Number Systems: A Comprehensive Guide for Class 9 Maths

The Family of Numbers: A Detailed Classification

Think of numbers as a big family with different branches. At the very beginning, we have:

• Natural Numbers (N): These are your counting numbers: {1, 2, 3, 4, ...}. They are the numbers we use for counting objects.
• Whole Numbers (W): If we add zero (0) to the family of natural numbers, we get whole numbers: {0, 1, 2, 3, ...}. So, all natural numbers are whole numbers, but not all whole numbers are natural numbers (because of 0).
• Integers (Z): Now, let's introduce the negative counterparts. Integers include all whole numbers and their negative versions: {..., -3, -2, -1, 0, 1, 2, 3, ...}. The letter 'Z' comes from the German word 'Zahlen', which means 'to count'.
• Rational Numbers (Q): This is a much bigger family. A rational number is any number that can be written in the form p/q, where 'p' and 'q' are integers and 'q' is not equal to zero. This includes all integers (since 3 can be written as 3/1), fractions (like 1/2, 7/8), and terminating or recurring decimals (like 0.5 or 0.333...).

Introducing Irrational and Real Numbers

Irrational Numbers (P or Q')

A number is called irrational if it cannot be written in the form p/q, where p and q are integers and q ≠ 0. Their decimal expansion is non-terminating and non-recurring. Famous examples include √2, √3, and π (pi).

Real Numbers (R)

This is the complete set of all rational and irrational numbers combined. If you pick any point on the number line, it will correspond to a unique real number. The collection of all these numbers forms the set of Real Numbers.

Worked Examples: From Theory to Practice

• Example 1: Find five rational numbers between 3/5 and 4/5. Step 1: Identify the numbers. We need to find 5 rational numbers between a = 3/5 and b = 4/5. Step 2: Create a common denominator. Since we need 5 numbers, we can multiply the numerator and denominator of both fractions by a number greater than 5, like 6 (or 10 for simplicity). Let's use 10. 3/5 = (3 × 10) / (5 × 10) = 30/50 4/5 = (4 × 10) / (5 × 10) = 40/50 Step 3: List the numbers in between. Now we need to find five rational numbers between 30/50 and 40/50. We can easily pick any five integers between 30 and 40 as the numerator. Final Answer: Five rational numbers are 31/50, 32/50, 33/50, 34/50, and 35/50.
• Example 2: Represent √5 on the number line. Step 1: Understand the concept. We will use the Pythagoras theorem (a² + b² = c²). We need to find two numbers whose squares add up to 5. We can use 2² + 1² = 4 + 1 = 5. So, if we construct a right-angled triangle with a base of 2 units and a height of 1 unit, its hypotenuse will be √5. Step 2: Construct on the number line. Draw a number line. Mark point O at 0 and point A at 2. At point A, draw a perpendicular line segment AB of length 1 unit. Join O and B. The length of the line segment OB is the hypotenuse of the right-angled triangle OAB. Step 3: Find the length of the hypotenuse. By Pythagoras theorem, OB² = OA² + AB² = 2² + 1² = 4 + 1 = 5. So, OB = √5. Step 4: Mark on the number line. With O as the center and OB as the radius, draw an arc that intersects the number line at a point P. Since OB = OP (radii of the same arc), the point P represents √5 on the number line.

Laws of Exponents for Real Numbers

Exponents are a shorthand for repeated multiplication. When working with real numbers, especially irrational ones, these laws are essential. Let 'a' > 0 be a real number and 'p' and 'q' be rational numbers. Then, we have the following laws:

1. Product of Powers: a^p * a^q = a^(p+q)

Example: 2³ 2² = 2^(3+2) = 2⁵ = 32*

1. Power of a Power: (a^p)^q = a^(p*q)

Example: (3²)³ = 3^(23) = 3⁶ = 729*

1. Quotient of Powers: a^p / a^q = a^(p-q)

Example: 5⁷ / 5⁴ = 5^(7-4) = 5³ = 125

1. Power of a Product: (ab)^p = a^p b^p

Example: (23)² = 2² 3² = 4 9 = 36*

1. Negative Exponent: a^(-p) = 1 / a^p

Example: 7⁻² = 1 / 7² = 1/49

1. Fractional Exponent (Roots): a^(p/q) = (q√a)^p

Example: 8^(2/3) = (³√8)² = 2² = 4

Common Exam Mistakes to Avoid

In your exams, precision is key. Here are some common traps students fall into in the Number Systems chapter:

• Confusing All Roots with Irrational Numbers: A common mistake is to assume any number under a square root (√) is irrational. Remember that √4 = 2, √9 = 3, and √16 = 4. These are all rational numbers. A number √x is only irrational if x is a non-perfect square.
• Mixing up Recurring and Non-Recurring Decimals: A decimal like 0.121212... is non-terminating but recurring. It is a rational number (it can be written as 12/99). An irrational number like π (3.14159...) is non-terminating and non-recurring.
• Incorrectly Rationalizing the Denominator: When rationalizing a denominator like 1/(√a - √b), you must multiply the numerator and denominator by its conjugate, which is (√a + √b), not (√a - √b). Forgetting to change the sign is a frequent error.

Practice Questions with Solutions

• Q: Is zero a rational number? Can you write it in the form p/q, where p and q are integers and q ≠ 0? A: Step 1: Recall the definition of a rational number. A number 'r' is rational if it can be expressed as p/q, where p and q are integers and q is not zero. Step 2: Check if zero (0) fits this definition. We can write 0 as 0/1, 0/2, 0/-5, etc. Step 3: In all these cases, the numerator (p=0) is an integer, and the denominator (q=1, 2, -5) is a non-zero integer. Therefore, the condition is satisfied. Final answer: Yes, zero is a rational number.
• Q: Express 0.4777... in the form p/q. A: Step 1: Let x = 0.4777... Step 2: Multiply by 10 to isolate the repeating part after the decimal. 10x = 4.777... Step 3: Multiply by 100 to get another equation. 100x = 47.777... Step 4: Subtract the equation from Step 2 from the equation in Step 3. 100x - 10x = 47.777... - 4.777... 90x = 43 Step 5: Solve for x. x = 43/90 Final answer: 0.4777... can be expressed as 43/90.
• Q: Rationalise the denominator of 1 / (√7 - 2). A: Step 1: Identify the denominator and its conjugate. The denominator is (√7 - 2). Its conjugate is (√7 + 2). Step 2: Multiply the numerator and the denominator by the conjugate. [1 / (√7 - 2)] * [(√7 + 2) / (√7 + 2)] Step 3: Simplify the numerator and denominator. The numerator becomes (√7 + 2). The denominator becomes (√7)² - (2)², using the identity (a-b)(a+b) = a²-b². (√7)² - (2)² = 7 - 4 = 3. Step 4: Combine the results. The expression simplifies to (√7 + 2) / 3. Final answer: The rationalised form is (√7 + 2) / 3.
• Q: Simplify: (125)^(-1/3). A: Step 1: Recall the law of exponents a^(-p) = 1/a^p. So, (125)^(-1/3) = 1 / (125)^(1/3). Step 2: Recall the law a^(1/q) = q√a. So, (125)^(1/3) is the cube root of 125. Step 3: Find the cube root of 125. We know that 5 × 5 × 5 = 125, so ³√125 = 5. Step 4: Substitute this value back into the expression from Step 1. 1 / (125)^(1/3) = 1/5. Final answer: (125)^(-1/3) = 1/5.

Frequently Asked Questions