# Class 10 Maths Chapter 1 Real Numbers

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

###### (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.

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