NCERT Solutions for Real Numbers
Exercise 1.1
1. Use Euclid’s Division Algorithm to Find the HCF of:
i. 135 and 225
ii. 196 and 38220
iii. 867 and 255
Solution:
i. 135 and 225
Given two numbers are \(225\) and \(135\). Of these, the larger number is \(225\), and the smaller number is \(135\).
By Euclid’s Division Algorithm, we express the larger number as the quotient times the smaller number plus the remainder:
\(225 = 135 \times 1 + 90\)
The remainder is \(90\), which is not zero. We now apply the division algorithm to \(135\) and \(90\):
\(135 = 90 \times 1 + 45\)
The remainder is \(45\), which is not zero. Repeating the process with \(90\) and \(45\), we get:
\(90 = 45 \times 2 + 0\)
Since the remainder is now zero, the divisor in the last step, \(45\), is the HCF. Therefore,
\[ \text{HCF}(225, 135) = 45 \]
Hence, the HCF of \(225\) and \(135\) is \(45\).
ii. 196 and 38220
Given two numbers are \(38220\) and \(196\). Of these, the larger number is \(38220\), and the smaller number is \(196\).
By Euclid’s Division Algorithm, we express the larger number as the quotient times the smaller number plus the remainder:
\(38220 = 196 \times 195 + 0\)
Since the remainder is already zero, the divisor in this step, \(196\), is the HCF. Therefore,
\[ \text{HCF}(196, 38220) = 196 \]
Hence, the HCF of \(196\) and \(38220\) is \(196\).
iii. 867 and 255
Given two numbers are \(867\) and \(255\). Of these, the larger number is \(867\), and the smaller number is \(255\).
By Euclid’s Division Algorithm, we express the larger number as the quotient times the smaller number plus the remainder:
\(867 = 255 \times 3 + 102\)
The remainder is \(102\), which is not zero. Taking \(255\) as the new dividend and \(102\) as the divisor, we proceed:
\(255 = 102 \times 2 + 51\)
The remainder is \(51\), which is not zero. Repeating the process with \(102\) and \(51\), we get:
\(102 = 51 \times 2 + 0\)
Since the remainder is now zero, the divisor in the last step, \(51\), is the HCF. Therefore,
\[ \text{HCF}(867, 255) = 51 \]
Hence, the HCF of \(867\) and \(255\) is \(51\).
― ― ― ― solutions by ncertbooks.net ― ― ― ―
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:
Let \( a \) be any positive integer and \( b = 6 \). By Euclid’s Division Algorithm, we can express \( a \) as:
\( a = 6q + r \)
Here, \( q \) is an integer (\( q \geq 0 \)), and the remainder \( r \) satisfies \( 0 \leq r < 6 \). Therefore, \( r \) can take one of the following values: \( r = 0, 1, 2, 3, 4, 5 \).
Substituting these possible values of \( r \) into the equation \( a = 6q + r \), we get:
If \( r = 0 \), then \( a = 6q \)
If \( r = 1 \), then \( a = 6q + 1 \)
If \( r = 2 \), then \( a = 6q + 2 \)
If \( r = 3 \), then \( a = 6q + 3 \)
If \( r = 4 \), then \( a = 6q + 4 \)
If \( r = 5 \), then \( a = 6q + 5 \)
Now, observe that:
- If \( a = 6q, 6q + 2, \) or \( 6q + 4 \), the number \( a \) is divisible by \( 2 \), making it an even number.
- If \( a = 6q + 1, 6q + 3, \) or \( 6q + 5 \), the number \( a \) is not divisible by \( 2 \), making it an odd number.
Since any positive integer is either even or odd, we conclude that any positive odd integer must be of the form:
\( a = 6q + 1, \ 6q + 3, \ \text{or} \ 6q + 5 \)
where \( q \) is some integer.
― ― ― ― solutions by ncertbooks.net ― ― ― ―
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:
Given:
Number of army contingent members = \( 616 \)
Number of army band members = \( 32 \)
If the two groups have to march in the same column, we need to find the highest common factor (HCF) of \( 616 \) and \( 32 \).
By using Euclid’s Division Algorithm, we take \( 616 \) as the dividend and \( 32 \) as the divisor. We get the following:
\( 616 = 32 \times 19 + 8 \)
Since the remainder is \( 8 \neq 0 \), we take \( 32 \) as the new dividend and \( 8 \) as the new divisor:
\( 32 = 8 \times 4 + 0 \)
Now the remainder is \( 0 \). Therefore, the HCF of \( 616 \) and \( 32 \) is:
\( \text{HCF}(616, 32) = 8 \)
Hence, the maximum number of columns in which they can march is \( 8 \).
― ― ― ― solutions by ncertbooks.net ― ― ― ―
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.
Solution:
Let \( x \) be any positive integer, and let \( y = 3 \). According to Euclid’s Division Lemma, any integer \( x \) can be expressed as:
\( x = 3q + r \)
Here, \( q \) is an integer (\( q \geq 0 \)), and the remainder \( r \) satisfies \( 0 \leq r < 3 \). Therefore, \( r \) can take the values \( r = 0, 1, 2 \). Substituting these possible values of \( r \) into the equation \( x = 3q + r \), we get:
\( x = 3q \), \( x = 3q + 1 \), \( x = 3q + 2 \)
Now, we square each of these expressions to find \( x^2 \) and analyze the results:
Case 1: When \( x = 3q \)
Squaring \( x \), we get:
\( x^2 = (3q)^2 = 9q^2 \)
This can be written as:
\( x^2 = 3 \times (3q^2) \)
Let \( 3q^2 = m \), where \( m \) is an integer. Therefore:
\( x^2 = 3m \) ……………………..(1)
Case 2: When \( x = 3q + 1 \)
Squaring \( x \), we get:
\( x^2 = (3q + 1)^2 \)
We use the algebraic identity \( (a + b)^2 = a^2 + b^2 + 2ab \). Here, \( a = 3q \) and \( b = 1 \).
Applying the formula:
\( x^2 = (3q + 1)^2 = (3q)^2 + 1^2 + 2 \times 3q \times 1 \)
Simplifying, we have:
\( x^2 = 9q^2 + 1 + 6q \)
Factorizing, we can write:
\( x^2 = 3(3q^2 + 2q) + 1 \)
Let \( 3q^2 + 2q = m \), where \( m \) is an integer. Therefore:
\( x^2 = 3m + 1 \) ……………………………. (2)
Case 3: When \( x = 3q + 2 \)
Squaring \( x \), we get:
\( x^2 = (3q + 2)^2 = (3q)^2 + 2^2 + 2 \times 3q \times 2 \)
Simplifying, we have:
\( x^2 = 9q^2 + 4 + 12q \)
Factorizing, we can write:
\( x^2 = 3(3q^2 + 4q + 1) + 1 \)
Let \( 3q^2 + 4q + 1 = m \), where \( m \) is an integer. Therefore:
\( x^2 = 3m + 1 \) ……………………………. (3)
Conclusion: From equations (1), (2), and (3), we observe that the square of any positive integer is either of the form:
\( x^2 = 3m \) or \( x^2 = 3m + 1 \)
where \( m \) is some integer.
― ― ― ― solutions by ncertbooks.net ― ― ― ―
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:
Let \( x \) be any positive integer and let \( y = 3 \).
By Euclid’s Division Lemma, any integer \( x \) can be expressed as:
\( x = 3q + r \)
Here, \( q \) is an integer (\( q \geq 0 \)), and the remainder \( r \) satisfies \( 0 \leq r < 3 \). Therefore, \( r \) can take the values:
\( r = 0, 1, 2 \)
Substituting these possible values of \( r \) into the equation \( x = 3q + r \), we get:
\( x = 3q \), \( x = 3q + 1 \), \( x = 3q + 2 \)
Now, we take the cube of each case to find \( x^3 \) and analyze the results:
Case (i): When \( r = 0 \)
Cubing \( x \), we get:
\( x^3 = (3q)^3 = 27q^3 = 9(3q^3) \)
Let \( m = 3q^3 \). Therefore:
\( x^3 = 9m \)
Case (ii): When \( r = 1 \)
Cubing \( x \), we get:
\( x^3 = (3q + 1)^3 \)
Using the algebraic identity \( (a + b)^3 = a^3 + b^3 + 3a^2b + 3ab^2 \), we have:
\( x^3 = (3q)^3 + 1^3 + 3(3q)^2(1) + 3(3q)(1^2) \)
Simplifying, we get:
\( x^3 = 27q^3 + 1 + 27q^2 + 9q \)
Taking \( 9 \) as a common factor, we write:
\( x^3 = 9(3q^3 + 3q^2 + q) + 1 \)
Let \( m = 3q^3 + 3q^2 + q \). Therefore:
\( x^3 = 9m + 1 \)
Case (iii): When \( r = 2 \)
Cubing \( x \), we get:
\( x^3 = (3q + 2)^3 \)
Using the algebraic identity \( (a + b)^3 = a^3 + b^3 + 3a^2b + 3ab^2 \), we have:
\( x^3 = (3q)^3 + 2^3 + 3(3q)^2(2) + 3(3q)(2^2) \)
Simplifying, we get:
\( x^3 = 27q^3 + 8 + 54q^2 + 36q \)
Taking \( 9 \) as a common factor, we write:
\( x^3 = 9(3q^3 + 6q^2 + 4q) + 8 \)
Let \( m = 3q^3 + 6q^2 + 4q \). Therefore:
\( x^3 = 9m + 8 \)
Conclusion: From the above cases, we have shown that the cube of any positive integer is of the form:
\( x^3 = 9m \), \( x^3 = 9m + 1 \), or \( x^3 = 9m + 8 \)
where \( m \) is some integer.
― ― ― ― solutions by ncertbooks.net ― ― ― ―
Exercise 1.2
1. Express each number as a product of its prime factors:
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429
Solutions:
(i) 140
By taking the LCM of 140, we get the product of its prime factors:
\( 140 = 2 \times 2 \times 5 \times 7 = 2^2 \times 5 \times 7 \)
(ii) 156
By taking the LCM of 156, we get the product of its prime factors:
\( 156 = 2 \times 2 \times 13 \times 3 = 2^2 \times 13 \times 3 \)
(iii) 3825
By taking the LCM of 3825, we get the product of its prime factors:
\( 3825 = 3 \times 3 \times 5 \times 5 \times 17 = 3^2 \times 5^2 \times 17 \)
(iv) 5005
By taking the LCM of 5005, we get the product of its prime factors:
\( 5005 = 5 \times 7 \times 11 \times 13 \)
(v) 7429
By taking the LCM of 7429, we get the product of its prime factors:
\( 7429 = 17 \times 19 \times 23 \)
― ― ― ― solutions by ncertbooks.net ― ― ― ―
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:
(i) 26 and 91
Expressing 26 and 91 as a product of their prime factors:
\( 26 = 2 \times 13 \)
\( 91 = 7 \times 13 \)
Now, the LCM is the product of all prime factors with the highest powers:
\( \text{LCM}(26, 91) = 2 \times 7 \times 13 = 182 \)
The HCF is the product of common prime factors with the lowest powers:
\( \text{HCF}(26, 91) = 13 \)
Verification:
Product of 26 and 91 = \( 26 \times 91 = 2366 \)
Product of LCM and HCF = \( 182 \times 13 = 2366 \)
Hence, \( \text{LCM} \times \text{HCF} = \text{Product of the two numbers.} \)
(ii) 510 and 92
Expressing 510 and 92 as a product of their prime factors:
\( 510 = 2 \times 3 \times 5 \times 17 \)
\( 92 = 2 \times 2 \times 23 \)
Now, the LCM is the product of all prime factors with the highest powers:
\( \text{LCM}(510, 92) = 2^2 \times 3 \times 5 \times 17 \times 23 = 23460 \)
The HCF is the product of common prime factors with the lowest powers:
\( \text{HCF}(510, 92) = 2 \)
Verification:
Product of 510 and 92 = \( 510 \times 92 = 46920 \)
Product of LCM and HCF = \( 23460 \times 2 = 46920 \)
Hence, \( \text{LCM} \times \text{HCF} = \text{Product of the two numbers.} \)
(iii) 336 and 54
Expressing 336 and 54 as a product of their prime factors:
\( 336 = 2^4 \times 3 \times 7 \)
\( 54 = 2 \times 3^3 \)
Now, the LCM is the product of all prime factors with the highest powers:
\( \text{LCM}(336, 54) = 2^4 \times 3^3 \times 7 = 3024 \)
The HCF is the product of common prime factors with the lowest powers:
\( \text{HCF}(336, 54) = 2 \times 3 = 6 \)
Verification:
Product of 336 and 54 = \( 336 \times 54 = 18144 \)
Product of LCM and HCF = \( 3024 \times 6 = 18144 \)
Hence, \( \text{LCM} \times \text{HCF} = \text{Product of the two numbers.} \)
― ― ― ― solutions by ncertbooks.net ― ― ― ―
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:
(i) 12, 15 and 21
To find the LCM and HCF of 12, 15, and 21, we first express each number as a product of its prime factors:
\( 12 = 2 \times 2 \times 3 = 2^2 \times 3 \)
\( 15 = 5 \times 3 \)
\( 21 = 7 \times 3 \)
Step 1: Find the HCF
The HCF is the product of the prime factors common to all three numbers with the lowest powers. Here, the common prime factor is \( 3 \), and it occurs in all three numbers.
\( \text{HCF}(12, 15, 21) = 3 \)
Step 2: Find the LCM
The LCM is the product of all prime factors of the numbers with the highest powers:
\( \text{LCM}(12, 15, 21) = 2^2 \times 3 \times 5 \times 7 = 420 \)
Therefore, the HCF is \( 3 \), and the LCM is \( 420 \).
(ii) 17, 23 and 29
To find the LCM and HCF of 17, 23, and 29, we first express each number as a product of its prime factors:
\( 17 = 1 \times 17 \)
\( 23 = 1 \times 23 \)
\( 29 = 1 \times 29 \)
Step 1: Find the HCF
Since 17, 23, and 29 have no common factors other than 1, the HCF is:
\( \text{HCF}(17, 23, 29) = 1 \)
Step 2: Find the LCM
The LCM is the product of all the prime factors, as there are no common factors:
\( \text{LCM}(17, 23, 29) = 17 \times 23 \times 29 = 11339 \)
Therefore, the HCF is \( 1 \), and the LCM is \( 11339 \).
(iii) 8, 9 and 25
To find the LCM and HCF of 8, 9, and 25, we first express each number as a product of its prime factors:
\( 8 = 2 \times 2 \times 2 = 2^3 \)
\( 9 = 3 \times 3 = 3^2 \)
\( 25 = 5 \times 5 = 5^2 \)
Step 1: Find the HCF
Since 8, 9, and 25 have no common prime factors, the HCF is:
\( \text{HCF}(8, 9, 25) = 1 \)
Step 2: Find the LCM
The LCM is the product of all prime factors with the highest powers:
\( \text{LCM}(8, 9, 25) = 2^3 \times 3^2 \times 5^2 = 1800 \)
Therefore, the HCF is \( 1 \), and the LCM is \( 1800 \).
― ― ― ― solutions by ncertbooks.net ― ― ― ―
4. Given that HCF(306, 657) = 9, find LCM(306, 657).
Solution:
We know the relationship between HCF and LCM:
HCF × LCM = Product of the two numbers
Given:
HCF(306, 657) = 9
Substituting the values, we get:
\( 9 \times \text{LCM} = 306 \times 657 \)
Now, calculate the LCM:
\(\text{LCM} = \dfrac{306 \times 657}{9} = 34 \times 657 = 22338\)
Hence, the LCM of 306 and 657 is:
\( \text{LCM}(306, 657) = 22338 \)
― ― ― ― solutions by ncertbooks.net ― ― ― ―
5. Check whether \( 6^n \) can end with the digit 0 for any natural number \( n \).
Solution:
For \( 6^n \) to end with the digit 0, it must be divisible by 10. A number is divisible by 10 if it has both \( 2 \) and \( 5 \) as its prime factors.
Now, let us find the prime factorization of \( 6^n \):
\( 6^n = (2 \times 3)^n \)
From the prime factorization, we see that \( 6^n \) contains the factors \( 2 \) and \( 3 \), but it does not contain the prime factor \( 5 \).
Since \( 6^n \) does not have \( 5 \) as a factor, it cannot be divisible by 10.
Therefore, \( 6^n \) cannot end with the digit 0 for any natural number \( n \).
Hence, \( 6^n \) cannot end with the digit 0 for any natural number \( n \).
― ― ― ― solutions by ncertbooks.net ― ― ― ―
6. Explain why \( 7 \times 11 \times 13 + 13 \) and \( 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 \) are composite numbers.
Solution:
A number is composite if it has factors other than 1 and itself. We will check each given expression to verify if it is a composite number.
(i) \( 7 \times 11 \times 13 + 13 \)
Taking \( 13 \) as a common factor:
\( 7 \times 11 \times 13 + 13 = 13(7 \times 11 + 1) \)
Simplify the terms inside the parentheses:
\( 13(7 \times 11 + 1) = 13(77 + 1) = 13 \times 78 \)
Factorize \( 78 \):
\( 13 \times 78 = 13 \times 13 \times 3 \times 2 \)
Since the expression has factors other than 1 and itself, it is a composite number.
Hence, \( 7 \times 11 \times 13 + 13 \) is a composite number.
(ii) \( 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 \)
Taking \( 5 \) as a common factor:
\( 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 = 5(7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1) \)
Simplify the terms inside the parentheses:
\( 5(7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1) = 5(1008 + 1) = 5 \times 1009 \)
Since the expression has factors other than 1 and itself, it is a composite number.
Hence, \( 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 \) is a composite number.
― ― ― ― solutions by ncertbooks.net ― ― ― ―
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:
Since both Sonia and Ravi move in the same direction and start at the same time, the time when they will meet again at the starting point is given by the Least Common Multiple (LCM) of their times to complete one round.
First, we find the prime factorization of 18 and 12:
\( 18 = 2 \times 3 \times 3 = 2 \times 3^2 \)
\( 12 = 2 \times 2 \times 3 = 2^2 \times 3 \)
The LCM is the product of all prime factors with the highest powers:
\( \text{LCM}(18, 12) = 2^2 \times 3^2 = 4 \times 9 = 36 \)
Therefore, Sonia and Ravi will meet again at the starting point after 36 minutes.
Hence, Sonia and Ravi will meet again at the starting point after 36 minutes.
― ― ― ― solutions by ncertbooks.net ― ― ― ―
Exercise 1.3
1. Prove that \( \sqrt{5} \) is irrational.
Solution:
We will prove \( \sqrt{5} \) is irrational by contradiction. Let us assume that \( \sqrt{5} \) is a rational number.
This means we can write \( \sqrt{5} \) as:
\( \sqrt{5} = \dfrac{x}{y} \), where \( x \) and \( y \) are co-prime integers (i.e., their HCF is 1).
Rearranging the equation, we get:
\( y\sqrt{5} = x \)
Squaring both sides, we get:
\( (y\sqrt{5})^2 = x^2 \)
Simplify:
\( 5y^2 = x^2 \) ……………………. (1)
From equation (1), we observe that \( x^2 \) is divisible by 5. Therefore, \( x \) is also divisible by 5 (since the square of a number is divisible by 5 only if the number itself is divisible by 5).
Let \( x = 5k \), where \( k \) is an integer. Substituting \( x = 5k \) into equation (1), we get:
\( 5y^2 = (5k)^2 \)
Simplify:
\( 5y^2 = 25k^2 \)
Dividing both sides by 5:
\( y^2 = 5k^2 \)
This shows that \( y^2 \) is divisible by 5, which means \( y \) is also divisible by 5.
Since both \( x \) and \( y \) are divisible by 5, they have a common factor of 5. This contradicts our initial assumption that \( x \) and \( y \) are co-prime.
Thus, our assumption that \( \sqrt{5} \) is rational is incorrect.
Hence, \( \sqrt{5} \) is an irrational number.
― ― ― ― solutions by ncertbooks.net ― ― ― ―
2. Prove that \( 3 + 2\sqrt{5} \) is irrational.
Solution:
We will prove \( 3 + 2\sqrt{5} \) is irrational by contradiction. Let us assume that \( 3 + 2\sqrt{5} \) is a rational number.
This means we can write:
\( 3 + 2\sqrt{5} = \dfrac{x}{y} \), where \( x \) and \( y \) are co-prime integers (\( y \neq 0 \)).
Rearranging the equation, we get:
\( 2\sqrt{5} = \dfrac{x}{y} – 3 \)
Simplify the right-hand side:
\( 2\sqrt{5} = \dfrac{x – 3y}{y} \)
Divide through by 2:
\( \sqrt{5} = \dfrac{x – 3y}{2y} \)
Since \( x \), \( y \), and 2 are integers, \( \dfrac{x – 3y}{2y} \) is a rational number. This implies \( \sqrt{5} \) is rational.
However, this contradicts the fact that \( \sqrt{5} \) is irrational.
Therefore, our assumption that \( 3 + 2\sqrt{5} \) is rational is incorrect.
Hence, \( 3 + 2\sqrt{5} \) is irrational.
― ― ― ― solutions by ncertbooks.net ― ― ― ―
3. Prove that the following are irrationals:
(i) \( \dfrac{1}{\sqrt{2}} \)
(ii) \( 7\sqrt{5} \)
(iii) \( 6 + \sqrt{2} \)
Solution:
(i) \( \dfrac{1}{\sqrt{2}} \)
Let us assume \( \dfrac{1}{\sqrt{2}} \) is rational.
Then we can write:
\( \dfrac{1}{\sqrt{2}} = \dfrac{x}{y} \), where \( x \) and \( y \) are co-primes, \( y \neq 0 \).
Rearranging, we get:
\( \sqrt{2} = \dfrac{y}{x} \)
Since \( x \) and \( y \) are integers, \( \sqrt{2} \) is rational. This contradicts the fact that \( \sqrt{2} \) is irrational.
Hence, \( \dfrac{1}{\sqrt{2}} \) is irrational.
(ii) \( 7\sqrt{5} \)
Let us assume \( 7\sqrt{5} \) is rational.
Then we can write:
\( 7\sqrt{5} = \dfrac{x}{y} \), where \( x \) and \( y \) are co-primes, \( y \neq 0 \).
Rearranging, we get:
\( \sqrt{5} = \dfrac{x}{7y} \)
Since \( x \) and \( y \) are integers, \( \sqrt{5} \) is rational. This contradicts the fact that \( \sqrt{5} \) is irrational.
Hence, \( 7\sqrt{5} \) is irrational.
(iii) \( 6 + \sqrt{2} \)
Let us assume \( 6 + \sqrt{2} \) is rational.
Then we can write:
\( 6 + \sqrt{2} = \dfrac{x}{y} \), where \( x \) and \( y \) are co-primes, \( y \neq 0 \).
Rearranging, we get:
\( \sqrt{2} = \dfrac{x}{y} – 6 \)
Since \( x \) and \( y \) are integers, \( \dfrac{x}{y} – 6 \) is rational. This implies \( \sqrt{2} \) is rational, which contradicts the fact that \( \sqrt{2} \) is irrational.
Hence, \( 6 + \sqrt{2} \) is irrational.
― ― ― ― solutions by ncertbooks.net ― ― ― ―
Exercise 1.4
1. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:
(i) \( \dfrac{13}{3125} \) (ii) \( \dfrac{17}{8} \) (iii) \( \dfrac{64}{455} \) (iv) \( \dfrac{15}{1600} \) (v) \( \dfrac{29}{343} \) (vi) \( \dfrac{23}{2^{3}5^{2}} \) (vii) \( \dfrac{129}{2^{2}5^{7}7^{5}} \) (viii) \( \dfrac{6}{15} \) (ix) \( \dfrac{35}{50} \) (x) \( \dfrac{77}{210} \)
Solution:
Note: A rational number has a terminating decimal expansion if the denominator, after prime factorization, is in the form \( 2^m \times 5^n \). Otherwise, it has a non-terminating repeating decimal expansion.
(i) \( \dfrac{13}{3125} \)
Factorizing the denominator:
\( 3125 = 5 \times 5 \times 5 \times 5 \times 5 = 5^5 \)
The denominator has only \( 5 \) as its factor. Therefore, \( \dfrac{13}{3125} \) has a terminating decimal expansion.
(ii) \( \dfrac{17}{8} \)
Factorizing the denominator:
\( 8 = 2 \times 2 \times 2 = 2^3 \)
The denominator has only \( 2 \) as its factor. Therefore, \( \dfrac{17}{8} \) has a terminating decimal expansion.
(iii) \( \dfrac{64}{455} \)
Factorizing the denominator:
\( 455 = 5 \times 7 \times 13 \)
Since the denominator contains prime factors other than \( 2 \) and \( 5 \), \( \dfrac{64}{455} \) has a non-terminating repeating decimal expansion.
(iv) \( \dfrac{15}{1600} \)
Factorizing the denominator:
\( 1600 = 2^6 \times 5^2 \)
The denominator is in the form \( 2^m \times 5^n \). Therefore, \( \dfrac{15}{1600} \) has a terminating decimal expansion.
(v) \( \dfrac{29}{343} \)
Factorizing the denominator:
\( 343 = 7 \times 7 \times 7 = 7^3 \)
Since the denominator contains a factor other than \( 2 \) and \( 5 \), \( \dfrac{29}{343} \) has a non-terminating repeating decimal expansion.
(vi) \( \dfrac{23}{2^{3}5^{2}} \)
Factorizing the denominator:
Clearly, the denominator is in the form \( 2^m \times 5^n \).
Therefore, \( \dfrac{23}{2^{3}5^{2}} \) has a terminating decimal expansion.
(vii) \( \dfrac{129}{2^{2}5^{7}7^{5}}\)
Since the denominator is not in the form \( 2^m \times 5^n \), \( \dfrac{129}{2^{2}5^{7}7^{5}} \) has a non-terminating repeating decimal expansion.
(viii) \( \dfrac{6}{15} \)
Reducing the fraction:
\( \dfrac{6}{15} = \dfrac{2}{5} \)
Since the denominator has only \( 5 \) as its factor, \( \dfrac{6}{15} \) has a terminating decimal expansion.
(ix) \( \dfrac{35}{50} \)
Reducing the fraction:
\( \dfrac{35}{50} = \dfrac{7}{10} \)
Factorizing the denominator:
\( 10 = 2 \times 5 \)
The denominator is in the form \( 2^m \times 5^n \). Therefore, \( \dfrac{35}{50} \) has a terminating decimal expansion.
(x) \( \dfrac{77}{210} \)
Reducing the fraction:
\( \dfrac{77}{210} = \dfrac{11}{30} \)
Factorizing the denominator:
\( 30 = 2 \times 3 \times 5 \)
Since the denominator contains a factor other than \( 2 \) and \( 5 \), \( \dfrac{77}{210} \) has a non-terminating repeating decimal expansion.
― ― ― ― solutions by ncertbooks.net ― ― ― ―
2. Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.
Solution:
Note: A rational number has a terminating decimal expansion if the denominator, after prime factorization, is in the form \( 2^m \times 5^n \). Otherwise, it has a non-terminating repeating decimal expansion.
(i) \( \dfrac{13}{3125} \)
The long division for \( \dfrac{13}{3125} \):
0.00416
----------------
3125)13.00000
0
--------
13000
-12500
--------
5000
-3125
--------
18750
-18750
--------
0
The decimal expansion is ( 0.00416 ), which is terminating.
(ii) \( \dfrac{17}{8} \)
The long division for \( \frac{17}{8} \):
2.125
----------------
8)17.000
16
-----
10
8
-----
20
16
-----
40
40
-----
0
The decimal expansion is \( 2.125 \), which is terminating.
(iii) \( \dfrac{64}{455} \)
Factorizing the denominator:
455 = 5 × 7 × 13
Since the denominator contains prime factors other than \( 2 \) and \( 5 \), \( \dfrac{64}{455} \) has a non-terminating repeating decimal expansion.
(iv) \( \dfrac{15}{1600} \)
The long division for \( \frac{15}{1600} \):
0.009375
-----------------
1600)15.000000
0
---------
15000
-14400
---------
6000
-5600
---------
4000
-4000
---------
0The decimal expansion is \( 0.009375 \), which is terminating.
(v) \( \dfrac{29}{343} \)
Factorizing the denominator:
343 = 7 × 7 × 7 = 7³
Since the denominator contains a factor other than \( 2 \) and \( 5 \), \( \dfrac{29}{343} \) has a non-terminating repeating decimal expansion.
(vi) \( \dfrac{23}{2^3 \times 5^2} \)
Factorizing the denominator:
2³ × 5² = 8 × 25 = 200
Since the denominator is in the form \( 2^m \times 5^n \), \( \dfrac{23}{200} \) has a terminating decimal expansion.
(vii) \( \dfrac{129}{2^2 \times 5^7 \times 7^5} \)
Factorizing the denominator:
2² × 5⁷ × 7⁵ = Contains factor \( 7 \), not in \( 2^m \times 5^n \)
Since the denominator contains a factor \( 7 \), \( \dfrac{129}{2^2 \times 5^7 \times 7^5} \) has a non-terminating repeating decimal expansion.
(viii) \( \dfrac{6}{15} \)
Simplifying the fraction:
6/15 = 2/5
Since the denominator \( 5 \) is in \( 2^m \times 5^n \), \( \dfrac{6}{15} \) has a terminating decimal expansion.
(ix) \( \dfrac{35}{50} \)
Simplifying the fraction:
35/50 = 7/10
Since the denominator \( 10 \) is in \( 2^m \times 5^n \), \( \dfrac{35}{50} \) has a terminating decimal expansion.
(x) \( \dfrac{77}{210} \)
Simplifying the fraction:
77/210 = 11/30
Factorizing the denominator:
30 = 2 × 3 × 5
Since the denominator contains \( 3 \), \( \dfrac{77}{210} \) has a non-terminating repeating decimal expansion.
― ― ― ― solutions by ncertbooks.net ― ― ― ―
3. The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form \( \frac{p}{q} \), what can you say about the prime factors of \( q \)?
Solution:
(i) \( 43.123456789 \)
This number has a terminating decimal expansion. A number with a terminating decimal expansion is always a rational number and can be expressed in the form \( \frac{p}{q} \), where:
– \( p \) and \( q \) are integers.- \( q \) has prime factors of only \( 2 \) and/or \( 5 \).
Therefore, \( 43.123456789 \) is a rational number, and \( q \) has prime factors of \( 2 \) and \( 5 \) only.
(ii) \( 0.120120012000120000\ldots \)
This number has a non-terminating and non-repeating decimal expansion. A number with a non-terminating, non-repeating decimal expansion is always an irrational number.
Therefore, \( 0.120120012000120000\ldots \) is an irrational number.
(iii) Non-terminating but repeating decimal expansion
A number with a non-terminating but repeating decimal expansion is always a rational number and can be expressed in the form \( \frac{p}{q} \), where:
– \( p \) and \( q \) are integers.- \( q \) has prime factors other than \( 2 \) and \( 5 \).
Therefore, a number with a non-terminating but repeating decimal expansion is a rational number, and \( q \) has prime factors other than \( 2 \) and \( 5 \).
― ― ― ― solutions by ncertbooks.net ― ― ― ―