Numbers System

The Number System chapter in Class 9 Mathematics introduces and expands upon the concept of numbers, building on what you learned in previous grades.

Key Concepts

• Types of Numbers:
  
  • Natural Numbers: Counting numbers starting from 1 (1, 2, 3, …)
  • Whole Numbers: Includes 0 along with natural numbers (0, 1, 2, 3, …)
  • Integers: Positive and negative whole numbers, including zero (…, -2, -1, 0, 1, 2, …)
  • Rational Numbers: Numbers that can be expressed in the form p/q, where p and q are integers and q ≠ 0. These include terminating and recurring decimals.
  • Irrational Numbers: Numbers that cannot be expressed in the form p/q. They are non-terminating and non-repeating decimals (e.g., √2, π).
  • Real Numbers: The combination of rational and irrational numbers.
• Number Line: A visual representation of numbers.
  
• Representing Numbers on the Number Line: Plotting different types of numbers on the number line.
  
• Operations on Real Numbers: Addition, subtraction, multiplication, and division of real numbers.
  
• Properties of Real Numbers: Commutative, associative, distributive, and identity properties.
  
• Rationalization: Converting a denominator with a radical into a rational number.
  
• Laws of Radicals: Rules for simplifying expressions with radicals.
  

By understanding these concepts and properties, you can perform various mathematical operations and solve problems involving numbers effectively.

Exercise 9.1

1. Is zero a rational number? Can you write it in the form p/q ,where p and q are integers and q ≠0?

Ans : 

Yes, zero is a rational number.

For zero, we can write it as:

• 0/1
• 0/2
• 0/3
• … and so on

In all these cases, the numerator is an integer (0) and the denominator is a non-zero integer. 

Hence, zero is indeed a rational number.

2. Find six rational numbers between 3 and 4.

Ans : 

To find six rational numbers between 3 and 4, we can express 3 and 4 as fractions with a common denominator.

Since we need six numbers, let’s convert 3 and 4 into fractions with a denominator of 7.

• 3 = 21/7
• 4 = 28/7

Now, we can easily find six rational numbers between 21/7 and 28/7:

• 22/7
• 23/7
• 24/7
• 25/7
• 26/7
• 27/7

3. Find five rational numbers between 3/5 and ⅘

Ans : 

We need to express both fractions with a common denominator.

Let’s use a common denominator of 30.

• 3/5 = (3 * 6) / (5 * 6) = 18/30
• 4/5 = (4 * 6) / (5 * 6) = 24/30

Now we can easily find five rational numbers between 18/30 and 24/30:

• 19/30
• 20/30
• 21/30
• 22/30
• 23/30

Therefore, five rational numbers between 3/5 and 4/5 are 19/30, 20/30, 21/30, 22/30, and 23/30.

4. State whether the following statements are true or false. Give reasons for your answers.

(i) Every natural number is a whole number.

(ii) Every integer is a whole number.

(iii) Every rational number is a whole number.

Ans : 

(i)True.

• Natural numbers start from 1 (1, 2, 3, …)

(ii)False.

• Integers include negative numbers (…, -2, -1, 0, 1, 2, …)
• Whole numbers are only non-negative (0, 1, 2, 3, …)
• Therefore, not all integers are whole numbers.

(iii)False.

• Whole numbers are a subset of rational numbers where the denominator q is always 1.
• Many rational numbers like 1/2, 3/4, -2/5 are not whole numbers.

Exercise 1.2

1. State whether the following statements are true or false. Justify your answers.

(i) Every irrational number is a real number.

(ii) Every point on the number line is of the form √m , where m is a natural number.

(iii) Every real number is an irrational number.

Ans : 

(i) True.

• Because the set of all rational numbers combined with all irrational numbers constitutes the group of real numbers.

(ii) False.

• Because no natural number can have a negative square root.

(iii) False.

• Because rational numbers are included within the set of real numbers.

2. Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational number.

Ans : No, because if we take a positive integer, like 9, its square root is 3, which is a rational number.

3. Show how √5 can be represented on the number line.

Ans : 

Draw a number line: Draw a horizontal line and mark points at integer intervals.

Construct a right-angled triangle:

• From point 2 on the number line, draw a perpendicular line segment of length 1 unit upwards. Label the endpoint as B.
• Join point A (0 on the number line) to point B. This line segment AB is the hypotenuse of the right-angled triangle ABC, where C is the point where the perpendicular intersects the number line.

Apply Pythagoras theorem:

• In right-angled triangle ABC, AB² = AC² + BC²
• AB² = 2² + 1² = 5
• AB = √5

Transfer the length to the number line:

• With point A as the center and AB as the radius, draw an arc that intersects the number line at point D.
• Point D represents √5 on the number line.

Exercise 1.3

Write the following in decimal form and say what kind of decimal expansion each has

Ans : 

i) 36/100

• Decimal form: 0.36
• Type: Terminating decimal

ii) 1/11

• Decimal form: 0.090909…
• Type: Non-terminating recurring decimal

iii) 4 1/8

• Convert to improper fraction: (4*8 + 1)/8 = 33/8
• Decimal form: 4.125
• Type: Terminating decimal

iv) 3/13

• Decimal form: 0.230769230769…
• Type: Non-terminating recurring decimal

v) 2/11

• Decimal form: 0.181818…
• Type: Non-terminating recurring decimal

vi) 329/400

• Decimal form: 0.8225
• Type: Terminating decimal

2. You know that 1/7 = 0.142857  Can you predict what the decimal expansions of 2/7 , 13/7 , 4/7 , 5/7 , 6/7 are , without actually doing the long division? If so, how?

Ans : 

2/7: This is simply double the value of 1/7. So, we double each digit in the decimal expansion of 1/7.

• 2/7 = 0.285714 (bar on top of 285714)

3/7: Similarly, we triple each digit in the decimal expansion of 1/7.

• 3/7 = 0.428571 (bar on top of 428571)

4/7: Quadruple each digit in the decimal expansion of 1/7.

• 4/7 = 0.571428 (bar on top of 571428)

5/7: Multiply each digit in the decimal expansion of 1/7 by 5.

• 5/7 = 0.714285 (bar on top of 714285)

6/7: Multiply each digit in the decimal expansion of 1/7 by 6.

• 6/7 = 0.857142 (bar on top of 857142)

3. Express the following in the form p/q where p and q are integers and q ≠ 0.

(i) 0.6

(ii) 0.47

(iii) 0.001

Ans : 

i) 0.6¯¯¯¯

Let x = 0.666… (Equation 1) Multiplying both sides by 10, we get: 10x = 6.666… (Equation 2)

Subtracting Equation 1 from Equation 2: 9x = 6 x = 2/3

Therefore, 0.6¯¯¯¯ = 2/3

ii) 0.47¯¯¯¯

Let x = 0.474747… (Equation 1) Multiplying both sides by 100, we get: 100x = 47.474747… (Equation 2)

Subtracting Equation 1 from Equation 2: 99x = 47 x = 47/99

Therefore, 0.47¯¯¯¯ = 47/99

iii) 0.001¯¯¯¯¯¯¯¯¯¯

Let x = 0.001001001… (Equation 1) Multiplying both sides by 1000, we get: 1000x = 1.001001… (Equation 2)

Subtracting Equation 1 from Equation 2: 999x = 1 x = 1/999

Therefore, 0.001¯¯¯¯¯¯¯¯¯¯ = 1/999

4. Express 0.99999… in the form p/q Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.

Ans : 

Expressing 0.999… as a Fraction

Let x = 0.999…

Multiplying both sides by 10, we get:

• 10x = 9.999…

Subtracting the first equation from the second:

• 10x – x = 9.999… – 0.999…
• 9x = 9
• x = 1

Therefore, 0.999… = 1/1 = 1

5. What can the maximum number of digits be in the repeating block of digits in the decimal expansion of 1/17? Perform the division to check your answer.

Ans : 

Therefore, the maximum number of digits in the repeating block of the decimal expansion of 1/17 is 16, and this is indeed the case.

6. Look at several examples of rational numbers in the form p/q(q ≠ 0). Where, p and q are integers with no common factors other that 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?

Ans : 

7. Write three numbers whose decimal expansions are non-terminating non-recurring.

Ans : 

√2: The square root of 2 is approximately 1.41421356… and continues without any repeating pattern.

π (pi): A well-known irrational number, approximately equal to 3.14159265…

√7: The square root of 7 is approximately 2.64575131… and continues without any repeating pattern.

8. Find three different irrational numbers between the rational numbers 5/7and 9/11

Ans : 

To find irrational numbers between 5/7 and 9/11, we can follow these steps:

1. Convert the given rational numbers to decimals:
   • 5/7 ≈ 0.714285
   • 9/11 ≈ 0.818181
2. Identify a range: We need to find irrational numbers between 0.714285 and 0.818181.
3. Construct irrational numbers: Since irrational numbers have non-repeating and non-terminating decimal expansions, we can create numbers within the desired range by combining random digits.

Here are three examples of irrational numbers between 5/7 and 9/11:

• 0.723105926…
• 0.75182463…
• 0.79435127…

9. Classify the following numbers as rational or irrational

(i) 23

(ii) 225

(iii) 0.3796

(iv) 7.478478…..

(v) 1.101001000100001………

Ans  : 

(i) 23 – This is a natural number, and all natural numbers are rational. So, 23 is a rational number. 

(ii) 225 – This is a perfect square (15 * 15) and can be expressed as 225/1, which is of the form p/q. Hence, it’s a rational number. 

(iii) 0.3796 – This is a terminating decimal, so it can be expressed as 3796/10000, which is of the form p/q. Therefore, it’s a rational number. 

(iv) 7.478478… – This is a non-terminating but recurring decimal, so it can be expressed as a fraction (using the method for recurring decimals). Hence, it’s a rational number. 

(v) 1.101001000100001… – This is a non-terminating and non-repeating decimal. Therefore, it’s an irrational number.

Exercise 1.4

1. Classify the following numbers as rational or irrational.

Ans : 

i) 2 – √5

• 2 is a rational number, and √5 is an irrational number.
• So, 2 – √5 is irrational.

ii) (3 + √23) – √23

• √23 cancels out, leaving 3.
• 3 can be expressed as 3/1, which is of the form p/q.
• So, (3 + √23) – √23 is rational.

iii) 2√7 / 7√7

• The √7 terms cancel out, leaving 2/7.
• 2/7 is in the form p/q.
• So, 2√7 / 7√7 is rational.

iv) 1/√2

• The denominator is irrational (√2).
• Rationalizing the denominator would result in an irrational number in the numerator.
• So, 1/√2 is irrational.

v) 2π

• π is an irrational number.
• The product of a rational number (2) and an irrational number (π) is always irrational.
• So, 2π is irrational.

2. Simplify each of the following expressions

Ans : 

(i) (3 + √3)(2 + √2)

Using the distributive property (FOIL), we get:

• 3 * 2 + 3 * √2 + √3 * 2 + √3 * √2
• 6 + 3√2 + 2√3 + √6

Therefore, (3 + √3)(2 + √2) simplifies to 6 + 3√2 + 2√3 + √6.

(ii) (3 + √3)(3 – √3)

This is a product of conjugates. Using the formula (a + b)(a – b) = a² – b², we get:

• 3² – (√3)²
• 9 – 3
• 6

Therefore, (3 + √3)(3 – √3) simplifies to 6.

(iii) (√5 + √2)²

Using the formula (a + b)² = a² + 2ab + b², we get:

• (√5)² + 2 * √5 * √2 + (√2)²
• 5 + 2√10 + 2
• 7 + 2√10

Therefore, (√5 + √2)² simplifies to 7 + 2√10.

(iv) (√5 – √2)(√5 + √2)

This is another product of conjugates. Using the formula (a + b)(a – b) = a² – b², we get:

• (√5)² – (√2)²
• 5 – 2
• 3

3. Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is π = c/d. This seems to contradict the fact that n is irrational. How will you resolve this contradiction?

Ans : When we measure the length of a line using a scale or any other device, we only obtain an approximate rational value, meaning both c and d are irrational. Therefore, cdcdcd is irrational, and hence π is irrational.

 Thus, there is no contradiction in stating that π is irrational.

4. Represent 9.3 on the number line

Ans : 

5. Rationalise the denominator of the following

Ans : 

i) 1/√7

(1/√7) * (√7/√7) = √7/(√7 * √7) 

= √7/7

ii) 1/(√7-√6)

To rationalize the denominator, we multiply both the numerator and the denominator by the conjugate of the denominator, which is (√7+√6):  

(1/(√7-√6)) * ((√7+√6)/(√7+√6)) = (√7+√6)/((√7-√6)(√7+√6)) = (√7+√6)/(7-6) 

= √7+√6

iii) 1/(√5+√2)

To rationalize the denominator, we multiply both the numerator and the denominator by the conjugate of the denominator, which is (√5-√2):  

(1/(√5+√2)) * ((√5-√2)/(√5-√2)) = (√5-√2)/((√5+√2)(√5-√2)) = (√5-√2)/(5-2) 

= (√5-√2)/3

iv) 1/(√7-2)

To rationalize the denominator, we multiply both the numerator and the denominator by the conjugate of the denominator, which is (√7+2):  

(1/(√7-2)) * ((√7+2)/(√7+2)) = (√7+2)/((√7-2)(√7+2)) = (√7+2)/(7-4)

 = (√7+2)/3