NCERT Solutions for Class 10 Mathematics Chapter 1: Real Numbers (Complete Guide)
Number systems form the foundation of senior school mathematics, providing the theoretical underpinnings for algebra, functional calculus, and modern analytic algorithms. In the CBSE Class 10 syllabus, Chapter 1: Real Numbers establishes the rigorous structural properties of the set of positive integers ($\mathbb{Z}^+$). Transitioning from elementary arithmetic to formal algebraic proofs, this chapter focuses on two principal mathematical concepts: The Fundamental Theorem of Arithmetic and the analytical verification of irrationality using proof by contradiction.
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+---------------------------------------------------------------------------------------+
| REAL NUMBERS SYSTEM |
| |
| +---------------------------------------------------+ |
| | REAL NUMBERS (R) | |
| +-------------------------+-------------------------+ |
| | |
| +-------------------------+-------------------------+ |
| | | |
| +--------------v---------------+ +---------------v---------------+ |
| | RATIONAL NUMBERS (Q) | | IRRATIONAL NUMBERS (Q') | |
| | Form: p/q (q != 0, p,q e Z)| | Non-terminating, | |
| +--------------+---------------+ | non-recurring decimals | |
| | | Examples: v2, v3, v5, p | |
| +--------------v---------------+ +-------------------------------+ |
| | INTEGERS (Z) | |
| | {..., -3, -2, -1, 0, ...} | |
| +--------------+---------------+ |
| | |
| +--------------v---------------+ |
| | WHOLE NUMBERS (W) | |
| | {0, 1, 2, 3, 4, ...} | |
| +--------------+---------------+ |
| | |
| +--------------v---------------+ |
| | NATURAL NUMBERS (N) | |
| | {1, 2, 3, 4, ...} | |
| +------------------------------+ |
+---------------------------------------------------------------------------------------+
The Fundamental Theorem of Arithmetic (also known as the Unique Factorization Theorem) asserts that every composite integer greater than $1$ can be uniquely factored into a product of prime numbers, up to the permutation of factors. This theorem provides an efficient method for determining the Highest Common Factor ($\text{HCF}$) and Least Common Multiple ($\text{LCM}$) of sets of integers. For any two positive integers $a$ and $b$, this yields the fundamental relation:
$$\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$$
Note that this identity applies strictly to pairs of integers and cannot be generalized directly to three or more numbers without modification.
The second major concept is the analytical study of irrational numbers ($\mathbb{R} \setminus \mathbb{Q}$). Through formal proofs by contradiction, students verify the irrationality of radical expressions ($\sqrt{p}$, where $p$ is prime) along with their linear combinations with rational quantities ($a \pm b\sqrt{p}$ and $a\sqrt{p}$). This requires working with coprime integers, examining quadratic divisibility ($p \mid a^2 \implies p \mid a$), and using modular arithmetic properties.
According to the official CBSE assessment scheme, solving number-theoretic proofs requires clear and rigorous steps. Every proof by contradiction must state the initial assumption of rationality, define variables as coprime integers, show divisibility on both sides of the equation, identify the resulting algebraic contradiction, and state the final non-rationality conclusion. Missing statements—such as omitting $\text{HCF}(a, b) = 1$—will result in point deductions under CBSE board evaluation criteria.
| Mathematical Concept | Standard Formula / Definition | Essential CBSE Constraints & Conditions | Analytical Importance |
| Fundamental Theorem of Arithmetic | $x = p_1^{k_1} p_2^{k_2} \cdots p_n^{k_n}$ | $p_1 < p_2 < \dots < p_n$ are distinct primes, $k_i \in \mathbb{N}$ | Guarantees unique prime factorization for every composite number. |
| HCF (Prime Factorisation) | $\prod p_i^{\min(a_i, b_i)}$ | Common prime factors only | Largest positive integer dividing given inputs without remainder. |
| LCM (Prime Factorisation) | $\prod p_i^{\max(a_i, b_i)}$ | All prime factors involved | Smallest positive integer divisible by all given inputs. |
| Two-Number Identity | $\text{HCF}(a,b) \times \text{LCM}(a,b) = a \cdot b$ | Valid only for $n = 2$ numbers; fails for $n \ge 3$ | Solves for an unknown term given three known parameters. |
| Prime Divisibility Property | If $p \mid a^2$, then $p \mid a$ | $p$ must be a prime number; $a \in \mathbb{Z}^+$ | Core theorem used to prove the irrationality of radical numbers. |
| Irrational Linear Combinations | $q \pm r\sqrt{p}$ or $q \cdot \sqrt{p}$ | $q \in \mathbb{Q}, q \neq 0$, $p \in \mathbb{P}$ (prime) | Shows that algebraic operations on rationals and irrationals yield irrationals. |
The Fundamental Theorem of Arithmetic – Examples & Core Problems (Page No. 2-6)
Example 1. Consider the numbers $4^n$, where $n$ is a natural number. Check whether there is any value of $n$ for which $4^n$ ends with the digit zero. (Page No. 4) [BOARD EXAM FAVORITE / CBSE 2019, 2023]
Answer:
A positive integer ends with the digit zero if and only if it is divisible by $10$. This means its prime factorisation must contain both $2$ and $5$ as prime factors (since $10 = 2 \times 5$).
We are given the expression:
$$4^n = (2^2)^n = 2^{2n}$$
The only prime factor in the prime factorisation of $4^n$ is $2$.
According to the uniqueness part of the Fundamental Theorem of Arithmetic, there are no other prime factors in the factorisation of $4^n$. Since the prime factor $5$ is absent, $4^n$ cannot be divisible by $5$ for any natural number $n$.
Therefore, there is no natural number $n$ for which $4^n$ ends with the digit zero.
Example 2. Find the LCM and HCF of $6$ and $20$ by the prime factorisation method. (Page No. 4)
Answer: To find the $\text{HCF}$ and $\text{LCM}$, we first write the prime factorisations of both numbers:
$$\begin{aligned} 6 &= 2^1 \times 3^1 \\ 20 &= 2 \times 2 \times 5 = 2^2 \times 5^1 \end{aligned}$$
Finding the $\text{HCF}$ and $\text{LCM}$:
- $\text{HCF}(6, 20) =$ Product of the smallest power of each common prime factor involved in the numbers: $$\text{HCF}(6, 20) = 2^1 = 2$$
- $\text{LCM}(6, 20) =$ Product of the greatest power of each prime factor involved in the numbers: $$\text{LCM}(6, 20) = 2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60$$
Verification:
$$\begin{aligned} \text{HCF}(6, 20) \times \text{LCM}(6, 20) &= 2 \times 60 = 120 \\ \text{Product of numbers } (6 \times 20) &= 120 \end{aligned}$$
Thus, $\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$ is verified.
Example 3. Find the HCF of $96$ and $404$ by the prime factorisation method. Hence, find their LCM. (Page No. 4)
Answer: We write the prime factorisation of each integer:
$$\begin{aligned} 96 &= 2 \times 48 = 2 \times 2 \times 24 = 2^5 \times 3^1 \\ 404 &= 2 \times 202 = 2^2 \times 101^1 \end{aligned}$$
The common prime factor is $2$, and its smallest exponent is $2$.
$$\text{HCF}(96, 404) = 2^2 = 4$$
Using the relation between the $\text{HCF}$ and $\text{LCM}$ of two numbers:
$$\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b \implies \text{LCM}(a, b) = \frac{a \times b}{\text{HCF}(a, b)}$$
Substituting the values:
$$\begin{aligned} \text{LCM}(96, 404) &= \frac{96 \times 404}{4} \\ &= 96 \times 101 \\ &= 9696 \end{aligned}$$
Therefore, $\text{HCF}(96, 404) = 4$ and $\text{LCM}(96, 404) = 9696$.
Example 4. Find the HCF and LCM of $6$, $72$ and $120$, using the prime factorisation method. (Page No. 5)
Answer: We begin by expressing each number as a product of its prime factors:
$$\begin{aligned} 6 &= 2^1 \times 3^1 \\ 72 &= 8 \times 9 = 2^3 \times 3^2 \\ 120 &= 8 \times 15 = 2^3 \times 3^1 \times 5^1 \end{aligned}$$
Step 1: Finding HCF Identify the common prime factors in all three numbers: $2$ and $3$. Take the smallest power of each common factor:
$$\text{HCF}(6, 72, 120) = 2^1 \times 3^1 = 6$$
Step 2: Finding LCM Take the highest power of every prime factor present ($2$, $3$, and $5$):
$$\begin{aligned} \text{LCM}(6, 72, 120) &= 2^3 \times 3^2 \times 5^1 \\ &= 8 \times 9 \times 5 \\ &= 360 \end{aligned}$$
Important Analytical Note on Three Numbers: The product formula $\text{HCF}(a, b, c) \times \text{LCM}(a, b, c) \neq a \times b \times c$.
$$\begin{aligned} \text{HCF} \times \text{LCM} &= 6 \times 360 = 2160 \\ a \times b \times c &= 6 \times 72 \times 120 = 51840 \\ 2160 &\neq 51840 \end{aligned}$$
This confirms that the two-number product identity does not extend directly to three numbers.
Revisiting Irrational Numbers – Examples & Proofs (Page No. 6-9)
Theorem 1.2. Let $p$ be a prime number. If $p$ divides $a^2$, then $p$ divides $a$, where $a$ is a positive integer. (Page No. 6)
Proof:
Let the prime factorisation of the positive integer $a$ be given by:
$$a = p_1 p_2 \cdots p_n$$
where $p_1, p_2, \dots, p_n$ are prime numbers, not necessarily distinct.
Squaring both sides of the equation:
$$a^2 = (p_1 p_2 \cdots p_n)(p_1 p_2 \cdots p_n) = p_1^2 p_2^2 \cdots p_n^2$$
We are given that the prime number $p$ divides $a^2$. According to the Fundamental Theorem of Arithmetic, the prime factorisation of $a^2$ is unique. Therefore, the only prime factors of $a^2$ are $p_1, p_2, \dots, p_n$.
Since $p$ is a prime factor of $a^2$, $p$ must be one of the primes $p_1, p_2, \dots, p_n$.
Because $a = p_1 p_2 \cdots p_n$, it follows directly that $p$ divides $a$.
Theorem 1.3 / Example. Prove that $\sqrt{2}$ is irrational. (Page No. 6) [BOARD EXAM FAVORITE / CBSE 2020, 2022]
Proof: We prove this theorem using the method of proof by contradiction.
Step 1: Assumption Assume, to the contrary, that $\sqrt{2}$ is a rational number. Then there exist coprime positive integers $a$ and $b$ ($b \neq 0$) such that:
$$\sqrt{2} = \frac{a}{b} \quad \text{where } \text{HCF}(a, b) = 1$$
Step 2: Squaring and algebraic rearrangement
Rearranging the equation gives $a = b\sqrt{2}$. Squaring both sides:
$$a^2 = 2b^2 \quad \text{— (Equation 1)}$$
This equation shows that $2$ divides $2b^2$, which means $2 \mid a^2$. By Theorem 1.2, since $2$ is prime and divides $a^2$, $2$ must also divide $a$.
Step 3: Substituting a second variable
Since $2$ divides $a$, we can express $a$ as:
$$a = 2c \quad \text{for some integer } c$$
Substitute this expression for $a$ back into Equation 1:
$$\begin{aligned} (2c)^2 &= 2b^2 \\ 4c^2 &= 2b^2 \\ 2c^2 &= b^2 \implies b^2 = 2c^2 \end{aligned}$$
This shows that $2$ divides $b^2$. Applying Theorem 1.2 again, $2$ must also divide $b$.
Step 4: Deducing the contradiction From Steps 2 and 3, $2$ divides both $a$ and $b$. This means $a$ and $b$ share a common factor of at least $2$.
However, this contradicts our initial statement that $a$ and $b$ are coprime ($\text{HCF}(a, b) = 1$).
This contradiction arose because we assumed that $\sqrt{2}$ is rational.
Therefore, we conclude that $\sqrt{2}$ is irrational.
Example 5. Prove that $\sqrt{3}$ is irrational. (Page No. 7) [BOARD EXAM FAVORITE / CBSE 2018, 2023]
Proof: We use a proof by contradiction.
Assume, to the contrary, that $\sqrt{3}$ is rational. Then there exist coprime integers $a$ and $b$ ($b \neq 0$) such that:
$$\sqrt{3} = \frac{a}{b} \quad \text{where } \text{HCF}(a, b) = 1$$
Rearranging and squaring both sides:
$$\begin{aligned} a &= b\sqrt{3} \\ a^2 &= 3b^2 \quad \text{— (Equation 1)} \end{aligned}$$
From Equation 1, $3$ divides $a^2$. Since $3$ is a prime number, it follows that $3$ divides $a$.
We can therefore write $a = 3c$ for some integer $c$. Substituting this into Equation 1:
$$\begin{aligned} (3c)^2 &= 3b^2 \\ 9c^2 &= 3b^2 \\ 3c^2 &= b^2 \implies b^2 = 3c^2 \end{aligned}$$
This implies that $3$ divides $b^2$, which means $3$ divides $b$.
Thus, $3$ is a common factor of both $a$ and $b$, which contradicts the fact that $a$ and $b$ are coprime.
This contradiction shows that our initial assumption was incorrect.
Therefore, $\sqrt{3}$ is irrational.
Example 6. Show that $5 – \sqrt{3}$ is irrational. (Page No. 7) [BOARD EXAM FAVORITE / CBSE 2019, 2021]
Answer: Let us assume, to the contrary, that $5 – \sqrt{3}$ is rational.
Then we can find coprime integers $a$ and $b$ ($b \neq 0$) such that:
$$5 – \sqrt{3} = \frac{a}{b} \quad \text{where } \text{HCF}(a, b) = 1$$
Rearranging the terms to isolate the radical component $\sqrt{3}$:
$$\begin{aligned} \sqrt{3} &= 5 – \frac{a}{b} \\ \sqrt{3} &= \frac{5b – a}{b} \end{aligned}$$
Since $a$, $b$, and $5$ are integers and $b \neq 0$, the expression $\frac{5b – a}{b}$ is a rational number.
This equality implies that $\sqrt{3}$ must also be a rational number.
However, this contradicts the known fact that $\sqrt{3}$ is irrational.
This contradiction arose from our incorrect assumption that $5 – \sqrt{3}$ is rational.
Therefore, we conclude that $5 – \sqrt{3}$ is irrational.
Example 7. Show that $3\sqrt{2}$ is irrational. (Page No. 7) [BOARD EXAM FAVORITE / CBSE 2020]
Answer: Assume, to the contrary, that $3\sqrt{2}$ is rational.
Then there exist coprime integers $a$ and $b$ ($b \neq 0$) such that:
$$3\sqrt{2} = \frac{a}{b} \quad \text{where } \text{HCF}(a, b) = 1$$
Rearranging the equation to isolate $\sqrt{2}$:
$$\sqrt{2} = \frac{a}{3b}$$
Since $a$, $b$, and $3$ are integers with $b \neq 0$, the quantity $\frac{a}{3b}$ is a rational number.
This implies that $\sqrt{2}$ must also be rational.
However, this contradicts the established fact that $\sqrt{2}$ is irrational.
This contradiction confirms that our assumption was false.
Consequently, $3\sqrt{2}$ is irrational.
Chapter-End Exercises – Exercise 1.1 (Page No. 5-6)
Question 1. Express each number as a product of its prime factors: (Page No. 5) (i) $140$ (ii) $156$ (iii) $3825$ (iv) $5005$ (v) $7429$
Answer:
(i) Factorisation of $140$:
We divide $140$ successively by the smallest possible prime factors:
$$\begin{aligned} 140 \div 2 &= 70 \\ 70 \div 2 &= 35 \\ 35 \div 5 &= 7 \\ 7 \div 7 &= 1 \end{aligned}$$
Writing this in exponential form:
$$140 = 2 \times 2 \times 5 \times 7 = \underline{2^2 \times 5 \times 7}$$
(ii) Factorisation of $156$:
Using successive prime division:
$$\begin{aligned} 156 \div 2 &= 78 \\ 78 \div 2 &= 39 \\ 39 \div 3 &= 13 \\ 13 \div 13 &= 1 \end{aligned}$$
In exponential prime notation:
$$156 = 2 \times 2 \times 3 \times 13 = \underline{2^2 \times 3 \times 13}$$
(iii) Factorisation of $3825$:
The sum of the digits ($3+8+2+5=18$) is divisible by $3$, so we begin with $3$:
$$\begin{aligned} 3825 \div 3 &= 1275 \\ 1275 \div 3 &= 425 \\ 425 \div 5 &= 85 \\ 85 \div 5 &= 17 \\ 17 \div 17 &= 1 \end{aligned}$$
Expressing as a product of prime powers:
$$3825 = 3 \times 3 \times 5 \times 5 \times 17 = \underline{3^2 \times 5^2 \times 17}$$
(iv) Factorisation of $5005$:
The number ends in $5$, so we divide by $5$:
$$\begin{aligned} 5005 \div 5 &= 1001 \\ 1001 \div 7 &= 143 \\ 143 \div 11 &= 13 \\ 13 \div 13 &= 1 \end{aligned}$$
Expressing in prime factor form:
$$5005 = \underline{5 \times 7 \times 11 \times 13}$$
(v) Factorisation of $7429$:
Testing primes systematically reveals that the smallest prime divisor is $17$:
$$\begin{aligned} 7429 \div 17 &= 437 \\ 437 \div 19 &= 23 \\ 23 \div 23 &= 1 \end{aligned}$$
Expressing as a product of primes:
$$7429 = \underline{17 \times 19 \times 23}$$
Question 2. Find the LCM and HCF of the following pairs of integers and verify that $\text{LCM} \times \text{HCF} = \text{product of the two numbers}$. (Page No. 5) (i) $26$ and $91$ (ii) $510$ and $92$ (iii) $336$ and $54$
Answer:
(i) For $26$ and $91$:
Find the prime factorisations:
$$\begin{aligned} 26 &= 2^1 \times 13^1 \\ 91 &= 7^1 \times 13^1 \end{aligned}$$
- $\text{HCF}(26, 91) = 13^1 = 13$
- $\text{LCM}(26, 91) = 2^1 \times 7^1 \times 13^1 = 14 \times 13 = 182$
Verification:
$$\begin{aligned} \text{HCF} \times \text{LCM} &= 13 \times 182 = 2366 \\ \text{Product of Numbers} &= 26 \times 91 = 2366 \end{aligned}$$
Since $\text{HCF} \times \text{LCM} = \text{Product of Numbers} = 2366$, the identity is verified.
(ii) For $510$ and $92$:
Find the prime factorisations:
$$\begin{aligned} 510 &= 2 \times 3 \times 5 \times 17 = 2^1 \times 3^1 \times 5^1 \times 17^1 \\ 92 &= 2 \times 2 \times 23 = 2^2 \times 23^1 \end{aligned}$$
- $\text{HCF}(510, 92) = 2^1 = 2$
- $\text{LCM}(510, 92) = 2^2 \times 3^1 \times 5^1 \times 17^1 \times 23^1 = 4 \times 3 \times 5 \times 17 \times 23 = 23460$
Verification:
$$\begin{aligned} \text{HCF} \times \text{LCM} &= 2 \times 23460 = 46920 \\ \text{Product of Numbers} &= 510 \times 92 = 46920 \end{aligned}$$
Since $\text{HCF} \times \text{LCM} = \text{Product of Numbers} = 46920$, the identity is verified.
(iii) For $336$ and $54$:
Find the prime factorisations:
$$\begin{aligned} 336 &= 2^4 \times 3^1 \times 7^1 \\ 54 &= 2^1 \times 3^3 \end{aligned}$$
- $\text{HCF}(336, 54) = 2^1 \times 3^1 = 6$
- $\text{LCM}(336, 54) = 2^4 \times 3^3 \times 7^1 = 16 \times 27 \times 7 = 3024$
Verification:
$$\begin{aligned} \text{HCF} \times \text{LCM} &= 6 \times 3024 = 18144 \\ \text{Product of Numbers} &= 336 \times 54 = 18144 \end{aligned}$$
Since $\text{HCF} \times \text{LCM} = \text{Product of Numbers} = 18144$, the identity is verified.
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 $25$
Answer:
(i) For $12$, $15$, and $21$:
Find the prime factorisations:
$$\begin{aligned} 12 &= 2^2 \times 3^1 \\ 15 &= 3^1 \times 5^1 \\ 21 &= 3^1 \times 7^1 \end{aligned}$$
- $\text{HCF}(12, 15, 21) = 3^1 = \underline{3}$
- $\text{LCM}(12, 15, 21) = 2^2 \times 3^1 \times 5^1 \times 7^1 = 4 \times 3 \times 5 \times 7 = \underline{420}$
(ii) For $17$, $23$, and $29$:
Each of these numbers is prime:
$$\begin{aligned} 17 &= 17^1 \\ 23 &= 23^1 \\ 29 &= 29^1 \end{aligned}$$
- $\text{HCF}(17, 23, 29) = \underline{1}$ (no common prime factors)
- $\text{LCM}(17, 23, 29) = 17 \times 23 \times 29 = \underline{11339}$
(iii) For $8$, $9$, and $25$:
Find the prime factorisations:
$$\begin{aligned} 8 &= 2^3 \\ 9 &= 3^2 \\ 25 &= 5^2 \end{aligned}$$
- $\text{HCF}(8, 9, 25) = \underline{1}$ (the numbers are pairwise coprime)
- $\text{LCM}(8, 9, 25) = 2^3 \times 3^2 \times 5^2 = 8 \times 9 \times 25 = \underline{1800}$
Question 4. Given that $\text{HCF}(306, 657) = 9$, find $\text{LCM}(306, 657)$. [BOARD EXAM FAVORITE / CBSE 2021, 2023]
Answer:
Given Data:
- First integer, $a = 306$
- Second integer, $b = 657$
- $\text{HCF}(a, b) = 9$
Formula Required:
$$\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b \implies \text{LCM}(a, b) = \frac{a \times b}{\text{HCF}(a, b)}$$
Step-by-Step Calculation:
$$\begin{aligned} \text{LCM}(306, 657) &= \frac{306 \times 657}{9} \\ &= \frac{306}{9} \times 657 \\ &= 34 \times 657 \\ &= 22338 \end{aligned}$$
Final Statement:
The $\text{LCM}(306, 657)$ is $22338$.
Question 5. Check whether $6^n$ can end with the digit $0$ for any natural number $n$. [BOARD EXAM FAVORITE / CBSE 2018, 2020, 2022]
Answer:
A number ends with the digit $0$ if and only if it is divisible by $10$, which means its prime factorisation must contain both $2$ and $5$.
Consider the prime factorisation of $6^n$:
$$6^n = (2 \times 3)^n = 2^n \times 3^n$$
The prime factors of $6^n$ are only $2$ and $3$.
By the uniqueness of the Fundamental Theorem of Arithmetic, no other prime factors can appear in the prime factorisation of $6^n$. Since the factor $5$ is absent, $6^n$ is not divisible by $5$ for any natural number $n$.
Therefore, $6^n$ cannot end with the digit $0$ for any natural number $n$.
Question 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. [BOARD EXAM FAVORITE / CBSE 2019, 2023]
Answer:
A composite number is a positive integer greater than $1$ that has at least one positive divisor other than $1$ and itself.
Part 1: Analyzing $N_1 = 7 \times 11 \times 13 + 13$
Factor out the common term $13$:
$$\begin{aligned} N_1 &= 13 \times (7 \times 11 + 1) \\ &= 13 \times (77 + 1) \\ &= 13 \times 78 \\ &= 13 \times (2 \times 3 \times 13) \\ &= 2 \times 3 \times 13^2 \end{aligned}$$
Since $N_1$ can be factored into a product of multiple prime numbers ($2$, $3$, and $13$), it has factors other than $1$ and itself. Therefore, $N_1$ is a composite number.
Part 2: Analyzing $N_2 = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$
Factor out the common term $5$:
$$\begin{aligned} N_2 &= 5 \times (7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1) \\ &= 5 \times (1008 + 1) \\ &= 5 \times 1009 \end{aligned}$$
$1009$ is a prime number. Since $N_2$ is expressed as the product of two primes ($5$ and $1009$), it has divisors other than $1$ and itself. Therefore, $N_2$ 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? [BOARD EXAM FAVORITE / CBSE 2019, 2022]
Answer:
Let the total time taken to meet again at the starting point be $T$ minutes.
Sonia returns to the starting point at multiples of $18$ minutes: $18, 36, 54, \dots$
Ravi returns to the starting point at multiples of $12$ minutes: $12, 24, 36, 48, \dots$
They will meet at the starting point at a time that is a common multiple of both $18$ and $12$. To find the first time they meet again, we compute the Least Common Multiple ($\text{LCM}$) of $18$ and $12$.
Prime Factorisations:
$$\begin{aligned} 18 &= 2 \times 3^2 \\ 12 &= 2^2 \times 3 \end{aligned}$$
Calculating the LCM:
$$\begin{aligned} \text{LCM}(18, 12) &= 2^{\max(1, 2)} \times 3^{\max(2, 1)} \\ &= 2^2 \times 3^2 \\ &= 4 \times 9 \\ &= 36 \end{aligned}$$
Therefore, Sonia and Ravi will meet again at the starting point after $36$ minutes (during which Sonia completes $2$ rounds and Ravi completes $3$ rounds).
Chapter-End Exercises – Exercise 1.2 (Page No. 9)
Question 1. Prove that $\sqrt{5}$ is irrational. (Page No. 9) [BOARD EXAM FAVORITE / CBSE 2019, 2020, 2023]
Answer: We prove this by contradiction.
Step 1: Hypothesis
Assume, to the contrary, that $\sqrt{5}$ is rational.
Then there exist coprime positive integers $a$ and $b$ ($b \neq 0$) such that:
$$\sqrt{5} = \frac{a}{b} \quad \text{where } \text{HCF}(a, b) = 1$$
Step 2: Squaring and analyzing divisibility
Rearranging gives $a = b\sqrt{5}$. Squaring both sides:
$$a^2 = 5b^2 \quad \text{— (Equation 1)}$$
This shows that $5$ divides $a^2$.
Since $5$ is prime, by Theorem 1.2, $5$ divides $a$.
Step 3: Setting the second substitution
Since $5$ divides $a$, let $a = 5c$ for some integer $c$.
Substituting this into Equation 1:
$$\begin{aligned} (5c)^2 &= 5b^2 \\ 25c^2 &= 5b^2 \\ 5c^2 &= b^2 \implies b^2 = 5c^2 \end{aligned}$$
This shows that $5$ divides $b^2$.
By Theorem 1.2, $5$ divides $b$.
Step 4: Contradiction and Conclusion
From Steps 2 and 3, $5$ is a common factor of both $a$ and $b$.
This contradicts our initial assumption that $a$ and $b$ are coprime ($\text{HCF}(a, b) = 1$).
This contradiction shows that our assumption that $\sqrt{5}$ is rational was false.
Therefore, $\sqrt{5}$ is irrational.
Question 2. Prove that $3 + 2\sqrt{5}$ is irrational. (Page No. 9) [BOARD EXAM FAVORITE / CBSE 2020, 2022]
Answer:
Assume, to the contrary, that $3 + 2\sqrt{5}$ is rational.
Then there exist coprime integers $a$ and $b$ ($b \neq 0$) such that:
$$3 + 2\sqrt{5} = \frac{a}{b} \quad \text{where } \text{HCF}(a, b) = 1$$
Rearrange the equation to isolate the radical $\sqrt{5}$:
$$\begin{aligned} 2\sqrt{5} &= \frac{a}{b} – 3 \\ 2\sqrt{5} &= \frac{a – 3b}{b} \\ \sqrt{5} &= \frac{a – 3b}{2b} \end{aligned}$$
Since $a$ and $b$ are integers and $b \neq 0$, the expression $\frac{a – 3b}{2b}$ is a rational number.
This implies that $\sqrt{5}$ must be rational.
However, this contradicts the fact that $\sqrt{5}$ is irrational.
This contradiction arises because of our incorrect assumption that $3 + 2\sqrt{5}$ is rational.
Therefore, $3 + 2\sqrt{5}$ is irrational.
Question 3. Prove that the following are irrationals: (Page No. 9) (i) $\frac{1}{\sqrt{2}}$ (ii) $7\sqrt{5}$ (iii) $6 + \sqrt{2}$
Answer:
(i) Proof for $\frac{1}{\sqrt{2}}$:
Assume, to the contrary, that $\frac{1}{\sqrt{2}}$ is rational.
Then there exist coprime integers $a$ and $b$ ($a \neq 0, b \neq 0$) such that:
$$\frac{1}{\sqrt{2}} = \frac{a}{b} \quad \text{where } \text{HCF}(a, b) = 1$$
Inverting both sides:
$$\sqrt{2} = \frac{b}{a}$$
Since $a$ and $b$ are integers with $a \neq 0$, $\frac{b}{a}$ is rational.
This implies that $\sqrt{2}$ must be rational, which contradicts the fact that $\sqrt{2}$ is irrational.
Therefore, $\frac{1}{\sqrt{2}}$ is irrational.
(ii) Proof for $7\sqrt{5}$:
Assume, to the contrary, that $7\sqrt{5}$ is rational.
Then there exist coprime integers $a$ and $b$ ($b \neq 0$) such that:
$$7\sqrt{5} = \frac{a}{b} \quad \text{where } \text{HCF}(a, b) = 1$$
Rearranging the equation to isolate $\sqrt{5}$:
$$\sqrt{5} = \frac{a}{7b}$$
Since $a$, $b$, and $7$ are integers ($b \neq 0$), the right-hand side $\frac{a}{7b}$ is a rational number.
This implies that $\sqrt{5}$ is rational, which contradicts the fact that $\sqrt{5}$ is irrational.
Therefore, $7\sqrt{5}$ is irrational.
(iii) Proof for $6 + \sqrt{2}$:
Assume, to the contrary, that $6 + \sqrt{2}$ is rational.
Then there exist coprime integers $a$ and $b$ ($b \neq 0$) such that:
$$6 + \sqrt{2} = \frac{a}{b} \quad \text{where } \text{HCF}(a, b) = 1$$
Rearranging the equation to isolate $\sqrt{2}$:
$$\sqrt{2} = \frac{a}{b} – 6 = \frac{a – 6b}{b}$$
Since $a$, $b$, and $6$ are integers ($b \neq 0$), the quantity $\frac{a – 6b}{b}$ is rational.
This implies that $\sqrt{2}$ is rational, which contradicts the fact that $\sqrt{2}$ is irrational.
Therefore, $6 + \sqrt{2}$ is irrational.
Mathematical Foundations & Proof Mechanics
Plaintext
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PROOF BY CONTRADICTION FLOWCHART (v p)
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[Step 1: Set Hypothesis] --> Assume vp = a/b where a, b are Coprime Integers, b != 0
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[Step 2: Clear Radicals] --> Square both sides: a^2 = p * b^2
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[Step 3: Prime Division 1] --> p divides a^2 ==> p divides a ==> Let a = p*c
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[Step 4: Prime Division 2] --> Substitute: (p*c)^2 = p*b^2 ==> b^2 = p*c^2
==> p divides b^2 ==> p divides b
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[Step 5: Identify Flaw] --> Common factor p divides BOTH a and b
Contradicts Coprime statement (HCF(a, b) = 1)
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[Step 6: State Conclusion] --> Assumption is False ==> vp is IRRATIONAL.
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15 High-Yield Board Exam FAQs
FAQ 1. Can two numbers have $18$ as their $\text{HCF}$ and $380$ as their $\text{LCM}$? Explain with reasons.
Answer:
No. The $\text{HCF}$ of two numbers must always be a divisor of their $\text{LCM}$.
Dividing $380$ by $18$:
$$\frac{380}{18} = \frac{190}{9} = 21.111\dots$$
Since $18$ does not divide $380$ completely (leaving a remainder of $2$), two numbers cannot have an $\text{HCF}$ of $18$ and an $\text{LCM}$ of $380$.
FAQ 2. If $p$ and $q$ are two distinct prime numbers, what is their $\text{HCF}$ and $\text{LCM}$?
Answer:
Because $p$ and $q$ are distinct primes, their only common positive divisor is $1$.
- $\text{HCF}(p, q) = 1$
- $\text{LCM}(p, q) = p \times q$
FAQ 3. What is the smallest composite number and the smallest prime number? Find their $\text{HCF}$.
Answer:
- Smallest prime number $= 2$
- Smallest composite number $= 4 = 2^2$
The common prime factor is $2$, with a minimum exponent of $1$:
$$\text{HCF}(2, 4) = 2^1 = \underline{2}$$
FAQ 4. If $\text{HCF}(a, b) = 12$ and $a \times b = 1800$, find $\text{LCM}(a, b)$.
Answer:
Using the two-number identity:
$$\begin{aligned} \text{LCM}(a, b) &= \frac{a \times b}{\text{HCF}(a, b)} \\ &= \frac{1800}{12} \\ &= 150 \end{aligned}$$
The $\text{LCM}(a, b) = 150$.
FAQ 5 (HOTS). Prove that $\sqrt{p} + \sqrt{q}$ is an irrational number, where $p$ and $q$ are distinct primes.
Answer:
Assume, to the contrary, that $\sqrt{p} + \sqrt{q} = x$, where $x$ is rational ($x \neq 0$).
Isolate $\sqrt{p}$ and square both sides:
$$\begin{aligned} \sqrt{p} &= x – \sqrt{q} \\ (\sqrt{p})^2 &= (x – \sqrt{q})^2 \\ p &= x^2 – 2x\sqrt{q} + q \\ 2x\sqrt{q} &= x^2 + q – p \\ \sqrt{q} &= \frac{x^2 + q – p}{2x} \end{aligned}$$
Since $x$ is rational ($x \neq 0$) and $p, q$ are integers, the right side $\frac{x^2 + q – p}{2x}$ is rational.
This implies that $\sqrt{q}$ is rational, which contradicts the fact that the square root of any prime number is irrational.
Therefore, $\sqrt{p} + \sqrt{q}$ is irrational.
FAQ 6. Write the prime factorisation of the denominator of the rational number in simplest form that decides if its decimal expansion terminates.
Answer:
A rational number $\frac{p}{q}$ in simplest form ($\text{HCF}(p, q) = 1$) has a terminating decimal expansion if and only if the prime factorisation of the denominator $q$ is of the form:
$$q = 2^m \times 5^n \quad \text{where } m, n \in \mathbb{W} \text{ (non-negative integers)}$$
If $q$ contains any prime factor other than $2$ or $5$, the expansion is non-terminating repeating.
FAQ 7. Show that any positive odd integer cannot be of the form $4q + 2$ or $4q + 0$.
Answer:
Any positive integer $n$ can be expressed in the form $4q + r$, where $r \in \{0, 1, 2, 3\}$:
- If $n = 4q = 2(2q)$, it is divisible by $2$, hence even.
- If $n = 4q + 2 = 2(2q + 1)$, it is divisible by $2$, hence even.
- If $n = 4q + 1 = 2(2q) + 1$, it is not divisible by $2$, hence odd.
- If $n = 4q + 3 = 2(2q + 1) + 1$, it is not divisible by $2$, hence odd.
Therefore, odd integers can only be written in the form $4q + 1$ or $4q + 3$, and cannot be of the form $4q$ or $4q + 2$.
FAQ 8 (Assertion-Reasoning).
- Assertion (A): The number $12^n$ cannot end with the digit $0$ for any natural number $n$.
- Reason (R): The prime factorisation of $12$ contains only the primes $2$ and $3$, and does not contain $5$.Answer:Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A). Since $12^n = (2^2 \times 3)^n = 2^{2n} \times 3^n$, the prime factor $5$ is missing, so $12^n$ cannot end in $0$.
FAQ 9 (HOTS). Find the largest number that divides $2053$ and $367$, leaving remainders of $5$ and $7$ respectively.
Answer:
Subtract the required remainders from each number:
$$\begin{aligned} 2053 – 5 &= 2048 \\ 367 – 7 &= 360 \end{aligned}$$
The required number is the $\text{HCF}(2048, 360)$.
Prime factorisations:
$$\begin{aligned} 2048 &= 2^{11} \\ 360 &= 36 \times 10 = (2^2 \times 3^2) \times (2 \times 5) = 2^3 \times 3^2 \times 5^1 \end{aligned}$$
Computing the $\text{HCF}$:
$$\text{HCF}(2048, 360) = 2^3 = 8$$
The largest number is $8$.
FAQ 10. Three bells toll together at intervals of $9, 12, 15$ minutes respectively. If they start tolling together, after what time interval will they toll together next?
Answer:
The time interval is given by the $\text{LCM}(9, 12, 15)$.
Prime factorisations:
$$\begin{aligned} 9 &= 3^2 \\ 12 &= 2^2 \times 3^1 \\ 15 &= 3^1 \times 5^1 \end{aligned}$$
Computing the $\text{LCM}$:
$$\text{LCM}(9, 12, 15) = 2^2 \times 3^2 \times 5^1 = 4 \times 9 \times 5 = 180\text{ minutes}$$
The bells will toll together next after $180\text{ minutes}$ (or $3\text{ hours}$).
FAQ 11. If $\text{HCF}(a, b) = 1$, what can you say about the integers $a$ and $b$?
Answer:
The integers $a$ and $b$ are coprime (or relatively prime), meaning they share no common positive factors other than $1$.
FAQ 12 (Case-Study Application). A sweet seller has $420$ Kaju barfis and $130$ Badam barfis. She wants to stack them so that each stack has the same number of barfis and takes up the least area on the tray. What is the maximum number of barfis that can be placed in each stack?
Answer:
To minimize the area occupied on the tray, the number of barfis in each stack must be maximized. This maximum number is given by the $\text{HCF}(420, 130)$.
Prime factorisations:
$$\begin{aligned} 420 &= 42 \times 10 = (2 \times 3 \times 7) \times (2 \times 5) = 2^2 \times 3^1 \times 5^1 \times 7^1 \\ 130 &= 13 \times 10 = 2^1 \times 5^1 \times 13^1 \end{aligned}$$
Finding the $\text{HCF}$:
$$\text{HCF}(420, 130) = 2^1 \times 5^1 = 10$$
The maximum number of barfis per stack is $10$.
FAQ 13. Prove that the product of two consecutive positive integers is divisible by $2$.
Answer:
Let the two consecutive positive integers be $n$ and $n + 1$.
- Every positive integer $n$ is either even ($2k$) or odd ($2k + 1$) for some integer $k$.
- Case 1: If $n = 2k$, then $n(n + 1) = 2k(2k + 1) = 2[k(2k + 1)]$, which is divisible by $2$.
- Case 2: If $n = 2k + 1$, then $n + 1 = 2k + 2 = 2(k + 1)$. The product is $(2k + 1) \cdot 2(k + 1) = 2[(2k + 1)(k + 1)]$, which is also divisible by $2$.
Thus, the product of two consecutive integers is always divisible by $2$.
FAQ 14 (HOTS). If $n$ is an odd integer, prove that $n^2 – 1$ is divisible by $8$.
Answer:
Any odd positive integer can be written in the form $n = 4k + 1$ or $n = 4k + 3$ for some integer $k$.
Case 1: $n = 4k + 1$
$$\begin{aligned} n^2 – 1 &= (4k + 1)^2 – 1 \\ &= 16k^2 + 8k + 1 – 1 \\ &= 8(2k^2 + k) \implies \text{Divisible by } 8 \end{aligned}$$
Case 2: $n = 4k + 3$
$$\begin{aligned} n^2 – 1 &= (4k + 3)^2 – 1 \\ &= 16k^2 + 24k + 9 – 1 \\ &= 16k^2 + 24k + 8 \\ &= 8(2k^2 + 3k + 1) \implies \text{Divisible by } 8 \end{aligned}$$
Therefore, $n^2 – 1$ is divisible by $8$ for any odd integer $n$.
FAQ 15. Explain why the sum of a rational number and an irrational number is always irrational.
Answer:
Let $r$ be a rational number and $x$ be an irrational number. Assume, to the contrary, that their sum is rational:
$$r + x = s \quad \text{where } s \in \mathbb{Q}$$
Rearranging gives:
$$x = s – r$$
Since the set of rational numbers is closed under subtraction, $s – r$ must be rational. This implies that $x$ is rational, which contradicts our premise that $x$ is irrational.
Therefore, the sum of a rational number and an irrational number is always irrational.
To score full marks on Real Numbers questions in the CBSE Class 10 Board Examinations, present every step clearly and systematically. In prime factorisation problems, write the full prime factorization in ascending order using proper exponential notation before computing the $\text{HCF}$ and $\text{LCM}$. For proofs by contradiction, state all initial assumptions explicitly, define variables as coprime integers ($\text{HCF}(a, b) = 1$), show divisibility on both sides of the equation, and clearly identify the contradiction. When applying the two-number formula $\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$, write the general formula first before substituting values, and remember that this relationship applies only to pairs of numbers.