Class 10 Maths Chapter 1 Real Numbers# Class 10 Maths Real Numbers

In mathematics, real numbers refer to the set of numbers that includes rational numbers (such as integers and fractions) and irrational numbers (numbers that cannot be expressed as a simple fraction). Real numbers are used to represent quantities that can be any value on the number line, and they form a continuous and uncountable set.

The real numbers are often represented on a number line, where the position of a number represents its magnitude relative to other numbers. The real numbers can also be represented using a coordinate system, where each real number corresponds to a point on a coordinate plane.

There are different types of real numbers, such as positive and negative numbers, whole numbers, integers, and decimals. Additionally, real numbers have certain properties, such as the associative, commutative, and distributive properties, which allow for arithmetic operations to be performed on them.

**Exercise 1.1**

**Question1**. **Use Euclid’s division algorithm to find the HCF of:**

**(i) 135 and 225 (ii) 196 and 38220**

**(iii) 867 and 255**

**Solution 1**.

###### (i) Applying Euclid’s division algorithm on 135 and 225, to

get

225 = 1 × 135 + 90

Again, applying Euclid’s division algorithm on 90 and

135, to get

135 = 1 × 90 + 45

Again, applying Euclid’s division algorithm on 45 and

90, to get

90 = 2 × 45 + 0.

Now HCF (225, 135) = HCF (135, 90) = HCF (90, 45)

= 45.

(ii) Applying Euclid’s division algorithm on 196 and 38220,

to get

38220 = 195 × 196 + 0

∴ HCF (38220, 196) = 196.

(iii) Applying Euclid’s division algorithm on 867 and 225, to

get

867 = 3 × 255 + 102

Again, applying Euclid’s division algorithm on 102 and

255, to get

255 = 2 × 102 + 51

Again, applying Euclid’s division algorithm on 51 and

102, to get

102 = 2 × 51 + 0

Now HCF (867, 255) = HCF (255, 102)

= HCF (102, 51) = 51.

**Question 2**. **Show that any positive odd integer is of the form 6q + 1,**

**or 6q + 3, or 6q + 5, where q is some integer.**

**Solution 2**.

Consider a, a positive integer, we apply division algorithm

with q and b = 6, 0 ≤ r < 6 such that

a = 6q + r

As 0 ≤ r < 6. Therefore, possible remainders are 0, 1, 2, 3,4, 5.

Number a can be 6q, 6q + 1, 6q + 2, 6q + 3, 6q + 4 or 6q + 5.

6q = 2(3q) = 2m; 6q + 1 = 2(3q) + 1 = 2m + 1;

6q + 2 = 2(3q + 1) = 2n; 6q + 3 = 2(3q + 1) + 1 = 2n + 1;

6q + 4 = 2(3q + 2) = 2t; 6q + 5 = 2(3q + 2) + 1 = 2t + 1.

We note 6q, 6q + 2 and 6q + 4 are of the form 2r, r N,

which are even numbers; and 6q + 1, 6q + 3 and 6q + 5

are of the form 2r + 1, r € N, which are odd numbers.

Hence, 6q + 1, 6q + 3 and 6q + 5 are positive and odd

integers.

**Question 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?**

**Solution 3**.

To find maximum number of columns, we have to find

HCF of 616 and 32.

Using Euclid’s division algorithm, we have

616 = 19 × 32 + 8

Now applying Euclid’s division algorithm on 32 and 8, we have

32 = 4 × 8 + 0

∴ HCF (616, 32) = HCF (32, 8) = 8

Hence, maximum number of columns is 8.

**Question 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.**

[Hint: Let x be any positive integer then it is of the form 3q,

3q + 1 or 3q + 2. Now square each of these and show that

they can be rewritten in the form 3m or 3m + 1.]

**Solution 4.**

Consider positive integer x, with q and b = 3, 0 r < 3

x can be of the form 3q, 3q + 1 and 3q + 2 because

r = 0, 1, 2

There exist three cases:

**Case I**: x = 3q

x² = (3q)²= 9q² = 3m, m = 3q²…(i)

**Case II**: x = 3q + 1

x² = (3q + 1)² = 9q²+ 6q + 1 = 3(3q² + 2q) + 1

= 3m + 1, m = 3q² + 2q …(ii)

**Case III:** x = 3q + 2

x²= (3q + 2)² = 9q² + 12q + 4

x²= 3(3q² + 4q + 1) + 1

x²= 3m + 1, m = 3q² + 4q + 1 …(iii)

Hence, from (i), (ii) and (iii), the square of any positive

integer is either of the form 3m or 3m + 1.

**Question 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.**

**Solution 5**.

Consider positive integer a, with q and b = 3, 0 r < 3.

a can be of the form 3q, 3q + 1 or 3q + 2 because r = 0,1, 2.

Consider a = 3q

a³ = 27q³ = 9(3q³) = 9m, m = 3q³.

Consider a = 3q + 1

a³ = (3q + 1)³ = 27q³ + 27q² + 9q + 1

a³ = 9(3q³+ 3q² + q) + 1 = 9m + 1,

m = 3q³ + 3q² + q

And consider a = 3q + 2

a³ = (3q + 2)³ = 27q³ + 54q² + 36q + 8

= 9(3q3 + 6q2 + 4q) + 8 = 9m + 8,

m = 3q³ + 6q² + 4q

Hence, cube of any positive integer is of the form 9m, 9m + 1

or 9m + 8.

**Exercise 1.2**

**Question 1. Express each number as a product of its prime factors:**

**(i) 140 (ii) 156 (iii) 3825**

**(iv) 5005 (v) 7429**

**Solution.1**

**Question 2**.** Find the LCM and HCF of the following pairs of integers**

**and verify that LCM × HCF = product of the two numbers.**

**(i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54**

**Solution 2.**

(i) 26 = 2 × 13, 91 = 7 × 13

∴ HCF = 13 and LCM = 2 × 7 × 13 = 182

Now, HCF × LCM = 13 × 182 = 2366 …(i)

Product of the numbers = 26 × 91 = 2366 …(ii)

From (i) and (ii), we get the result.

**Question 3**. **Find the LCM and HCF of the following integers by**

**applying the prime factorisation method.**

**(i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25**

**Solution 3**.

(i) 12 = 22 × 3

15 = 3 × 5

21 = 3 × 7

∴ HCF (12, 15, 21) = 3;

LCM (12, 15, 21) = 22 × 3 × 5 × 7 = 420.

(ii) 17 = 1 × 17

23 = 1 × 23

29 = 1 × 29

∴ HCF (17, 23, 29) = 1,

LCM (17, 23, 29) = 1 × 17 × 23 × 29 = 11339.

(iii) 8 = 23

9 = 32

25 = 52

∴ HCF (8, 9, 25) = 1,

LCM (8, 9, 25) = 23 × 32 × 52 = 1800

**Question 4. Given that HCF (306, 657) = 9, find LCM (306, 657).**

**Solution 4**. We know for two numbers a and b HCF (a, b) × LCM

(a, b) = a × b

∴ LCM (306, 657) =306× 657/HCF (306, 657)

=306 × 657/9

= 22338.

**Question 5. Check whether 6n can end with the digit 0 for any natural**

**number n.**

**Solution 5.**

If 6n for n € N, ends with digit zero, then

6n = 5 × m + 0, for some m € N.

⇒Prime factorisation of 6n contains 5 as a prime factor.

Only prime factorisation of 6 is 3 and 2 and by Uniqueness

of Fundamental Theorem of Arithmetic, 5 is not a prime factor.

Therefore, it contradicts our supposition that 5 is a factor of

6n. Hence, 5 is not a factor of 6n, i.e., 6n does not end with

digit zero.

**Question 6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2**

**× 1 + 5 are composite numbers.**

**Solution 6.**

Composite numbers have more than two factors.

Number 7 × 11 × 13 + 13 = 13 (7 × 11 + 1) has factors

7 × 11 + 1, 1 and 13. As it has more than 2 factors it is a

composite number, therefore, the number is a composite

number.

Similarly, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 (7 × 6 × 4 × 3

× 2 × 1 + 1)

⇒The number has factors 1, 5 and (7 × 6 × 4 × 3 × 2× 1 + 1).

As it has more than 2 factors, therefore, the number is a composite number.

**Question 7. There is a circular path around a sports field. Sonia takes**

**18 minutes to drive one round of the field, while Ravi takes**

**12 minutes for the same. Suppose they both start at the**

**same point and at the same time, and go in the same**

**direction. After how many minutes will they meet again at**

**the starting point?**

**Solution 7.**

If we have to find after how many minutes they again

meet, we have to find the least common multiple of 18 and 12.

Now, 18 = 2 × 32 and 12 = 22 × 3

∴ LCM = 22 × 32 = 4 × 9 = 36.

Hence, after 36 minutes they will meet again at the

starting point.

Exerise 1.3

**Question 1. Prove that √5 is irrational**.

**Solution 1**

. Let, if possible,√5 is a rational number.

√5=p/q,

q € Z, q ≠ 0; p and q are coprime.

⇒ p² = 5q² …(i)

⇒ 5 divides p² ⇒5 divides p.

Let p = 5m, m € Z…………………………………………………………………….(ii)

∴ (5m)² = 5q² ⇒q2 = 5m2 [From (i)]

⇒ 5 divides q² ⇒5 divides q

∴ q = 5n, n € Z……………………………………………………………………….(iii)

From (ii) and (iii), we notice

p and q have 5 as common factor.

∴p and q are not co-prime.

Hence, our supposition,√5 is rational, is wrong.

Hence, is an irrational number.

**Question 2. Prove that 3 + 2√5 is irrational.**

**Solution 2**. Let if possible, 3 + 2√5 is a rational number.

3 + 2√5=a/b, a, b € Z, b ≠ 0; a and b are coprime.