Real Numbers

Express each number as a product of its prime factors :

(i) 140

(ii) 156

(iii) 3825

(iv) 5005

(v) 7429

Answer

(i) On prime factorizing, we get :

140
= 2 x 70
= 2 x 2 x 35
= 2 x 2 x 5 x 7
= 2² × 5 × 7

Hence, 140 = 2² × 5 × 7.

(ii) On prime factorizing, we get :

156
= 2 x 78
= 2 x 2 x 39
= 2 x 2 x 3 x 13
= 2² × 3 × 13

Hence, 156 = 2² × 3 × 13.

(iii) On prime factorizing, we get :

3825
= 3 x 1275
= 3 x 3 x 425
= 3 x 3 x 5 x 85
= 3 x 3 x 5 x 5 x 17
= 3² × 5² × 17

Hence, 3825 = 3² × 5² × 17.

(iv) On prime factorizing, we get :

5005
= 5 x 1001
= 5 x 7 x 143
= 5 x 7 x 11 x 13

Hence, 5005 = 5 × 7 × 11 × 13.

(v) On prime factorizing, we get :

7429
= 17 x 437
= 17 × 19 × 23

Hence, 7429 = 17 × 19 × 23.

Find the L.C.M. and H.C.F. of the following pairs of integers and verify that L.C.M. × H.C.F. = product of two numbers.

(i) 26 and 91

(ii) 510 and 92

(iii) 336 and 54

Answer

We know that,

H.C.F. = Product of smallest power of each common prime factor in the numbers.

L.C.M. = Product of the greatest power of each prime factor, involved in the numbers.

(i) On prime factorizing, we get :

26
= 2 × 13

91
= 7 × 13

H.C.F. (26, 91) = 13

L.C.M. (26, 91) = 2 × 7 × 13 = 182.

Product of numbers = 26 × 91 = 2366

Product of H.C.F. and L.C.M. = 13 × 182 = 2366.

∴ Product of the two numbers = Product of H.C.F. and L.C.M.

Hence, H.C.F. = 13 and L.C.M. = 182.

(ii) On prime factorizing, we get :

510
= 2 × 255
= 2 × 3 × 85
= 2 × 3 × 5 × 17

92
= 2 × 46
= 2 × 2 × 23
= 2² × 23

H.C.F. (510, 92) = 2

L.C.M. (510, 92) = 2² × 3 × 5 × 17 × 23 = 23460.

Product of numbers = 510 × 92 = 46920

Product of H.C.F. and L.C.M. = 2 × 23460 = 46920.

∴ Product of the two numbers = Product of H.C.F. and L.C.M.

Hence, H.C.F. = 2 and L.C.M. = 23460.

(iii) On prime factorizing, we get :

336
= 2 × 168
= 2 × 2 × 84
= 2 × 2 × 2 × 42
= 2 × 2 × 2 × 2 × 21
= 2 × 2 × 2 × 2 × 3 × 7
= 2⁴ × 3 × 7

54
= 2 × 27
= 2 × 3 × 9
= 2 × 3 × 3 × 3
= 3³ × 2

H.C.F. (336, 54) = 2 × 3 = 6

L.C.M. (336, 54) = 2⁴ × 3³ × 7 = 3024.

Product of numbers = 336 × 54 = 18144

Product of H.C.F. and L.C.M. = 6 × 3024 = 18144.

∴ Product of numbers = Product of H.C.F. and L.C.M.

Hence, H.C.F. = 6 and L.C.M. = 3024.

Find the L.C.M. and H.C.F. 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) We have :

12 = 2² × 3

15 = 3 × 5

21 = 3 × 7.

Here, 3¹ are the smallest powers of the common factors 3, respectively.

So, HCF (12, 15, 21) = 3.

2², 3¹, 5¹ and 7¹ are the greatest powers of the prime factors 2, 3, 5 and 7 respectively.

So, LCM (12, 15, 21) = 2² × 3¹ × 5¹ × 7¹ = 4 × 3 × 5 × 7 = 420.

Hence, L.C.M. = 420 and H.C.F. = 3.

(ii) We have :

17 = 1 × 17

23 = 1 × 23

29 = 1 × 29.

Here, 1 is the only common factor.

So, HCF (17, 23, 29) = 1.

1¹, 17¹, 23¹ and 29¹ are the greatest powers of the prime factors 1, 17, 23 and 29 respectively.

So, LCM (17, 23, 29) = 1 × 17¹ × 23¹ × 29¹ = 1 × 17 × 23 × 29 = 11339.

Hence, L.C.M. = 11339 and H.C.F. = 1.

(iii) We have :

8 = 2³

9 = 3²

25 = 5².

Here, 1 is the only common factor.

So, HCF (8, 9, 25) = 1.

2³, 3², 5² are the greatest powers of the prime factors 2, 3, 5 respectively.

So, LCM (8, 9, 25) = 2³ × 3² × 5² = 8 × 9 × 25 = 1800.

Hence, L.C.M. = 1800 and H.C.F. = 1.

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

Answer

We know that,

Product of numbers = Product of H.C.F. and L.C.M.

306 × 657 = 9 × L.C.M.

L.C.M. = $\dfrac{306 \times 657}{9} = \dfrac{201042}{9}$ = 22338.

Hence, L.C.M. = 22338.

Check whether 6ⁿ can end with the digit 0 for any natural number n.

Answer

If the number 6ⁿ, for any natural number n, ends with digit 0, then it would be divisible by 5. That is, the prime factorization of 6ⁿ would contain the prime number 5. This is not possible because 6ⁿ = (2 x 3)ⁿ; so the only primes in the factorization of 6ⁿ are 2 and 3 and the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorization of 6ⁿ. So, there is no natural number n for which 6ⁿ ends with the digit zero.

Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

Answer

It can be observed that,

7 × 11 × 13 + 13 = 13 (7 × 11 + 1)

= 13(77 + 1)

= 13 × 78

= 13 × 13 × 6 × 1

= 13 × 13 × 2 × 3 × 1

The given number has 2, 3, 13, and 1 as its factors.

For the number to be prime, it should have only two factors — 1 and the number itself.

As the given number has more that two factors,

∴ 7 × 11 × 13 + 13 is a composite number.

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

= 5 × (1008 + 1)

= 5 × 1009 × 1

1009 cannot be factorized further.

The given number has 1, 5, and 1009 as its factors.

For the number to be prime, it should have only two factors — 1 and the number itself.

As the given number has more that two factors,

∴ 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is a composite number.

Hence, proved that 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

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

From the question, it is given that Ravi and Sonia start from the same point but take different time to complete the round. So, now we have to find the time taken by both of them to meet again at the starting point.

That means we have to find the LCM of 18 and 12.

Now we will find the LCM of 18 and 12 by prime factorization method:

18 = 2 × 3²

12 = 2² × 3

Now, prime factor 2 and 3 are common here. So, take prime factors with highest power i.e. 2² and 3².

L.C.M. = 2² × 3² = 4 × 9 = 36 minutes.

Hence, time taken by Ravi and Sonia to meet again at the starting point is 36 minutes.

Prove that $\sqrt{5}$ is irrational.

Answer

Let us assume, to the contrary, that $\sqrt{5}$ is a rational number.

If $\sqrt{5}$ is rational, that means it can be written in the form of $\dfrac{a}{b}$, where a and b are integers that have no common factor other than 1 and b ≠ 0. i.e., a and b are co-prime numbers.

$\Rightarrow \dfrac{\sqrt{5}}{1} = \dfrac{a}{b} \\[1em] \Rightarrow \sqrt{5}b = a$

Squaring both sides,

⇒ 5b² = a² ...........(1)

⇒ $b^2 = \dfrac{a^2}{5}$

This means 5 divides a².

From this, 5 also divides a.

Then a = 5c, for some integer 'c'.

On squaring, we get

⇒ a² = 25c²

Substituting above value of a² in equation (1)

⇒ 5b² = 25c²

⇒ b² = $\dfrac{25c^2}{5}$

⇒ b² = 5c²

⇒ c² = $\dfrac{b^2}{5}$

This means b² is divisible by 5 and so b is also divisible by 5. Therefore, a and b have 5 as common factor but this contradicts the fact that a and b are co-prime. This contradiction has arisen because of our incorrect assumption that $\sqrt{5}$ is a rational number.

Hence, proved that $\sqrt{5}$ is irrational.

Prove that 3 + $2\sqrt{5}$ is irrational.

Answer

Let us assume, to the contrary, that 3 + $2\sqrt{5}$ is a rational number.

So, it can be written in the form $\dfrac{a}{b}$.

3 + $2\sqrt{5} = \dfrac{a}{b}$ ..........(1)

Here a and b are coprime numbers and b ≠ 0.

Solving equation (1), we get :

$\Rightarrow 3 + 2\sqrt{5} = \dfrac{a}{b} \\[1em] \Rightarrow 2\sqrt{5} = \dfrac{a}{b} - 3 \\[1em] \Rightarrow 2\sqrt{5} = \dfrac{a - 3b}{b} \\[1em] \Rightarrow \sqrt{5} = \dfrac{a - 3b}{2b}.$

This shows that $\dfrac{a - 3b}{2b}$ is a rational number, but we know that $\sqrt{5}$ is an irrational number.

∴ Our assumption of $3 + 2\sqrt{5}$ is a rational number is incorrect.

Hence, proved that $3 + 2\sqrt{5}$ is irrational.

Prove that the following are irrationals :

(i) $\dfrac{1}{\sqrt{2}}$

(ii) $7\sqrt{5}$

(iii) 6 + $\sqrt{2}$

Answer

(i) Let us assume, to the contrary, that $\dfrac{1}{\sqrt{2}}$ is a rational number.

Then, $\dfrac{1}{\sqrt{2}} = \dfrac{a}{b}$, where a and b have no common factors other than 1.

Solving above equation,

$\Rightarrow b = \sqrt{2}a \\[1em] \Rightarrow \sqrt{2} = \dfrac{b}{a}$

Since b and a are integers, $\dfrac{b}{a}$ is a rational number and so, $\sqrt{2}$ is rational.

We know that $\sqrt{2}$ is irrational. So, our assumption was wrong.

Hence, proved that $\dfrac{1}{\sqrt{2}}$ is an irrational number.

(ii) Let us assume, to the contrary, that $7\sqrt{5}$ is a rational number.

Then, $7\sqrt{5} = \dfrac{a}{b}$, where a and b have no common factors other than 1.

$\Rightarrow 7\sqrt{5}b = a \\[1em] \Rightarrow \sqrt{5} = \dfrac{a}{7b}$

Since, a, 7, and b are integers, so, $\dfrac{a}{7b}$ is a rational number. This means $\sqrt{5}$ is rational but this contradicts the fact that $\sqrt{5}$ is irrational. So, our assumption was wrong.

Hence, proved that $7\sqrt{5}$ is an irrational number.

(iii) Let us assume, to the contrary, that $6 + \sqrt{2}$ is rational.

Then, $6 + \sqrt{2} = \dfrac{a}{b}$, where a and b have no common factors other than 1.

$\sqrt{2} = \dfrac{a}{b} - 6$

Since, a, b, and 6 are integers, so, $\dfrac{a}{b} - 6$ is a rational number. This means $\sqrt{2}$ is also a rational number.

This contradicts the fact that $\sqrt{2}$ is irrational. So, our assumption was wrong.

Hence, proved that $6 + \sqrt{2}$ is an irrational number.