

That are,

1250, 9375, 15625

For HCF of 1250, 9375, 15625

Let p = 15625,

q = 9375

Using Euclid’s algorithm, p = qm+ r

15625 = 9375 × 1 + 6250

9375 = 6250 × 1 + 3125

6250 = 3125 × 2 + 0

Therefore,

HCF (15625, 9375) = 3125

Let a = 1250 and

b = 3125

By Euclid’s division algorithm, b = am + r

3125 = 1250 × 2 + 625

1250 = 625 × 2 + 0

Therefore, the HCF of (15625, 1250 and 9375) is 625.

10. Prove that (√3 + √5) is irrational. 

Solution

Let us consider (√3 + √5) is a rational number that can be written as 

From the above expression, RHS comes out to be rational but we know that LHS = √3 is irrational, which contradicts our fact. So, (√3 + √5) is irrational. 

11. Show that 12ⁿ cannot end with the digit 0 or 5 for any natural number n.

Solution

Number ending at 0 or 5 must be divisible by 5. 

So, 

(12)ⁿ  = (2 × 2 × 3)ⁿ  
= 2ⁿ × 3ⁿ  
(12)ⁿ  does not have 5 as its factor. So, 12ⁿ   can never end with 5 and zero.

12. On a morning walk, three persons, step off together and their steps measure 40cm,42 cm and 45cm respectively. What is the minimum distance each should walk, so that each can cover the same distance in complete steps?

Solution

Then minimum distance = LCM of covered steps.

40 = 2³ × 5
42 = 2 × 3 ×7
45 = 3² × 5
LCM of (40, 42, 45) = 2³ × 3² × 5 × 7 = 2520 cm 

Hence, the minimum distance each should walk is 2520 cm. 

13. Write the denominator of rational number 257/5000 in the form of 2m × 5n , where m, n are non-negative integers. Hence, write its decimal expansion, without actual division.

Solution

We have, denominator of the rational number 27/5000 = 5000
Therefore, 

5000 = 2³ × 5⁴ 
It is of the form 2^m × 5ⁿ 

Here m = 3 and n =  4 
So, 

257/5000 = (257 × 2)/(5000 × 2) = 0.0514

14. Prove that (√p + √q) is irrational, where p and q are primes. 

Solution

Let us take (√p + √q) = rational and can be represented as (√P + √q) = x  

As q is prime so √q is no rational but in above solution we had √q = rational because a, p, q are non - zero integers which contradicts our fact. 

So, (√p + √q) is irrational.