Real Numbers

Exercise 1.1

Express each number as a product of its prime factors:

Question 1 (i): 140

To express 140 as a product of its prime factors, we break it down step by step:

Step 1: Start dividing by the smallest prime number (2)

140 ÷ 2 = 70 70 ÷ 2 = 35
(2 is no longer a factor)

Step 2: Try the next prime number (3)

35 ÷ 3 = not a whole number (skip)

Step 3: Try 5

35 ÷ 5 = 7 (7 is a prime number)

Final result: 140 = 2 × 2 × 5 × 7 Or in exponential form: 140 = 2² × 5 × 7

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Question 1 (ii): 156

Solution:

To express 156 as a product of its prime factors, follow these steps:

Prime Factorization of 156

Step 1: 156 is even, so divide by 2: 156 ÷ 2 = 78

Step 2: 78 is also even, divide by 2 again: 78 ÷ 2 = 39

Step 3: 39 is divisible by 3: 39 ÷ 3 = 13 (13 is a prime number)

Final Answer: 156 = 2² × 3 × 13

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Question 1 (iii): 3825

Solution:

Prime Factorization of 3825

Step 1: Divide by the smallest prime (start with 3)

3825 ÷ 3 = 1275

1275 ÷ 3 = 425

425 is not divisible by 3.

Step 2: Try the next prime number (5)

425 ÷ 5 = 85

85 ÷ 5 = 17

17 is a prime number.

Step 3: List all prime factors

So, 3825 = 3 × 3 × 5 × 5 × 17 Or in exponential form: 3825 = 3² × 5² × 17

_____________________________________________________________________________________Question 1 (iv): 5005

Solution:

Prime Factorization of 5005

Step 1: Start dividing by the smallest prime number.

5005 ÷ 5 = 1001 → 5 is a prime factor

Step 2: Factor 1001

1001 ÷ 7 = 143 → 7 is a prime factor

Step 3: Factor 143

143 ÷ 11 = 13 → 11 is a prime factor (13 is already a prime number)

Final Answer: 5005 = 5 × 7 × 11 × 13

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Question 1 (v): 7429

Solution:

Prime Factorization of 7429

We want to express 7429 as a product of its prime factors .

Step 1: Check divisibility by small prime numbers:

7429 is odd , so not divisible by 2.

7 + 4 + 2 + 9 = 22 → not divisible by 3.

It does not end in 0 or 5 → not divisible by 5.

7429 ÷ 7 = 1061.29 → not divisible by 7.

Try 17: 7429 ÷ 17 = 437 → this is a whole number

Step 2: Now factor 437 :

437 is odd, so not divisible by 2.

4 + 3 + 7 = 14 → not divisible by 3.

Does not end in 5 or 0 → not divisible by 5.

437 ÷ 19 = 23

Now, check if 23 is prime → Yes it is a prime number.

Final Prime Factorization:

7429 = 17 x 19 x 23

_____________________________________________________________________________________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

Solution:

Find the LCM and HCF of 26 and 91

Step 1: Prime Factorization

26 = 2 × 13

91 = 7 × 13

Step 2: HCF (Highest Common Factor)

The common prime factor is 13

So, HCF(26, 91) = 13

Step 3: LCM (Lowest Common Multiple)

Take the highest power of all prime factors : 2 (from 26), 7 (from 91), 13 (common in both)

So, LCM(26, 91) = 2 × 7 × 13 = 182

Step 4: Verification, According to the property:

LCM × HCF = Product of the two numbers

LCM × HCF = 182 × 13 = 2366

26 × 91 = 2366

Final Answer:

HCF = 13, LCM = 182

LCM × HCF = 2366 = 26 × 91

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(ii) 510 and 92

Solution:

Step 1: Find the HCF (Highest Common Factor)

We use prime factorization or division method.

92 = 2 × 2 × 23

510 = 2 × 3 × 5 × 17

Common factor: 2

So, HCF = 2

Step 2: Find the LCM (Lowest Common Multiple)

LCM = (product of both numbers) ÷ HCF

LCM = (510 × 92) ÷ 2

LCM = 46920 ÷ 2 = 23460

Step 3: Verify that LCM × HCF = product of the two numbers

LCM × HCF = 23460 × 2 = 46920

Product of the two numbers = 510 × 92 = 46920

Verified: LCM × HCF = product of the two numbers

_____________________________________________________________________________________Q2 (iii) 336 and 54

Solution:

Step 1: Find the HCF (using prime factorization)

Prime factorization of 336: 336 = 2 × 2 × 2 × 2 × 3 × 7 = 2⁴ × 3 × 7

Prime factorization of 54: 54 = 2 × 3 × 3 × 3 = 2 × 3³

Common prime factors: 2 and 3

HCF = 2 × 3 = 6

Step 2: Find the LCM

Take the highest powers of all prime factors:

LCM = 2⁴ × 3³ × 7 = 16 × 27 × 7 = 3024

Step 3: Verify that LCM × HCF = product of the numbers

LCM × HCF = 3024 × 6 = 18144

336 × 54 = 18144

Conclusion: LCM × HCF = 336 × 54 = 18144

_____________________________________________________________________________________Question 3:

Find the LCM and HCF of the following integers by applying the prime factorisation method.

(i) 12, 15 and 21

Solution:

Step 1: Prime factorization

12 = 2 × 2 × 3 = 2² × 3

15 = 3 × 5 = 3 × 5

21 = 3 × 7 = 3 × 7

Step 2: Find the HCF

Common prime factor in all three numbers is 3

HCF = 3

Step 3: Find the LCM

Take the highest powers of all prime factors present in any number:

LCM = 2² × 3 × 5 × 7 = 4 × 3 × 5 × 7 = 420

Conclusion:

HCF = 3

LCM = 420

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Q3: (ii) 17, 23 and 29

Solution:

Step 1: Prime Factorisation

17 is a prime number.

23 is a prime number.

29 is a prime number.

Step 2: Find HCF

Since all three numbers are prime and different, they have no common prime factors.

Therefore, HCF = 1.

Step 3: Find LCM

Since the numbers are prime and different, the LCM is the product of the numbers.

LCM = 17 × 23 × 29

First, multiply:

17 × 23 = 391

Then, 391 × 29 = 11339

Thus, LCM = 11339.

Final Answer: HCF = 1 ; LCM = 11339

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Q3 (iii) 8, 9, and 25

Solution:

Step 1: Prime Factorisation

8 = 2 × 2 × 2 = 2³

9 = 3 × 3 = 3²

25 = 5 × 5 = 5²

Step 2: Find the HCF (Highest Common Factor)

The HCF is found by taking the product of the lowest powers of common prime factors.

There are no common prime factors among 8, 9, and 25.

Thus, HCF = 1

Step 3: Find the LCM (Lowest Common Multiple)

The LCM is found by taking the product of the highest powers of all prime factors.

LCM = 2³ × 3² × 5²

LCM = 8 × 9 × 25

LCM = 72 × 25

Thus, LCM = 1800

Final Answer: HCF = 1 ; LCM = 1800

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Question 4

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

Solution:

We are given: HCF(306, 657) = 9

We know the relation between HCF and LCM of two numbers:

HCF × LCM = Product of the two numbers

Thus, 9 × LCM(306, 657) = 306 × 657

First, find the product: 306 × 657 = 201042

Now, divide by 9 to find the LCM:

LCM(306, 657) = 201042 ÷ 9

LCM(306, 657) = 22338

Final Answer: LCM(306, 657) = 22338

_____________________________________________________________________________________Question 5:

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

Solution:

For a number to end with the digit 0, it must be divisible by 10.

That is, the number must have both 2 and 5 as prime factors.

We know: 6 = 2 × 3

Thus, 6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ

Clearly, 6ⁿ has the factors 2 and 3, but it does not have the factor 5.

Since there is no factor 5 in 6ⁿ, it cannot be divisible by 10.

Thus, 6ⁿ can never end with the digit 0 for any natural number n.

Answer: No, 6ⁿ can never end with the digit 0 for any natural number n.

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Question 6:

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

Solution:

First expression: 7 × 11 × 13 + 13
Step 1: Factor out 13.
7 × 11 × 13 + 13 = 13 × (7 × 11 + 1)
= 13 × (77 + 1)
= 13 × 78

Step 2: Since 13 and 78 are both greater than 1, their product is composite.
Thus, 7 × 11 × 13 + 13 is a composite number.

Second expression: 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
Step 1: Simplify the factorial part.
7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040

Thus, the expression becomes:
5040 + 5 = 5045

Step 2: Check for factors.
5045 ÷ 5 = 1009

Thus,
5045 = 5 × 1009

Step 3: Since 5 and 1009 are both greater than 1, their product is composite.
Thus, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is a composite number.

Answer:
Both expressions are composite numbers because they can be expressed as the product of two natural numbers greater than 1.

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Question 7:

There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?

Solution:

They will meet again at the starting point after a time equal to the Least Common Multiple (LCM) of their times.

Sonia’s time = 18 minutes

Ravi’s time = 12 minutes

Now, find the LCM of 18 and 12.

Prime factors:

18 = 2 × 3 × 3

12 = 2 × 2 × 3

LCM is the product of the highest powers of all prime factors:

LCM = 2² × 3² = 4 × 9 = 36

Thus, they will meet again at the starting point after 36 minutes.

Answer:

They will meet again at the starting point after 36 minutes.

_____________________________________________________________________________________ EXERCISE 1.2

Question 1

Prove that √5 is irrational.

Solution:

Let us prove by contradiction.

Assume that √5 is rational.

Then, √5 can be written as a fraction a/b, where a and b are integers, b ≠ 0, and the fraction is in its lowest terms (no common factors).

Thus,

√5 = a/b

Squaring both sides,

5 = a²/b²

Multiplying both sides by b²,

5b² = a²

This means that a² is divisible by 5.

Thus, a must also be divisible by 5.

Let a = 5k, where k is an integer.

Substituting into the equation,

5b² = (5k)²

5b² = 25k²

Dividing both sides by 5,

b² = 5k²

This shows that b² is divisible by 5.

Thus, b must also be divisible by 5.

But this means that both a and b are divisible by 5, which contradicts our original assumption that a/b is in the lowest terms.

Therefore, our assumption is wrong.

Hence, √5 is irrational.

Answer: √5 is irrational.

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Question 2: Prove that 3 + 2√5 is irrational.

Solution:

Assume, for contradiction, that 3 + 2√5 is rational.

Then there exists a rational number r such that: r = 3 + 2√5

Subtract 3 from both sides: r - 3 = 2√5

Since r and 3 are rational, r - 3 is also rational.

Now divide both sides by 2: (r - 3) / 2 = √5

This implies that √5 is rational, which is false because √5 is irrational.

This contradiction shows that our assumption is false.

Therefore, 3 + 2√5 is irrational.

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Question: 3

(i) Prove that 1 / √2 is irrational.

Solution:

Assume, for contradiction, that 1 / √2 is rational.

Then we can write: 1 / √2 = a / b,

where a and b are integers, and b ≠ 0, with the fraction in lowest terms.

Now multiply both sides by √2: 1 = (a√2) / b

Multiply both sides by b: b = a√2

Now divide both sides by a: b / a = √2

This means √2 is rational, which is false because √2 is irrational.

This contradiction shows that our assumption is false.

Therefore, 1 / √2 is irrational.

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Question: 3 (ii) Prove that 7√5 is irrational.

Solution:

Assume, for contradiction, that 7√5 is rational.

Then we can write: 7√5 = r, where r is a rational number.

Now divide both sides by 7: √5 = r / 7

Since r is rational and 7 is rational, r / 7 is also rational.

This means √5 is rational, which is false because √5 is irrational.

This contradiction shows that our assumption is false.

Therefore, 7√5 is irrational.

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Question: 3 (iii) Prove that 6 + √2 is irrational.

Solution:

Assume, for contradiction, that 6 + √2 is rational.

Then we can write: 6 + √2 = r, where r is a rational number.

Now subtract 6 from both sides: √2 = r - 6

Since r and 6 are rational, r - 6 is also rational.

This means √2 is rational, which is false because √2 is irrational.

This contradiction shows that our assumption is false.

Therefore, 6 + √2 is irrational.

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