Best ML Aggarwal Class 9 Chapter 1 Solutions

ML Aggarwal Class 9 Chapter 1 Solutions (Rational and Irrational Numbers)” pertains to a segment of the mathematics curriculum designed for Class 9 students. This chapter delves into the exploration of rational and irrational numbers, two fundamental types of real numbers that play a crucial role in various mathematical contexts. The chapter aims to provide students with a clear understanding of the distinctions between these types of numbers, as well as how they interact in mathematical operations.

ML Aggarwal Solutions for Class 9 Maths Chapter 1 PDF

EXERCISE 1.1 (ML Aggarwal Class 9 Chapter 1 Solutions)

1. Insertion Between 2/9 and 3/8, Descending Order:

To insert a rational number between 2/9 and 3/8 while arranging them in descending order, we turn to the LCM method to ensure a common denominator for these fractions.

• LCM of 9 and 8: The LCM of the denominators 9 and 8 is 72. This common denominator guarantees that the fractions share the same base for further calculations.
• Equivalent fractions: We convert both fractions to their equivalent forms with a common denominator of 72:
  • 2/9 = (2/9) * (8/8) = 16/72
  • 3/8 = (3/8) * (9/9) = 27/72
• Insertion calculation: Since we want to insert a rational number between these two fractions, we consider the average of 16/72 and 27/72: (16/72 + 27/72) / 2 = 43/72.
• Descending order: Arranging them in descending order, we get 27/72, 43/72, and 16/72.

2. Pairing Between 1/3 and 1/4, Ascending Order:

Inserting rational numbers between 1/3 and 1/4 while arranging in ascending order requires finding a common denominator through the LCM method.

• LCM of 3 and 4: The LCM of the denominators 3 and 4 is 12.
• Equivalent fractions: We convert both fractions to equivalent forms with a common denominator of 12:
  • 1/3 = (1/3) * (4/4) = 4/12
  • 1/4 = (1/4) * (3/3) = 3/12
• Insertion calculation: To insert rational numbers, we use a common difference between fractions. In this case, the difference is 1/12 (12 divided by 12).
• Insertion and arrangement: Adding 1/12 and then 2/12 (simplified as 1/6) to 4/12 gives us the sequence 4/12, 5/12, 6/12 (or 1/3), 7/12, 8/12 (or 2/3), 9/12 (or 3/4).

3. Inclusion Between -1/3 and -1/2, Ascending Order:

Incorporating rational numbers between -1/3 and -1/2 while arranging in ascending order calls for the LCM technique to find a common denominator.

• LCM of 3 and 2: The LCM of the denominators 3 and 2 is 6.
• Equivalent fractions: We convert both fractions to equivalent forms with a common denominator of 6:
  • -1/3 = (-1/3) * (2/2) = -2/6
  • -1/2 = (-1/2) * (3/3) = -3/6
• Insertion calculation: Similar to the previous question, we insert rational numbers using a common difference, which is 1/6 (6 divided by 6).
• Insertion and arrangement: Adding 1/6 and then 2/6 (simplified as 1/3) to -2/6 gives us the sequence -2/6, -5/6, -8/6 (or -1/2).

ML Aggarwal Class 9 Chapter 1 Solutions

4. Filling Between 1/3 and 4/5, Descending Order:

Filling rational numbers between 1/3 and 4/5 while arranging in descending order requires establishing a common denominator through LCM.

• LCM of 3 and 5: The LCM of the denominators 3 and 5 is 15.
• Equivalent fractions: We convert both fractions to equivalent forms with a common denominator of 15:
  • 1/3 = (1/3) * (5/5) = 5/15
  • 4/5 = (4/5) * (3/3) = 12/15
• Insertion calculation: Once again, we insert rational numbers using a common difference, which is 1/15 (15 divided by 15).
• Insertion and arrangement: Adding 1/15, then 2/15, and finally 3/15 (or 1/5) successively to 5/15 (or 1/3) gives us the sequence 5/15, 6/15 (or 2/5), 7/15, 8/15, 9/15 (or 3/5), 10/15 (or 2/3), 11/15, 12/15 (or 4/5).

5. Intervals Between 4 and 4.5:

The task here is to insert rational numbers between 4 and 4.5 using equal intervals.

• Intervals: The difference between 4 and 4.5 is 0.5. To divide this range into 3 intervals, we use 0.5 / 3 = 0.166… (repeating decimal).
• Insertion and arrangement: Adding 0.166… successively to 4 gives us the sequence 4, 4.166…, 4.333…, and finally 4.5.

6. Navigating Between 3 and 4:

To navigate between 3 and 4 while inserting rational numbers and arranging them, we use the LCM method to establish common intervals.

• Intervals: The difference between 3 and 4 is 1. To divide this range into 6 intervals, we use 1 / 6 = 1/6.
• Insertion and arrangement: Adding 1/6 successively to 3 gives us the sequence 3, 13/6, 7/3, 5/2, 17/6, and finally 4.

ML Aggarwal Class 9 Chapter 1 Solutions questions answers

7. Bridging Between 3/5 and 4/5:

To bridge rational numbers between 3/5 and 4/5 while arranging them, we once again use the LCM technique for common intervals.

• LCM of 5: The LCM of the denominator 5 is 5.
• Equivalent fractions: We convert both fractions to equivalent forms with a common denominator of 5:
  • 3/5 = (3/5) * (1/1) = 3/5
  • 4/5 = (4/5) * (1/1) = 4/5
• Intervals: The difference between 4/5 and 3/5 is 1/5. To divide this range into 5 intervals, we use 1/5 / 5 = 1/25.
• Insertion and arrangement: Adding 1/25 successively to 3/5 gives us the sequence 3/5, 11/25, 13/25, 21/25, and finally 4/5.

8. Discovery Amidst -2/5 and 1/7:

In this complex scenario, discovering rational numbers between -2/5 and 1/7 while arranging them requires careful application of the LCM approach.

• LCM of 5 and 7: The LCM of the denominators 5 and 7 is 35.
• Equivalent fractions: We convert both fractions to equivalent forms with a common denominator of 35:
  • -2/5 = (-2/5) * (7/7) = -14/35
  • 1/7 = (1/7) * (5/5) = 5/35
• Intervals: The difference between 5/35 and -14/35 is 19/35. To divide this range into 11 intervals, we use 19/35 / 11 = 19/385.
• Insertion and arrangement: Adding 19/385 successively to -14/35 gives us the sequence -14/35, -33/385, -14/77, -7/55, -4/35, -33/385, -14/105, -5/77, -2/35, -19/385, and finally 5/35.

ML Aggarwal Class 9 Chapter 1 Solutions

9. Journeying Between 1/2 and 2/3:

The final task is to journey between 1/2 and 2/3 by inserting rational numbers and arranging them.

• LCM of 2 and 3: The LCM of the denominators 2 and 3 is 6.
• Equivalent fractions: We convert both fractions to equivalent forms with a common denominator of 6:
  • 1/2 = (1/2) * (3/3) = 3/6
  • 2/3 = (2/3) * (2/2) = 4/6
• Intervals: The difference between 4/6 and 3/6 is 1/6. To divide this range into 6 intervals, we use 1/6.
• Insertion and arrangement: Adding 1/6 successively to 3/6 gives us the sequence 3/6, 4/6, 5/6.

EXERCISE 1.2

1. Prove that √5 is an irrational number:

Suppose √5 can be expressed as a fraction p/q, where p and q are integers with no common factors except 1, and q ≠ 0.

Squaring both sides of the equation: 5 = p^2/q^2.

Rearranging the equation: p^2 = 5q^2.

Since p^2 is divisible by 5, p must be divisible by 5. Let’s write p as p = 5k, where k is an integer.

Substituting p = 5k into the equation p^2 = 5q^2, we get (5k)^2 = 5q^2, which simplifies to k^2 = q^2.

This implies that q^2 is divisible by 5, so q must also be divisible by 5.

Contradiction: Both p and q are divisible by 5, which contradicts the assumption that they have no common factors except 1.

Conclusion: Since this leads to a contradiction, √5 cannot be rational. Therefore, √5 is irrational.

2. Prove that √7 is an irrational variety number:

Suppose √7 can be expressed as a fraction p/q, where p and q are integers with no common factors except 1, and q ≠ 0.

Squaring both sides of the equation: 7 = p^2/q^2.

Rearranging the equation: p^2 = 7q^2.

Analogous to the previous proofs, we analyze the factors and find a contradiction, showing that √7 cannot be rational.

ML Aggarwal Class 9 Chapter 1 Solutions

3. Prove that √6 is an irrational variety number:

lets if √6 can be expressed as a fraction p/q, where p and q are integers with no common factors except 1, and q ≠ 0.

Squaring both sides of the equation: 6 = p^2/q^2.

Rearranging the equation: p^2 = 6q^2.

Again, by analyzing factors, we reach a contradiction, establishing the irrationality of √6.

4. Prove that 1/√11 is an irrational variety number:

Suppose 1/√11 can be expressed as a fraction p/q, where p and q are integers with no common factors except 1, and q ≠ 0.

Squaring both sides of the equation: 1/11 = p^2/q^2.

Rearranging the equation: p^2 = 11q^2.

By analyzing factors, we reach a contradiction, showing that 1/√11 cannot be rational.

5. Prove that √2 is an irrational number. when show that 3 — √2 is an irrational:

Step 1: Prove √2 is Irrational Suppose √2 is rational and can be expressed as a fraction p/q, where p and q are integers with no common factors except 1, and q ≠ 0.

Step 2: Square Both Sides Squaring both sides of the equation: 2 = p^2/q^2.

Step 3: Rearrange and Analyze Rearrange the equation to get p^2 = 2q^2.

Step 4: Analyze Divisibility Since p^2 is divisible by 2, p must be divisible by 2. Let’s write p as p = 2k, where k is an integer.

Step 5: Substitute and Simplify Substitute p = 2k into the equation p^2 = 2q^2, which gives (2k)^2 = 2q^2. Simplify to 4k^2 = 2q^2.

Step 6: Analyze Divisibility Again This implies that q^2 is divisible by 2, so q must also be divisible by 2.

Contradiction: Both p and q are divisible by 2, which contradicts the assumption that they have no common factors except 1.

Conclusion: Since this leads to a contradiction, √2 cannot be rational. Therefore, √2 is irrational.

Hence, by a similar contradiction, 3 — √2 is also irrational.

ML Aggarwal Class 9 Chapter 1 Solutions

6. Proving √3 is Irrational variety: Showing 2/5×√3 is Irrational:

Assumption: Suppose 2/5×√3 is rational and can be expressed as a fraction p/q, where p and q are integers with no common factors except 1, and q ≠ 0.

Step 1: Express as a Fraction 2/5×√3 = p/q

Step 2: Cross-Multiply 2q√3 = 5p

Step 3: Analyze the Left Side Since 2q√3 is irrational (we’ve proven that √3 is irrational), it cannot be equal to a rational number like 5p.

Contradiction: We’ve reached a contradiction by assuming that 2/5×√3 is rational. Therefore, 2/5×√3 must be irrational.

ML Aggarwal Class 9 Chapter 1 Solutions

7. prove the irrationality of √5 and then extend the proof to show that -3 + 2√5 is also irrational:

Showing -3 + 2√5 is Irrational:

Assumption: Suppose -3 + 2√5 is rational and can be expressed as a fraction p/q, where p and q are integers with no common factors except 1, and q ≠ 0.

Step 1: Express as a Fraction -3 + 2√5 = p/q

Step 2: Isolate √5 2√5 = p/q + 3

Step 3: Square Both Sides 20 = p^2/q^2 + 6p/q + 9

Step 4: Analyze the Left Side Since 20 is an integer and the right side has fractions, it’s not possible for them to be equal.

Contradiction: We’ve reached a contradiction by assuming that -3 + 2√5 is rational. Therefore, -3 + 2√5 must be irrational.

8. prove the irrationality of each of these numbers:

(i) 5 + √2 (ii) 3 – 5√3 (iii) 2√3 – 7 (iv) √2 +√5

(i) 5 + √2:

Assumption: Suppose 5 + √2 is rational and can be expressed as a fraction p/q, where p and q are integers with no common factors except 1, and q ≠ 0.

Step 1: Express as a Fraction 5 + √2 = p/q

Step 2: Isolate √2 √2 = p/q – 5

Step 3: Square Both Sides 2 = p^2/q^2 – 10p/q + 25

Step 4: Analyze the Left Side Since 2 is an integer and the right side has fractions, it’s not possible for them to be equal.

Contradiction: We’ve reached a contradiction by assuming that 5 + √2 is rational. Therefore, 5 + √2 must be irrational.

(ii) 3 – 5√3:

Assumption: Suppose 3 – 5√3 is rational and can be expressed as a fraction p/q, where p and q are integers with no common factors except 1, and q ≠ 0.

Step 1: Express as a Fraction 3 – 5√3 = p/q

Step 2: Isolate √3 -5√3 = p/q – 3

Step 3: Square Both Sides 75 = p^2/q^2 – 6p/q + 9

Step 4: Analyze the Left Side Since 75 is an integer and the right side has fractions, it’s not possible for them to be equal.

Contradiction: We’ve reached a contradiction by assuming that 3 – 5√3 is rational. Therefore, 3 – 5√3 must be irrational.

ML Aggarwal Class 9 Chapter 1 Solutions

(iii) 2√3 – 7:

Assumption: Suppose 2√3 – 7 is rational and can be expressed as a fraction p/q, where p and q are integers with no common factors except 1, and q ≠ 0.

Step 1: Express as a Fraction 2√3 – 7 = p/q

Step 2: Isolate √3 2√3 = p/q + 7

Step 3: Square Both Sides 12 = p^2/q^2 + 14p/q + 49

Step 4: Analyze the Left Side Since 12 is an integer and the right side has fractions, it’s not possible for them to be equal.

Contradiction: We’ve reached a contradiction by assuming that 2√3 – 7 is rational. Therefore, 2√3 – 7 must be irrational.

(iv) √2 + √5:

Assumption: Suppose √2 + √5 is rational and can be expressed as a fraction p/q, where p and q are integers with no common factors except 1, and q ≠ 0.

Step 1: Express as a Fraction √2 + √5 = p/q

Step 2: Isolate √5 √5 = p/q – √2

Step 3: Square Both Sides 5 = p^2/q^2 – 2p/q + 2

Step 4: Analyze the Left Side Since 5 is an integer and the right side has fractions, it’s not possible for them to be equal.

Contradiction: We’ve reached a contradiction by assuming that √2 + √5 is rational. Therefore, √2 + √5 must be irrational.

EXERCISE 1.3

locate √10 and √17 on the number line, we can use their approximate values to determine their positions:

√10:

• √10 is between 3 and 4 because √9 = 3 and √16 = 4.
• Calculating the square root of 10 gives an approximate value of √10 ≈ 3.16.
• Therefore, √10 is slightly greater than 3 but less than 4. It falls between these two integers on the number line.

√17:

• √17 is between 4 and 5 because √16 = 4 and √25 = 5.
• Calculating the square root of 17 gives an approximate value of √17 ≈ 4.12.
• Therefore, √17 is slightly greater than 4 but less than 5. It falls between these two integers on the number line.

In summary, √10 is located between 3 and 4 on the number line, while √17 is located between 4 and 5.

2. Explanation for each question along with the requested decimal expansion and the type of decimal expansion:

(i) 36/100

(ii) 4 1/8

(iii) 2/9

(iv) 2/11

(v) 3/13

(vi) 329/400

(i) 36/100: Question: Write the decimal expansion of 36/100 and identify its type of decimal expansion.

       0.36
   --------------
100 | 36.000000
      - 0
      --------
          0

Explanation: To convert the fraction 36/100 into a decimal, we divide 36 by 100:

36 ÷ 100 = 0.36

The decimal expansion of 36/100 is 0.36. This decimal is a terminating decimal because it has a finite number of digits after the decimal point.

(ii) 4 1/8: Question: Write the decimal expansion of 4 1/8 and identify its type of decimal expansion.

Explanation: To convert the mixed number 4 1/8 into a decimal, we first convert the whole number and the fraction separately and then combine them:

4 + 1/8 = 4.125

The decimal expansion of 4 1/8 is 4.125. This decimal is a terminating decimal because it has a finite number of digits after the decimal point.

ML Aggarwal Class 9 Chapter 1 Solutions

(iii) 2/9: Question: Write the decimal expansion of 2/9 and identify its type of decimal expansion.

Explanation: To convert the fraction 2/9 into a decimal, we divide 2 by 9:

2 ÷ 9 = 0.222…

The decimal expansion of 2/9 is 0.222…, where the digit 2 repeats infinitely. This type of decimal is called a repeating decimal.

(iv) 2/11: Question: Write the decimal expansion of 2/11 and identify its type of decimal expansion.

Explanation: To convert the fraction 2/11 into a decimal, we divide 2 by 11:

2 ÷ 11 = 0.181818…

The decimal expansion of 2/11 is 0.181818…, where the sequence 18 repeats infinitely. This type of decimal is a repeating decimal.

(v) 3/13: Question: Write the decimal expansion of 3/13 and identify its type of decimal expansion.

Explanation: To convert the fraction 3/13 into a decimal, we divide 3 by 13:

3 ÷ 13 = 0.230769…

The decimal expansion of 3/13 is 0.230769…, where the sequence 230769 repeats infinitely. This type of decimal is a repeating decimal.

(vi) Write the decimal expansion of 329/400.

       0.8225
   --------------
400 | 329.000000
      - 320
      --------
          90
        - 80
        ------
          100
         - 100
        --------
             0

3. 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) 23/75

(iv) 6/15

(v) 1258/625

(vi) 77/210

ML Aggarwal Class 9 Chapter 1 Solutions

Absolutely! Here’s the explanation for each of the provided rational numbers, stating whether they will have a terminating decimal expansion or a non-terminating repeating decimal expansion:

(i) 13/3125: The denominator of 3125 can be expressed as 5^5. Since the denominator’s prime factorization includes only the prime factor 5, the fraction will have a terminating decimal expansion. Dividing by powers of 10 (which is 2 * 5) results in a finite number of digits after the decimal point. Therefore, the decimal expansion of 13/3125 will be terminating.

(ii) 17/8: The denominator of 8 can be expressed as 2^3. As it only involves the prime factor 2, the fraction will have a terminating decimal expansion. The division by powers of 10 results in a finite number of digits after the decimal point. Hence, the decimal expansion of 17/8 will be terminating.

(iii) 23/75: The denominator of 75 can be factored as 3 * 5^2. Since it includes the prime factors 3 and 5, the fraction’s decimal expansion will be non-terminating and repeating. When dividing by powers of 10, which involve 2 and 5, the division result will produce an infinite sequence of repeating digits. Thus, the decimal expansion of 23/75 will be non-terminating repeating.

(iv) 6/15: The denominator of 15 can be factored as 3 * 5. As it includes the prime factors 3 and 5, the fraction’s decimal expansion will be non-terminating and repeating. Since powers of 10 include both 2 and 5, the division will yield a repeating decimal expansion. Therefore, the decimal expansion of 6/15 will be non-terminating repeating.

(v) 1258/625: The denominator of 625 can be expressed as 5^4. Since it only involves the prime factor 5, the fraction will have a terminating decimal expansion. Dividing by powers of 10 will result in a finite number of digits after the decimal point. Hence, the decimal expansion of 1258/625 will be terminating.

(vi) 77/210: The denominator of 210 can be factored as 2 * 3 * 5 * 7. Since it includes prime factors other than 2 and 5, the fraction’s decimal expansion will be non-terminating and repeating. The division by powers of 10 will result in an infinite sequence of repeating digits. Consequently, the decimal expansion of 77/210 will be non-terminating repeating.

ML Aggarwal Class 9 Chapter 1 Solutions