Exercise 1.1

1. Use Euclid’s division algorithm to find the HCF of:

i. 135 and 225

ii. 196 and 38220

iii. 867 and 255

1. Using Euclid’s Division Algorithm to Find the HCF:

i. HCF of 135 and 225: To find the highest common factor (HCF) of 135 and 225 using Euclid’s division algorithm:

• Start with the larger number, 225, and divide it by 135.
• 225 = 135 × 1 + 90.
• Now, take the divisor from the previous step, which is 135, and divide it by the remainder, 90.
• 135 = 90 × 1 + 45.
• Repeat the process with 90 and 45.
• 90 = 45 × 2 + 0.

Since we’ve reached a remainder of 0, the process stops. In the last step, the divisor was 45. Therefore, the HCF of 225 and 135 is 45.

ii. HCF of 196 and 38,220: To find the HCF of 196 and 38,220:

• Start with the larger number, 38,220, and divide it by 196.
• 38,220 = 196 × 195 + 0.

Since we obtained a remainder of 0 in the first step, the HCF of 196 and 38,220 is 196.

iii. HCF of 867 and 255: To find the HCF of 867 and 255:

• Start with the larger number, 867, and divide it by 255.
• 867 = 255 × 3 + 102.
• Now, divide 255 by the remainder, which is 102.
• 255 = 102 × 2 + 51.
• Again, divide 102 by 51.
• 102 = 51 × 2 + 0.

Since we reached a remainder of 0, the process stops. In the last step, the divisor was 51. Therefore, the HCF of 867 and 255 is 51.

2. Proof that Any Positive Odd Integer is of the Form 6q + 1, 6q + 3, or 6q + 5: To demonstrate that any positive odd integer can be expressed in one of the forms 6q + 1, 6q + 3, or 6q + 5, where q is some integer:

• Let a be any positive odd integer.
• Express it as a = 6q + r, where q is an integer and r is the remainder when dividing a by 6. Since a is odd, r can take values from 0 to 5.
• Depending on the value of r:
  • If r = 0, then a = 6q.
  • If r = 1, then a = 6q + 1.
  • If r = 2, then a = 6q + 2, which is even and not an odd integer.
  • If r = 3, then a = 6q + 3.
  • If r = 4, then a = 6q + 4, which is even.
  • If r = 5, then a = 6q + 5.

Since a must be odd, the valid forms are 6q + 1, 6q + 3, and 6q + 5, where q is any integer.

Show that any positive odd integer is of the form 6q + 1, 6q + 3, or 6q + 5, where q is some integer.

Solution:

Let’s consider any positive odd integer, denoted as ‘a,’ and set ‘b’ equal to 6. According to Euclid’s algorithm, we can express ‘a’ as:

a = 6q + r

Here, ‘q’ is some integer greater than or equal to 0, and ‘r’ can take values from 0 to 5 since it satisfies the condition 0 ≤ r < 6.

Now, let’s examine the possible values of ‘r’:

• If r = 0, then a = 6q.
• If r = 1, then a = 6q + 1.
• If r = 2, then a = 6q + 2, which is an even number and not a positive odd integer.
• If r = 3, then a = 6q + 3.
• If r = 4, then a = 6q + 4, which is an even number and not a positive odd integer.
• If r = 5, then a = 6q + 5.

Therefore, we can see that when r equals 0, 2, or 4, ‘a’ is an even number and does not meet the criteria for a positive odd integer. However, when r equals 1, 3, or 5, ‘a’ takes the forms 6q + 1, 6q + 3, and 6q + 5, respectively.

Since any positive integer can be expressed as either even or odd, we conclude that any positive odd integer must be in one of the forms 6q + 1, 6q + 3, or 6q + 5, where ‘q’ is an integer.

3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. Both groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

Solution:

Given: Number of army contingent members = 616 Number of army band members = 32

To determine the maximum number of columns in which both groups can march together, we need to find the highest common factor (HCF) between the two group sizes, which is denoted as HCF(616, 32).

We can use Euclid’s algorithm to find their HCF:

1. Since 616 is greater than 32, we start with the division: 616 = 32 × 19 + 8
2. As the remainder is 8 (which is not zero), we continue the process using 32 as the new divisor: 32 = 8 × 4 + 0

Since we have reached a remainder of 0, the HCF of 616 and 32 is 8.

Therefore, the maximum number of columns in which the army contingent and the army band can march together is 8.

This means that they can march in a maximum of 8 columns during the parade.

4.Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Solution:

Let x be any positive integer, and consider the divisor y = 3. According to Euclid’s division lemma, we can express x as:

x = 3q + r

Where q is an integer greater than or equal to 0, and r can be any value from 0 to 2 because it satisfies 0 ≤ r < 3.

Therefore, x can be written as:

• x = 3q
• x = 3q + 1
• x = 3q + 2

Now, as per the question, we square both sides of these equations to explore the forms of x^2:

1. For x = 3q: x^2 = (3q)^2 = 9q^2 Let 9q^2 = m Therefore, x^2 = 3m
2. For x = 3q + 1: x^2 = (3q + 1)^2 = 9q^2 + 2(3q) + 1 = 9q^2 + 6q + 1 Let 9q^2 + 6q + 1 = m Therefore, x^2 = 3m + 1
3. For x = 3q + 2: x^2 = (3q + 2)^2 = 9q^2 + 4(3q) + 4 = 9q^2 + 12q + 4 Notice that x^2 = 3(3q^2 + 4q + 1) + 1 Let 3q^2 + 4q + 1 = m Therefore, x^2 = 3m + 1

Hence, from these three cases (x^2 = 3m, x^2 = 3m + 1, x^2 = 3m + 1), we conclude that the square of any positive integer can indeed be expressed in either the form 3m or 3m + 1 for some integer m.

5.Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.

Solution:

Let’s consider a positive integer ‘x’ and set ‘y’ as 3. According to Euclid’s division lemma, we can express ‘x’ as:

x = 3q + r

Here, ‘q’ is a non-negative integer, and ‘r’ can only take values 0, 1, or 2 due to the constraint 0 ≤ r < 3.

Now, let’s explore the cube of ‘x’ for each of these cases:

When r = 0: x^3 = (3q)^3 = 27q^3 We can represent this as 9m, where m = 3q^3.

When r = 1: x^3 = (3q + 1)^3 = 27q^3 + 9q^2 + 3q + 1 Factoring out 9, we get: x^3 = 9(3q^3 + q^2 + q) + 1 This is of the form 9m + 1, where m = 3q^3 + q^2 + q.

When r = 2: x^3 = (3q + 2)^3 = 27q^3 + 54q^2 + 36q + 8 Factoring out 9, we get: x^3 = 9(3q^3 + 6q^2 + 4q) + 8 This is of the form 9m + 8, where m = 3q^3 + 6q^2 + 4q.

Therefore, in all three cases, we have shown that the cube of any positive integer ‘x’ can be expressed in one of the forms 9m, 9m + 1, or 9m + 8, where ‘m’ is an integer.

Exercise 1.2

1. Express each number as a product of its prime factors:

(i) 140

(ii) 156

(iii) 3825

(iv) 5005

(v) 7429

Solution

(i) 140:

To express 140 as a product of its prime factors:

140 = 2 × 2 × 5 × 7 = 2^2 × 5 × 7

(ii) 156:

To express 156 as a product of its prime factors:

156 = 2 × 2 × 13 × 3 = 2^2 × 13 × 3

(iii) 3825:

To express 3825 as a product of its prime factors:

3825 = 3 × 3 × 5 × 5 × 17 = 3^2 × 5^2 × 17

(iv) 5005:

To express 5005 as a product of its prime factors:

5005 = 5 × 7 × 11 × 13

(v) 7429:

To express 7429 as a product of its prime factors:

• 7 × 19 × 23

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

Solutions:

(i) 26 and 91: To find the LCM and HCF of 26 and 91:

First, express 26 and 91 as the product of their prime factors:

26 = 2 × 13 × 1 91 = 7 × 13 × 1

Now, calculate the LCM by taking the product of all unique prime factors raised to their highest power: LCM (26, 91) = 2 × 7 × 13 × 1 = 182

Next, determine the HCF by identifying the common prime factors with their lowest power: HCF(26, 91) = 13

Verification: Now, verify that LCM × HCF equals the product of 26 and 91: Product of 26 and 91 = 26 × 91 = 2366 Product of LCM and HCF = 182 × 13 = 2366

Hence, LCM × HCF equals the product of 26 and 91.

(ii) 510 and 92: To find the LCM and HCF of 510 and 92:

First, express 510 and 92 as the product of their prime factors:

510 = 2 × 3 × 17 × 5 × 1 92 = 2 × 2 × 23 × 1

Now, calculate the LCM by taking the product of all unique prime factors raised to their highest power: LCM (510, 92) = 2 × 2 × 3 × 5 × 17 × 23 = 23460

Next, determine the HCF by identifying the common prime factors with their lowest power: HCF (510, 92) = 2

Verification: Now, verify that LCM × HCF equals the product of 510 and 92: Product of 510 and 92 = 510 × 92 = 46920 Product of LCM and HCF = 23460 × 2 = 46920

Hence, LCM × HCF equals the product of 510 and 92.

(iii) 336 and 54: To find the LCM and HCF of 336 and 54:

First, express 336 and 54 as the product of their prime factors:

336 = 2 × 2 × 2 × 2 × 7 × 3 × 1 54 = 2 × 3 × 3 × 3 × 1

Now, calculate the LCM by taking the product of all unique prime factors raised to their highest power: LCM (336, 54) = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 1 = 3024

Next, determine the HCF by identifying the common prime factors with their lowest power: HCF (336, 54) = 2 × 3 = 6

Verification: Now, verify that LCM × HCF equals the product of 336 and 54: Product of 336 and 54 = 336 × 54 = 18144 Product of LCM and HCF = 3024 × 6 = 18144

Hence, LCM × HCF equals the product of 336 and 54.

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

Solutions:

(i) 12, 15 and 21

Writing the product of prime factors for all the three numbers, we get,

12=2×2×3

15=5×3

21=7×3

Therefore,

HCF(12,15,21) = 3

LCM(12,15,21) = 2 × 2 × 3 × 5 × 7 = 420

(ii) 17, 23 and 29

Writing the product of prime factors for all the three numbers, we get,

17=17×1

23=23×1

29=29×1

Therefore,

HCF (17,23,29) = 1

LCM (17,23,29) = 17 × 23 × 29 = 11339

(iii) 8, 9 and 25

Writing the product of prime factors for all the three numbers, we get,

8=2×2×2×1

9=3×3×1

25=5×5×1

Therefore,

HCF (8,9,25) =1

LCM (8,9,25) = 2×2×2×3×3×5×5 = 1800

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

Given that the highest common factor (HCF) of 306 and 657 is 9, we want to find the least common multiple (LCM) of these two numbers.

We can use the relationship between HCF, LCM, and the product of two numbers:

HCF × LCM = Product of the two given numbers

Substituting the values:

9 × LCM = 306 × 657

Now, solve for LCM:

LCM = (306 × 657) / 9 = 22338

Hence, the LCM of 306 and 657 is 22338.

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

Solution: To determine if 6^n can end with the digit 0 for any natural number n, we need to consider the divisibility by 5 because any number ending in 0 or 5 is divisible by 5.

We can express 6n as its prime factorization: 6^n = (2×3) ^n.

Notably, the prime factorization of 6^n doesn’t include the prime factor 5. In other words, it lacks the factors necessary for divisibility by 5.

Therefore, it is evident that for any natural number n, 6^n is not divisible by 5. Consequently, it proves that 6^n cannot end with the digit 0 for any natural number n.

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

Explanation 1:

To determine if 7 × 11 × 13 + 13 is a composite number, we need to check if it has factors other than 1 and itself.

We can factor out 13 from the expression:

7 × 11 × 13 + 13 = 13(77 + 1) = 13 × 78 = 13 × 3 × 2 × 13

This demonstrates that 7 × 11 × 13 + 13 has factors other than 1 and itself, specifically 13, 3, and 2 × 13.

Therefore, 7 × 11 × 13 + 13 is a composite number.

Explanation 2:

To determine if 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is a composite number, we must verify if it has factors other than 1 and itself.

By factoring out 5 from the expression:

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5(7 × 6 × 4 × 3 × 2 × 1 + 1) = 5(1008 + 1) = 5 × 1009

This demonstrates that 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 has factors other than 1 and itself, specifically 5 and 1009.

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

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?

Since Sonia and Ravi are moving in the same direction and starting at the same time, we can determine the time they will meet again at the starting point by finding the least common multiple (LCM) of their individual times.

LCM (18, 12) = 2 × 3 × 3 × 2 × 1 = 36

Hence, Sonia and Ravi will meet again at the starting point after 36 minutes.

Exercise 1.3

1.Prove that √5 is irrational.

Solution: Let’s assume, for the sake of contradiction, that √5 is a rational number. This means that we can express √5 as a fraction in its simplest form, where both the numerator and denominator share no common factors other than 1. We’ll denote this fraction as √5 = x/y, where x and y are co-prime integers.

Multiplying both sides by y, we get y√5 = x.

Now, squaring both sides of this equation:

(y√5) ^2 = x^2

This simplifies to:

5y^2 = x^2

From this equation, we can see that x^2 is divisible by 5, which implies that x itself must be divisible by 5.

So, we can express x as x = 5k, where k is some integer. Substituting this into the equation above:

5y^2 = (5k) ^2

This further simplifies to:

y^2 = 5k^2

From this, we can conclude that y^2 is divisible by 5, which means y must also be divisible by 5.

However, this contradicts our initial assumption that x and y are co-prime, meaning they have no common factors other than 1.

Thus, our assumption that √5 is a rational number must be incorrect. Therefore, √5 is proven to be an irrational number.

This proof demonstrates that the square root of 5 cannot be expressed as a fraction of two integers in their simplest form, confirming its irrationality.

2.Prove that 3+253+25​ is an irrational number.

To prove that 3 + 2√5 is irrational, we can use a proof by contradiction.

Assume that 3 + 2√5 is a rational number. This would mean it can be expressed as a fraction a/b, where a and b are integers with no common factors other than 1 (i.e., co-prime).

So, we have:

3 + 2√5 = a/b

Now, let’s isolate the irrational part, 2√5:

2√5 = a/b – 3

Squaring both sides to remove the square root:

4 * 5 = (a/b – 3) ^2

This simplifies to:

20 = (a/b – 3) ^2

Expanding the right side:

20 = a^2/b^2 – 6a/b + 9

Now, multiplying both sides by b^2 to get rid of the fractions:

20b^2 = a^2 – 6ab + 9b^2

Rearrange:

0 = a^2 – 6ab – 11b^2

Now, notice that the left side is 0. If 3 + 2√5 were rational, then the right side would also be 0, as both sides would be equal. However, we have a quadratic equation in the form of ax^2 + bx + c = 0, where a = 1, b = -6b, and c = -11b^2.

The discriminant of this quadratic equation (b^2 – 4ac) is:

(-6b) ^2 – 4(1) (-11b^2) = 36b^2 + 44b^2 = 80b^2

For the equation to have real roots, the discriminant must be a perfect square. However, in this case, the discriminant is 80b^2, which isn’t a perfect square unless b = 0. But since we assumed that a and b are co-prime, b cannot be zero.

Therefore, we have reached a contradiction. Our assumption that 3 + 2√5 is rational must be incorrect. Hence, 3 + 2√5 is indeed an irrational number.

3.Prove that the following are irrationals:

(i) 1/√2

(ii) 7√5

(iii) 6 + √2

Solutions:

(I) 1/√2

Let us assume 1/√2 is rational.

Then we can find co-prime x and y (y ≠ 0) such that 1/√2 = x/y

After Rearranging, we get,

√2 = y/x

Since, x and y are integers, thus, √2 is a rational number, which contradicts the fact that √2 is irrational.

Hence, we can conclude that 1/√2 is irrational.

(ii) 7√5

Let us assume 7√5 is a rational number.

Then we can find co-prime a and b (b ≠ 0) such that 7√5 = x/y

After Rearranging, we get,

√5 = x/7y

Since, x and y are integers, thus, √5 is a rational number, which contradicts the fact that √5 is irrational.

Hence, we can conclude that 7√5 is irrational.

(iii) 6 +√2

Let us assume 6 +√2 is a rational number.

Then we can find co-primes x and y (y ≠ 0) such that 6 +√2 = x/y⋅

After Rearranging, we get,

√2 = (x/y) – 6

Since, x and y are integers, thus (x/y) – 6 is a rational number and therefore, √2 is rational. This contradicts the fact that √2 is an irrational number.

Hence, we can conclude that 6 +√2 is irrational.

Exercise 1.4

1. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:

(i) 13/3125 (ii) 17/8 (iii) 64/455 (iv) 15/1600

(v) 29/343 (vi) 23/(2352) (vii) 129/(225775)

(viii) 6/15 (ix) 35/50 (x) 77/210

(i) 13/3125:

The denominator is 3125, which can be factored as 5⁵. This implies that the denominator has only factors of 5. Consequently, 13/3125 has a terminating decimal expansion since it can be expressed as a finite decimal.

(ii) 17/8:

The denominator is 8, which can be factored as 2³. So,This means the denominator contains only factors of 2. Therefore, 17/8 has a terminating decimal expansion because it can be expressed as a finite decimal.

(iii) 64/455:

The denominator is 455, which can be factored into 5, 7, and 13. It includes factors other than 2 and 5, namely 7 and 13. Hence, 64/455 has a non-terminating decimal expansion because it cannot be expressed as a finite decimal.

(iv) 15/1600:

The denominator is 1600, which can be factored as 2⁶ × 5². This indicates that the denominator is in the form of 2^m × 5ⁿ. Therefore, 15/1600 has a terminating decimal expansion as it can be expressed as a finite decimal.

(v) 29/343:

The denominator is 343, which can be factored into 7 and 7 again (7³). It contains factors other than 2 and 5, specifically 7. Thus, 29/343 has a non-terminating decimal expansion as it cannot be expressed as a finite decimal.

(vi) 23/(2352):

The denominator is 2352, which can be factored as 2⁴ × 3 × 7 × 7. This demonstrates that the denominator is in the form of 2^m × 5ⁿ, making 23/(2352) have a terminating decimal expansion as it can be expressed as a finite decimal.

(vii) 129/(225775):

The denominator is 225775 and contains factors other than 2 and 5, making it a non-terminating decimal expansion since it cannot be expressed as a finite decimal.

(viii) 6/15:

The denominator is 15, which is solely a factor of 5. Therefore, 6/15 has a terminating decimal expansion since it can be expressed as a finite decimal.

(ix) 35/50:

The denominator is 50, which can be factored as 2 × 5, signifying that it is in the form of 2^m × 5ⁿ. Thus, 35/50 has a terminating decimal expansion as it can be expressed as a finite decimal.

(x) 77/210:

The denominator is 210, which can be factored into 2, 3, 5, and 7. It contains factors other than 2 and 5, specifically 3 and 7. Consequently, 77/210 has a non-terminating decimal expansion as it cannot be expressed as a finite decimal.

2 .Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.

(i) 13/3125

         0.00416
     ______________
3125 | 13.00000
          - 0
         ____
            13000
          - 12500
         ______
             5000
           - 5000
         _______
                000

13/3125 = 0.00416

(ii) 17/8

     2.125
   ------------
8 | 17.000
   - 16
   ------
       10
       - 8
       ----
         20
         - 16
         -----
           40
           - 40
           -----
             0

17/8 = 2.125

(iii) 64/455 is a non-terminating decimal expansion

(iv)15/ 1600

          0.009375
     ________________
1600 | 15.000000
           - 0
     ________________
           15000
         - 14400
     ________________
             6000
           - 5760
     ________________
              2400
            - 2400
     ________________
                 0

So, the decimal expansion of 15/1600 is 0.009375.

(v) 29/ 343, a non terminating decimal expansion

(vi) 23/(2^3*5^2)

(23/ (2³5²) = 23/(8×25)= 23/200

              0.115
     _______________
200 | 23.0000
            - 0
     _______________
             2300
           - 2000
     _______________
              3000
           - 2300
     _______________
              7000
           - 6000
     _______________
              1000
            - 1000
     _______________
                 0

23/ (2³5²) = 0.115

(vii) 129/ (2²5⁷7⁵), a non terminating decimal expansion

     0.4
   -----
5 | 2.00
     - 0
   -----
        0

So, the decimal expansion of 2/5 is 0.4.

(ix) 35/50 = 7/10

    0.7
   -----
10 | 7.00
     - 0
   -----
       7000
     - 7000
   -----
          0

So, the decimal expansion of 7/10 is 0.7.

(x) 77/210, a non-terminating decimal expansion.

3. The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form, p q what can you say about the prime factors of q?

Solutions:

(i) 43.123456789

Since it has a terminating decimal expansion, it is a rational number in the form of p/q, where q can have factors other than just 2 and 5.

(ii) 0.120120012000120000. . .

Since, it has non-terminating and non- repeating decimal expansion, it is an irrational number.