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Chapter 1 Real Numbers
Welcome to the comprehensive solutions guide for Chapter 1, "Real Numbers," from the latest Class 10 NCERT mathematics textbook designed for the academic session 2024-25. This foundational chapter revisits and deepens the understanding of the real number system, focusing on key concepts of number theory that are essential for higher mathematics. It builds upon knowledge acquired in Class 9, emphasizing rigorous proofs and the fundamental structure of integers and rational numbers. These meticulously crafted solutions provide clear, step-by-step methods and logical explanations for all exercises, aiming to solidify understanding and build confidence in tackling number-theoretic problems.
A cornerstone of this chapter, extensively covered in these solutions, is the Fundamental Theorem of Arithmetic. This unique theorem states that every composite number can be expressed (factorized) as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur. The solutions provide thorough guidance on finding the prime factorization of integers. This fundamental skill is then systematically applied to determine the Highest Common Factor (HCF) and the Least Common Multiple (LCM) of two or more positive integers. The method involves listing the prime factors of each number and then identifying the lowest powers of common factors for HCF and the highest powers of all factors involved for LCM. Furthermore, the solutions demonstrate the application and verification of the important relationship connecting the HCF and LCM of two numbers ($a$ and $b$): HCF$(a, b) \times$ LCM$(a, b) = a \times b$. This relationship provides a useful tool for finding one quantity when the other three are known.
A significant portion of the chapter addresses the crucial distinction between rational and irrational numbers, focusing particularly on proving the irrationality of certain numbers. This often represents a conceptual challenge, requiring a shift towards deductive proof techniques. The solutions provide detailed, rigorous proofs, primarily using the method of contradiction, to establish the irrationality of numbers such as $\sqrt{2}$, $\sqrt{3}$, and $\sqrt{5}$. Building upon these fundamental proofs, the solutions also demonstrate how to prove the irrationality of combinations involving irrational numbers, such as $3 + 2\sqrt{5}$ or $5 - \sqrt{3}$, by leveraging the established irrationality of the base radical (like $\sqrt{5}$ or $\sqrt{3}$) and the properties of rational numbers under arithmetic operations.
Finally, the chapter revisits the nature of decimal expansions of rational numbers ($\frac{p}{q}$ form). The solutions explain the key theorems that connect the decimal representation of a rational number to the prime factorization of its denominator ($q$). Specifically, a rational number has a terminating decimal expansion if and only if the prime factorization of its denominator (in simplest form) is of the form $2^n 5^m$, where $n$ and $m$ are non-negative integers. Otherwise, the decimal expansion is non-terminating repeating. The solutions demonstrate how to apply this theorem to predict the nature of the decimal expansion without performing long division and, for terminating decimals, how to determine the number of decimal places after which it terminates (related to the higher power of 2 or 5 in the denominator's factorization).
Regarding the rationalized syllabus for 2024-25, Chapter 1, "Real Numbers," in the Class 10 NCERT textbook has been streamlined. Notably, Euclid's Division Lemma and Algorithm for finding HCF have been removed. The focus is now squarely on the Fundamental Theorem of Arithmetic for HCF/LCM, proofs of irrationality for $\sqrt{2}, \sqrt{3}, \sqrt{5}$ and related combinations, and determining the nature of decimal expansions based on the denominator's prime factors. By diligently studying these focused solutions, students can master these essential number theory techniques, develop crucial proof-writing skills, and gain a deeper insight into the properties that distinguish rational and irrational numbers within the real number system.
Example 1 to 4 (Before Exercise 1.1)
Example 1. Consider the numbers 4n , where n is a natural number. Check whether there is any value of n for which 4n ends with the digit zero.
Answer:
Given:
The number is of the form $4^n$, where $n$ is a natural number ($n \in \{1, 2, 3, ...\}$).
To Check:
Whether there exists any value of $n$ for which $4^n$ ends with the digit zero.
Solution:
A natural number ends with the digit zero if and only if its prime factorization contains both 2 and 5.
Let's find the prime factorization of the base, 4.
$4 = 2 \times 2 = 2^2$
[Prime factorization of 4]
Now, let's find the prime factorization of $4^n$:
$4^n = (2^2)^n$
$4^n = 2^{2n}$
... (i)
From the prime factorization in equation (i), we can see that the only prime factor of $4^n$ is 2.
For a number to end with the digit zero, its prime factorization must contain both 2 and 5.
Since the prime factorization of $4^n$ contains only the prime factor 2 and not the prime factor 5, $4^n$ cannot end with the digit zero for any natural number $n$. This is a direct consequence of the Fundamental Theorem of Arithmetic (Unique Factorisation Theorem), which states that every composite number can be expressed as a product of primes in a unique way (except for the order of the prime factors).
Therefore, there is no value of $n$ 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.
Answer:
Given:
The numbers are 6 and 20.
To Find:
LCM and HCF of 6 and 20 using the prime factorisation method.
Solution:
First, we find the prime factorization of each number.
Prime factorization of 6:
$\begin{array}{c|cc} 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$
So, $6 = 2 \times 3 = 2^1 \times 3^1$.
Prime factorization of 20:
$\begin{array}{c|cc} 2 & 20 \\ \hline 2 & 10 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$
So, $20 = 2 \times 2 \times 5 = 2^2 \times 5^1$.
Now, we find the HCF and LCM using the prime factorizations.
Finding the HCF:
The HCF is the product of the lowest powers of the common prime factors.
The prime factor common to both 6 and 20 is 2.
The powers of 2 are $2^1$ (in 6) and $2^2$ (in 20). The lowest power is $2^1$.
HCF$(6, 20) = 2^1 = 2$
... (i)
Finding the LCM:
The LCM is the product of the highest powers of all prime factors involved in the factorizations.
The prime factors involved are 2, 3, and 5.
The highest power of 2 is $2^2$ (from 20).
The highest power of 3 is $3^1$ (from 6).
The highest power of 5 is $5^1$ (from 20).
LCM$(6, 20) = 2^2 \times 3^1 \times 5^1$
... (ii)
Calculate the value:
LCM$(6, 20) = 4 \times 3 \times 5$
LCM$(6, 20) = 12 \times 5$
LCM$(6, 20) = 60$
... (iii)
The HCF of 6 and 20 is 2.
The LCM of 6 and 20 is 60.
Verification (Optional):
We know that for any two positive integers $a$ and $b$, HCF$(a, b) \times$ LCM$(a, b) = a \times b$.
LHS = HCF$(6, 20) \times$ LCM$(6, 20) = 2 \times 60 = 120$.
RHS = $6 \times 20 = 120$.
LHS = RHS. The results are verified.
Example 3. Find the HCF of 96 and 404 by the prime factorisation method. Hence, find their LCM.
Answer:
Given:
The numbers are 96 and 404.
To Find:
The HCF and LCM of 96 and 404 using the prime factorisation method, and then find the LCM using the calculated HCF.
Solution:
First, we find the prime factorization of each number.
Prime factorization of 96:
$\begin{array}{c|cc} 2 & 96 \\ \hline 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$
So, $96 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^5 \times 3^1$.
Prime factorization of 404:
$\begin{array}{c|cc} 2 & 404 \\ \hline 2 & 202 \\ \hline 101 & 101 \\ \hline & 1 \end{array}$
So, $404 = 2 \times 2 \times 101 = 2^2 \times 101^1$.
Finding the HCF:
The HCF is the product of the lowest powers of the common prime factors.
The only common prime factor is 2.
The powers of 2 in the factorizations are $2^5$ and $2^2$. The lowest power is $2^2$.
HCF$(96, 404) = 2^2 = 4$
... (i)
Finding the LCM (Hence):
We use the relationship between HCF and LCM of two positive integers $a$ and $b$:
HCF$(a, b) \times$ LCM$(a, b) = a \times b$
... (ii)
Substitute $a = 96$, $b = 404$, and HCF$(96, 404) = 4$ into equation (ii):
$4 \times$ LCM$(96, 404) = 96 \times 404$
... (iii)
Now, solve for LCM$(96, 404)$:
LCM$(96, 404) = \frac{96 \times 404}{4}$
... (iv)
Simplify the expression:
LCM$(96, 404) = 96 \times \frac{\cancel{404}^{101}}{\cancel{4}_1}$
... (v)
LCM$(96, 404) = 96 \times 101$
... (vi)
Calculate the product:
LCM$(96, 404) = 9696$
... (vii)
The HCF of 96 and 404 is 4.
The LCM of 96 and 404 is 9696.
Example 4. Find the HCF and LCM of 6, 72 and 120, using the prime factorisation method.
Answer:
Given:
The numbers are 6, 72, and 120.
To Find:
HCF and LCM of 6, 72, and 120 using the prime factorisation method.
Solution:
First, we find the prime factorization of each number.
Prime factorization of 6:
$\begin{array}{c|cc} 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$
So, $6 = 2 \times 3 = 2^1 \times 3^1$.
Prime factorization of 72:
$\begin{array}{c|cc} 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$
So, $72 = 2 \times 2 \times 2 \times 3 \times 3 = 2^3 \times 3^2$.
Prime factorization of 120:
$\begin{array}{c|cc} 2 & 120 \\ \hline 2 & 60 \\ \hline 2 & 30 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$
So, $120 = 2 \times 2 \times 2 \times 3 \times 5 = 2^3 \times 3^1 \times 5^1$.
Finding the HCF:
The HCF is the product of the lowest powers of the common prime factors among all three numbers.
The common prime factors are 2 and 3.
Lowest power of 2: $\min(2^1, 2^3, 2^3) = 2^1$
Lowest power of 3: $\min(3^1, 3^2, 3^1) = 3^1$
HCF$(6, 72, 120) = 2^1 \times 3^1$
... (i)
HCF$(6, 72, 120) = 2 \times 3 = 6$
... (ii)
Finding the LCM:
The LCM is the product of the highest powers of all prime factors involved in the factorizations of the three numbers.
The prime factors involved are 2, 3, and 5.
Highest power of 2: $\max(2^1, 2^3, 2^3) = 2^3$
Highest power of 3: $\max(3^1, 3^2, 3^1) = 3^2$
Highest power of 5: $\max(5^0, 5^0, 5^1) = 5^1$ (Note: $3^0=1, 5^0=1$ for numbers where the prime factor is not present)
LCM$(6, 72, 120) = 2^3 \times 3^2 \times 5^1$
... (iii)
Calculate the value:
LCM$(6, 72, 120) = 8 \times 9 \times 5$
... (iv)
LCM$(6, 72, 120) = 72 \times 5$
... (v)
LCM$(6, 72, 120) = 360$
... (vi)
The HCF of 6, 72, and 120 is 6.
The LCM of 6, 72, and 120 is 360.
Exercise 1.1
Question 1. Express each number as a product of its prime factors:
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429
Answer:
We express each number as a product of its prime factors using the prime factorization method.
(i) 140
$\begin{array}{c|cc} 2 & 140 \\ \hline 2 & 70 \\ \hline 5 & 35 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$
So, $140 = 2 \times 2 \times 5 \times 7 = 2^2 \times 5^1 \times 7^1$.
(ii) 156
$\begin{array}{c|cc} 2 & 156 \\ \hline 2 & 78 \\ \hline 3 & 39 \\ \hline 13 & 13 \\ \hline & 1 \end{array}$
So, $156 = 2 \times 2 \times 3 \times 13 = 2^2 \times 3^1 \times 13^1$.
(iii) 3825
$\begin{array}{c|cc} 3 & 3825 \\ \hline 3 & 1275 \\ \hline 5 & 425 \\ \hline 5 & 85 \\ \hline 17 & 17 \\ \hline & 1 \end{array}$
So, $3825 = 3 \times 3 \times 5 \times 5 \times 17 = 3^2 \times 5^2 \times 17^1$.
(iv) 5005
$\begin{array}{c|cc} 5 & 5005 \\ \hline 7 & 1001 \\ \hline 11 & 143 \\ \hline 13 & 13 \\ \hline & 1 \end{array}$
So, $5005 = 5 \times 7 \times 11 \times 13 = 5^1 \times 7^1 \times 11^1 \times 13^1$.
(v) 7429
We check for divisibility by small prime numbers (2, 3, 5, 7, 11, 13, 17, 19, 23, ...).
Not divisible by 2, 3, 5.
$7429 \div 7 = 1061.28...$ (Not divisible by 7)
$7429 \div 11 = 675.36...$ (Not divisible by 11)
$7429 \div 13 = 571.46...$ (Not divisible by 13)
$7429 \div 17 = 437$. (Divisible by 17)
$\begin{array}{c|cc} 17 & 7429 \\ \hline 19 & 437 \\ \hline 23 & 23 \\ \hline & 1 \end{array}$
So, $7429 = 17 \times 19 \times 23 = 17^1 \times 19^1 \times 23^1$.
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
Answer:
We are asked to find the LCM and HCF of each pair of integers using the prime factorisation method and then verify that the product of the LCM and HCF is equal to the product of the two numbers.
(i) 26 and 91
Given: Integers 26 and 91.
To Find: LCM and HCF, and verify LCM $\times$ HCF = $26 \times 91$.
Solution:
First, we find the prime factorisation of each number.
Prime factorisation of 26:
$\begin{array}{c|cc} 2 & 26 \\ \hline 13 & 13 \\ \hline & 1 \end{array}$
$26 = 2^1 \times 13^1$
Prime factorisation of 91:
$\begin{array}{c|cc} 7 & 91 \\ \hline 13 & 13 \\ \hline & 1 \end{array}$
$91 = 7^1 \times 13^1$
The HCF is the product of the lowest powers of the common prime factors.
The only prime factor common to 26 and 91 is 13. The lowest power of 13 is $13^1$.
HCF$(26, 91) = 13$.
The LCM is the product of the highest powers of all prime factors involved in the factorisations.
The prime factors involved are 2, 7, and 13. The highest power of 2 is $2^1$, the highest power of 7 is $7^1$, and the highest power of 13 is $13^1$.
LCM$(26, 91) = 2^1 \times 7^1 \times 13^1 = 2 \times 7 \times 13 = 14 \times 13 = 182$.
Now, we verify the relationship: HCF $\times$ LCM = Product of the two numbers.
Product of the two numbers $= 26 \times 91 = 2366$.
Product of the two numbers $= 2366$
... (i)
HCF $\times$ LCM $= 13 \times 182 = 2366$.
HCF $\times$ LCM $= 2366$
... (ii)
From (i) and (ii), we see that $2366 = 2366$. Thus, HCF $\times$ LCM = product of the two numbers is verified for 26 and 91.
(ii) 510 and 92
Given: Integers 510 and 92.
To Find: LCM and HCF, and verify LCM $\times$ HCF = $510 \times 92$.
Solution:
First, we find the prime factorisation of each number.
Prime factorisation of 510:
$\begin{array}{c|cc} 2 & 510 \\ \hline 3 & 255 \\ \hline 5 & 85 \\ \hline 17 & 17 \\ \hline & 1 \end{array}$
$510 = 2^1 \times 3^1 \times 5^1 \times 17^1$
Prime factorisation of 92:
$\begin{array}{c|cc} 2 & 92 \\ \hline 2 & 46 \\ \hline 23 & 23 \\ \hline & 1 \end{array}$
$92 = 2^2 \times 23^1$
The HCF is the product of the lowest powers of the common prime factors.
The only prime factor common to 510 and 92 is 2. The lowest power of 2 is $2^1$.
HCF$(510, 92) = 2^1 = 2$.
The LCM is the product of the highest powers of all prime factors involved.
The prime factors involved are 2, 3, 5, 17, and 23. The highest power of 2 is $2^2$, of 3 is $3^1$, of 5 is $5^1$, of 17 is $17^1$, and of 23 is $23^1$.
LCM$(510, 92) = 2^2 \times 3^1 \times 5^1 \times 17^1 \times 23^1 = 4 \times 3 \times 5 \times 17 \times 23 = 12 \times 5 \times 17 \times 23 = 60 \times 17 \times 23 = 1020 \times 23 = 23460$.
LCM$(510, 92) = 23460$.
Now, we verify the relationship: HCF $\times$ LCM = Product of the two numbers.
Product of the two numbers $= 510 \times 92 = 46920$.
Product of the two numbers $= 46920$
... (iii)
HCF $\times$ LCM $= 2 \times 23460 = 46920$.
HCF $\times$ LCM $= 46920$
... (iv)
From (iii) and (iv), we see that $46920 = 46920$. Thus, HCF $\times$ LCM = product of the two numbers is verified for 510 and 92.
(iii) 336 and 54
Given: Integers 336 and 54.
To Find: LCM and HCF, and verify LCM $\times$ HCF = $336 \times 54$.
Solution:
First, we find the prime factorisation of each number.
Prime factorisation of 336:
$\begin{array}{c|cc} 2 & 336 \\ \hline 2 & 168 \\ \hline 2 & 84 \\ \hline 2 & 42 \\ \hline 3 & 21 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$
$336 = 2^4 \times 3^1 \times 7^1$
Prime factorisation of 54:
$\begin{array}{c|cc} 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$
$54 = 2^1 \times 3^3$
The HCF is the product of the lowest powers of the common prime factors.
The common prime factors are 2 and 3. The lowest power of 2 is $2^1$. The lowest power of 3 is $3^1$.
HCF$(336, 54) = 2^1 \times 3^1 = 2 \times 3 = 6$.
The LCM is the product of the highest powers of all prime factors involved.
The prime factors involved are 2, 3, and 7. The highest power of 2 is $2^4$, the highest power of 3 is $3^3$, and the highest power of 7 is $7^1$.
LCM$(336, 54) = 2^4 \times 3^3 \times 7^1 = 16 \times 27 \times 7 = 432 \times 7 = 3024$.
LCM$(336, 54) = 3024$.
Now, we verify the relationship: HCF $\times$ LCM = Product of the two numbers.
Product of the two numbers $= 336 \times 54 = 18144$.
Product of the two numbers $= 18144$
... (v)
HCF $\times$ LCM $= 6 \times 3024 = 18144$.
HCF $\times$ LCM $= 18144$
... (vi)
From (v) and (vi), we see that $18144 = 18144$. Thus, HCF $\times$ LCM = product of the two numbers is verified for 336 and 54.
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) Find the LCM and HCF of 12, 15 and 21.
Given: Integers 12, 15, and 21.
To Find: LCM and HCF.
Solution: Apply the prime factorisation method.
Prime factorisation of 12:
$\begin{array}{c|cc} 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$$12 = 2 \times 2 \times 3 = 2^2 \times 3$
Prime factorisation of 15:
$\begin{array}{c|cc} 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$$15 = 3 \times 5$
Prime factorisation of 21:
$\begin{array}{c|cc} 3 & 21 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$$21 = 3 \times 7$
To find the HCF, we take the product of the lowest powers of the common prime factors.
The only common prime factor is 3. The lowest power of 3 in the factorisations is $3^1$.
HCF(12, 15, 21) $= 3^1 = 3$.
To find the LCM, we take the product of the highest powers of all prime factors involved.
The prime factors involved are 2, 3, 5, and 7. The highest power of 2 is $2^2$, the highest power of 3 is $3^1$, the highest power of 5 is $5^1$, and the highest power of 7 is $7^1$.
LCM(12, 15, 21) $= 2^2 \times 3^1 \times 5^1 \times 7^1 = 4 \times 3 \times 5 \times 7 = 12 \times 35 = 420$.
(ii) Find the LCM and HCF of 17, 23 and 29.
Given: Integers 17, 23, and 29.
To Find: LCM and HCF.
Solution: Apply the prime factorisation method.
Prime factorisation of 17: Since 17 is a prime number, its only prime factor is 17.
$17 = 17^1$
Prime factorisation of 23: Since 23 is a prime number, its only prime factor is 23.
$23 = 23^1$
Prime factorisation of 29: Since 29 is a prime number, its only prime factor is 29.
$29 = 29^1$
To find the HCF, we take the product of the lowest powers of the common prime factors.
Since 17, 23, and 29 are prime numbers and different from each other, there are no common prime factors other than 1.
HCF(17, 23, 29) $= 1$.
To find the LCM, we take the product of the highest powers of all prime factors involved.
The prime factors involved are 17, 23, and 29. The highest power of 17 is $17^1$, of 23 is $23^1$, and of 29 is $29^1$.
LCM(17, 23, 29) $= 17^1 \times 23^1 \times 29^1 = 17 \times 23 \times 29$.
$17 \times 23 = 391$.
$391 \times 29 = 11339$.
LCM(17, 23, 29) $= 11339$.
(iii) Find the LCM and HCF of 8, 9 and 25.
Given: Integers 8, 9, and 25.
To Find: LCM and HCF.
Solution: Apply the prime factorisation method.
Prime factorisation of 8:
$\begin{array}{c|cc} 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$$8 = 2 \times 2 \times 2 = 2^3$
Prime factorisation of 9:
$\begin{array}{c|cc} 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$$9 = 3 \times 3 = 3^2$
Prime factorisation of 25:
$\begin{array}{c|cc} 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$$25 = 5 \times 5 = 5^2$
To find the HCF, we take the product of the lowest powers of the common prime factors.
The prime factors of 8 are 2. The prime factors of 9 are 3. The prime factors of 25 are 5. There are no common prime factors among 8, 9, and 25 other than 1.
HCF(8, 9, 25) $= 1$.
To find the LCM, we take the product of the highest powers of all prime factors involved.
The prime factors involved are 2, 3, and 5. The highest power of 2 is $2^3$, the highest power of 3 is $3^2$, and the highest power of 5 is $5^2$.
LCM(8, 9, 25) $= 2^3 \times 3^2 \times 5^2 = 8 \times 9 \times 25$.
$8 \times 9 = 72$.
$72 \times 25 = 1800$.
LCM(8, 9, 25) $= 1800$.
Question 4. Given that HCF (306, 657) = 9, find LCM (306, 657).
Answer:
Given:
The two integers are 306 and 657.
HCF(306, 657) = 9.
To Find:
LCM(306, 657).
Solution:
We know that for any two positive integers, the product of their HCF and LCM is equal to the product of the numbers.
Let the two integers be $a$ and $b$. The relationship is:
$\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$
Substitute the given values into the formula:
$9 \times \text{LCM}(306, 657) = 306 \times 657$
Now, solve for LCM(306, 657):
$\text{LCM}(306, 657) = \frac{306 \times 657}{9}$
We can simplify the calculation by dividing 306 by 9:
$\frac{\cancel{306}^{34} \times 657}{\cancel{9}_{1}} = 34 \times 657$
Calculate the product:
$34 \times 657$
$34 \times 657 = 22338$
Therefore, LCM(306, 657) = 22338.
Answer:
The LCM of 306 and 657 is 22338.
Question 5. Check whether 6n can end with the digit 0 for any natural number n.
Answer:
We need to check if $6^n$ can end with the digit 0 for any natural number $n$.
A number ends with the digit 0 if and only if it is divisible by 10. A number is divisible by 10 if and only if its prime factorization contains both 2 and 5.
Let's find the prime factorization of $6^n$.
The base is 6. The prime factorization of 6 is $2 \times 3$.
So, the prime factorization of $6^n$ is:
$6^n = (2 \times 3)^n = 2^n \times 3^n$
The prime factors of $6^n$ are only 2 and 3.
For $6^n$ to end with the digit 0, its prime factorization must contain 5.
However, the prime factorization of $6^n$ ($2^n \times 3^n$) does not contain the prime factor 5.
According to the Fundamental Theorem of Arithmetic, the prime factorization of every composite number is unique.
Since the prime factorization of $6^n$ only consists of primes 2 and 3, it cannot have 5 as a prime factor.
Therefore, $6^n$ cannot end with the digit 0 for any natural number $n$.
Question 6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Answer:
A composite number is a natural number greater than 1 that is not a prime number. In other words, a composite number is a number that has at least one divisor other than 1 and itself.
We need to show that the given expressions can be expressed as a product of two numbers greater than 1.
Consider the first expression:
$7 \times 11 \times 13 + 13$
We can see that 13 is a common factor in both terms ($7 \times 11 \times 13$ and 13).
Factor out 13:
$13 \times (7 \times 11 + 1)$
Now calculate the value inside the parenthesis:
$7 \times 11 + 1 = 77 + 1 = 78$
So the expression becomes:
$13 \times 78$
Since the number can be expressed as a product of two integers, 13 and 78, both of which are greater than 1, it is a composite number.
Consider the second expression:
$7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$
We can see that 5 is a common factor in both terms ($7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$ and 5).
Factor out 5:
$5 \times (7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1)$
Now calculate the value inside the parenthesis:
$7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1 = 42 \times 24 + 1$
$42 \times 24 = 1008$
$1008 + 1 = 1009$
So the expression becomes:
$5 \times 1009$
Since the number can be expressed as a product of two integers, 5 and 1009, both of which are greater than 1, it 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?
Answer:
Given:
Time taken by Sonia to complete one round = 18 minutes.
Time taken by Ravi to complete one round = 12 minutes.
They start at the same point, at the same time, and go in the same direction.
To Find:
The time after which they will meet again at the starting point.
Solution:
To find the time after which they will meet again at the starting point, we need to find the least common multiple (LCM) of the time taken by Sonia and Ravi to complete one round.
LCM of 18 and 12:
First, find the prime factorisation of 18 and 12.
Prime factorisation of 18:
$\begin{array}{c|cc} 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$$18 = 2 \times 3 \times 3 = 2^1 \times 3^2$
Prime factorisation of 12:
$\begin{array}{c|cc} 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$$12 = 2 \times 2 \times 3 = 2^2 \times 3^1$
The LCM is the product of the highest powers of all prime factors involved.
Prime factors are 2 and 3.
Highest power of 2 is $2^2$.
Highest power of 3 is $3^2$.
LCM(18, 12) = $2^2 \times 3^2 = 4 \times 9 = 36$.
They will meet again at the starting point after 36 minutes.
Answer:
Sonia and Ravi will meet again at the starting point after 36 minutes.
Example 5 to 7 (Before Exercise 1.2)
Example 5. Prove that $\sqrt{3}$ is irrational.
Answer:
To Prove: $\sqrt{3}$ is irrational.
Proof:
We will prove this by the method of contradiction.
Assume that $\sqrt{3}$ is a rational number.
If $\sqrt{3}$ is rational, then it can be written in the form $\frac{p}{q}$, where $p$ and $q$ are integers, $q \neq 0$, and $p$ and $q$ are coprime (i.e., their greatest common divisor is 1).
So, we assume:
$\sqrt{3} = \frac{p}{q}$
Squaring both sides, we get:
$(\sqrt{3})^2 = \left(\frac{p}{q}\right)^2$
$3 = \frac{p^2}{q^2}$
Rearranging the equation, we get:
$p^2 = 3q^2$
This equation implies that $p^2$ is divisible by 3.
According to a theorem (if a prime number $a$ divides $b^2$, then $a$ divides $b$), since 3 is a prime number and 3 divides $p^2$, it must be that 3 divides $p$.
Since $p$ is divisible by 3, we can write $p$ as $p = 3k$ for some integer $k$.
Substitute this value of $p$ into the equation $p^2 = 3q^2$:
$(3k)^2 = 3q^2$
$9k^2 = 3q^2$
Divide both sides by 3:
$3k^2 = q^2$
This equation implies that $q^2$ is divisible by 3.
Using the same theorem as before, since 3 is a prime number and 3 divides $q^2$, it must be that 3 divides $q$.
We have now shown that both $p$ and $q$ are divisible by 3.
This means that $p$ and $q$ have a common factor of 3.
However, this contradicts our initial assumption that $p$ and $q$ are coprime (have no common factors other than 1).
Conclusion:
Since our assumption that $\sqrt{3}$ is rational leads to a contradiction, our assumption must be false.
Therefore, $\sqrt{3}$ is an irrational number.
Example 6. Show that 5 – $\sqrt{3}$ is irrational.
Answer:
To Show: $5 - \sqrt{3}$ is irrational.
Proof:
We will prove this by the method of contradiction.
Assume that $5 - \sqrt{3}$ is a rational number.
If $5 - \sqrt{3}$ is rational, then it can be written in the form $\frac{p}{q}$, where $p$ and $q$ are integers, $q \neq 0$, and $p$ and $q$ are coprime (i.e., their greatest common divisor is 1).
So, we assume:
$5 - \sqrt{3} = \frac{p}{q}$
Now, we rearrange the equation to isolate $\sqrt{3}$ on one side.
Subtract 5 from both sides:
$-\sqrt{3} = \frac{p}{q} - 5$
$-\sqrt{3} = \frac{p - 5q}{q}$
Multiply both sides by -1:
$\sqrt{3} = -\frac{p - 5q}{q}$
$\sqrt{3} = \frac{-(p - 5q)}{q}$
$\sqrt{3} = \frac{5q - p}{q}$
Since $p$ and $q$ are integers, $5q$ is an integer. Also, $5q - p$ is an integer, and $q$ is an integer and $q \neq 0$.
Therefore, the expression $\frac{5q - p}{q}$ is a rational number because it is in the form of an integer divided by a non-zero integer.
This implies that $\sqrt{3}$ is equal to a rational number, which means $\sqrt{3}$ is rational.
We know that $\sqrt{3}$ is an irrational number. This fact was proven in a previous example/theorem.
So, the statement "$ \sqrt{3}$ is rational" contradicts the known fact that $\sqrt{3}$ is irrational.
Conclusion:
Since our assumption that $5 - \sqrt{3}$ is rational leads to a contradiction, our assumption must be false.
Therefore, $5 - \sqrt{3}$ is an irrational number.
Example 7. Show that 3$\sqrt{2}$ is irrational.
Answer:
To Show: $3\sqrt{2}$ is irrational.
Proof:
We will prove this by the method of contradiction.
Assume that $3\sqrt{2}$ is a rational number.
If $3\sqrt{2}$ is rational, then it can be written in the form $\frac{p}{q}$, where $p$ and $q$ are integers, $q \neq 0$, and $p$ and $q$ are coprime (i.e., their greatest common divisor is 1).
So, we assume:
$3\sqrt{2} = \frac{p}{q}$
Now, we rearrange the equation to isolate $\sqrt{2}$ on one side.
Divide both sides by 3:
$\sqrt{2} = \frac{p}{3q}$
Since $p$ is an integer and $q$ is a non-zero integer, $3q$ is also a non-zero integer.
Therefore, the expression $\frac{p}{3q}$ is a rational number because it is in the form of an integer divided by a non-zero integer.
This implies that $\sqrt{2}$ is equal to a rational number, which means $\sqrt{2}$ is rational.
We know that $\sqrt{2}$ is an irrational number. This is a well-established fact (often proven by contradiction, similar to how we proved $\sqrt{3}$ is irrational).
So, the statement "$\sqrt{2}$ is rational" contradicts the known fact that $\sqrt{2}$ is irrational.
Conclusion:
Since our assumption that $3\sqrt{2}$ is rational leads to a contradiction, our assumption must be false.
Therefore, $3\sqrt{2}$ is an irrational number.
Exercise 1.2
Question 1. Prove that $\sqrt{5}$ is irrational.
Answer:
To Prove: $\sqrt{5}$ is irrational.
Proof:
We will prove this by the method of contradiction.
Assume that $\sqrt{5}$ is a rational number.
If $\sqrt{5}$ is rational, then it can be written in the form $\frac{p}{q}$, where $p$ and $q$ are integers, $q \neq 0$, and $p$ and $q$ are coprime (i.e., their greatest common divisor is 1).
So, we assume:
$\sqrt{5} = \frac{p}{q}$
Squaring both sides, we get:
$(\sqrt{5})^2 = \left(\frac{p}{q}\right)^2$
$5 = \frac{p^2}{q^2}$
Rearranging the equation, we get:
$p^2 = 5q^2$
This equation implies that $p^2$ is divisible by 5.
According to a theorem (if a prime number $a$ divides $b^2$, then $a$ divides $b$), since 5 is a prime number and 5 divides $p^2$, it must be that 5 divides $p$.
Since $p$ is divisible by 5, we can write $p$ as $p = 5k$ for some integer $k$.
Substitute this value of $p$ into the equation $p^2 = 5q^2$:
$(5k)^2 = 5q^2$
$25k^2 = 5q^2$
Divide both sides by 5:
$5k^2 = q^2$
This equation implies that $q^2$ is divisible by 5.
Using the same theorem as before, since 5 is a prime number and 5 divides $q^2$, it must be that 5 divides $q$.
We have now shown that both $p$ and $q$ are divisible by 5.
This means that $p$ and $q$ have a common factor of 5.
However, this contradicts our initial assumption that $p$ and $q$ are coprime (have no common factors other than 1).
Conclusion:
Since our assumption that $\sqrt{5}$ is rational leads to a contradiction, our assumption must be false.
Therefore, $\sqrt{5}$ is an irrational number.
Question 2. Prove that 3 + 2$\sqrt{5}$ is irrational.
Answer:
To Prove: $3 + 2\sqrt{5}$ is irrational.
Proof:
We will prove this by the method of contradiction.
Assume that $3 + 2\sqrt{5}$ is a rational number.
If $3 + 2\sqrt{5}$ is rational, then it can be written in the form $\frac{p}{q}$, where $p$ and $q$ are integers, $q \neq 0$, and $p$ and $q$ are coprime (i.e., their greatest common divisor is 1).
So, we assume:
$3 + 2\sqrt{5} = \frac{p}{q}$
Now, we rearrange the equation to isolate $\sqrt{5}$ on one side.
Subtract 3 from both sides:
$2\sqrt{5} = \frac{p}{q} - 3$
$2\sqrt{5} = \frac{p - 3q}{q}$
Divide both sides by 2:
$\sqrt{5} = \frac{p - 3q}{2q}$
Since $p$ and $q$ are integers, $3q$ is an integer, and $p - 3q$ is an integer. Also, $q$ is a non-zero integer, so $2q$ is a non-zero integer.
Therefore, the expression $\frac{p - 3q}{2q}$ is a rational number because it is in the form of an integer divided by a non-zero integer.
This implies that $\sqrt{5}$ is equal to a rational number, which means $\sqrt{5}$ is rational.
We know that $\sqrt{5}$ is an irrational number. This fact was proven in a previous question/example.
So, the statement "$\sqrt{5}$ is rational" contradicts the known fact that $\sqrt{5}$ is irrational.
Conclusion:
Since our assumption that $3 + 2\sqrt{5}$ is rational leads to a contradiction, our assumption must be false.
Therefore, $3 + 2\sqrt{5}$ is an irrational number.
Question 3. Prove that the following are irrationals :
(i) $\frac{1}{\sqrt{2}}$
(ii) 7$\sqrt{5}$
(iii) 6 + $\sqrt{2}$
Answer:
(i) Prove that $\frac{1}{\sqrt{2}}$ is irrational.
To Prove: $\frac{1}{\sqrt{2}}$ is irrational.
Proof:
We will prove this by the method of contradiction.
Assume that $\frac{1}{\sqrt{2}}$ is a rational number.
If $\frac{1}{\sqrt{2}}$ is rational, then it can be written in the form $\frac{p}{q}$, where $p$ and $q$ are integers, $q \neq 0$, and $p$ and $q$ are coprime (i.e., their greatest common divisor is 1).
So, we assume:
$\frac{1}{\sqrt{2}} = \frac{p}{q}$
Taking the reciprocal of both sides, we get:
$\sqrt{2} = \frac{q}{p}$
Since $p$ and $q$ are integers and $q \neq 0$, if $p \neq 0$, then $\frac{q}{p}$ is a rational number.
If $p=0$, then $\frac{1}{\sqrt{2}} = \frac{0}{q} = 0$, which is impossible. So $p \neq 0$.
Therefore, the expression $\frac{q}{p}$ is a rational number because it is in the form of an integer divided by a non-zero integer.
This implies that $\sqrt{2}$ is equal to a rational number, which means $\sqrt{2}$ is rational.
We know that $\sqrt{2}$ is an irrational number.
So, the statement "$\sqrt{2}$ is rational" contradicts the known fact that $\sqrt{2}$ is irrational.
Conclusion:
Since our assumption that $\frac{1}{\sqrt{2}}$ is rational leads to a contradiction, our assumption must be false.
Therefore, $\frac{1}{\sqrt{2}}$ is an irrational number.
(ii) Prove that $7\sqrt{5}$ is irrational.
To Prove: $7\sqrt{5}$ is irrational.
Proof:
We will prove this by the method of contradiction.
Assume that $7\sqrt{5}$ is a rational number.
If $7\sqrt{5}$ is rational, then it can be written in the form $\frac{p}{q}$, where $p$ and $q$ are integers, $q \neq 0$, and $p$ and $q$ are coprime (i.e., their greatest common divisor is 1).
So, we assume:
$7\sqrt{5} = \frac{p}{q}$
Now, we rearrange the equation to isolate $\sqrt{5}$ on one side.
Divide both sides by 7:
$\sqrt{5} = \frac{p}{7q}$
Since $p$ is an integer and $q$ is a non-zero integer, $7q$ is also a non-zero integer.
Therefore, the expression $\frac{p}{7q}$ is a rational number because it is in the form of an integer divided by a non-zero integer.
This implies that $\sqrt{5}$ is equal to a rational number, which means $\sqrt{5}$ is rational.
We know that $\sqrt{5}$ is an irrational number (proven in a previous question).
So, the statement "$\sqrt{5}$ is rational" contradicts the known fact that $\sqrt{5}$ is irrational.
Conclusion:
Since our assumption that $7\sqrt{5}$ is rational leads to a contradiction, our assumption must be false.
Therefore, $7\sqrt{5}$ is an irrational number.
(iii) Prove that $6 + \sqrt{2}$ is irrational.
To Prove: $6 + \sqrt{2}$ is irrational.
Proof:
We will prove this by the method of contradiction.
Assume that $6 + \sqrt{2}$ is a rational number.
If $6 + \sqrt{2}$ is rational, then it can be written in the form $\frac{p}{q}$, where $p$ and $q$ are integers, $q \neq 0$, and $p$ and $q$ are coprime (i.e., their greatest common divisor is 1).
So, we assume:
$6 + \sqrt{2} = \frac{p}{q}$
Now, we rearrange the equation to isolate $\sqrt{2}$ on one side.
Subtract 6 from both sides:
$\sqrt{2} = \frac{p}{q} - 6$
$\sqrt{2} = \frac{p - 6q}{q}$
Since $p$ and $q$ are integers, $6q$ is an integer, and $p - 6q$ is an integer. Also, $q$ is a non-zero integer.
Therefore, the expression $\frac{p - 6q}{q}$ is a rational number because it is in the form of an integer divided by a non-zero integer.
This implies that $\sqrt{2}$ is equal to a rational number, which means $\sqrt{2}$ is rational.
We know that $\sqrt{2}$ is an irrational number.
So, the statement "$\sqrt{2}$ is rational" contradicts the known fact that $\sqrt{2}$ is irrational.
Conclusion:
Since our assumption that $6 + \sqrt{2}$ is rational leads to a contradiction, our assumption must be false.
Therefore, $6 + \sqrt{2}$ is an irrational number.