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Number Systems - Irrational Numbers

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

πŸ”‘Concepts

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Definition of Irrational Numbers: An irrational number is a number that cannot be expressed in the form pq\frac{p}{q}, where pp and qq are integers and q≠0q \neq 0. These numbers form a set that, together with rational numbers, makes up the Real Number system.

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Decimal Expansion: Unlike rational numbers, the decimal expansion of an irrational number is non-terminating and non-recurring. For example, Ο€β‰ˆ3.14159265...\pi \approx 3.14159265... and 2β‰ˆ1.41421356...\sqrt{2} \approx 1.41421356... continue infinitely without any repeating pattern of digits.

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Square Roots of Non-Perfect Squares: If nn is a positive integer that is not a perfect square, then n\sqrt{n} is always an irrational number. For instance, 2,3,5,Β andΒ 7\sqrt{2}, \sqrt{3}, \sqrt{5}, \text{ and } \sqrt{7} are all irrational because their radicands are not squares of integers.

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Representing n\sqrt{n} on the Number Line: We can locate irrational numbers on a number line using the Pythagoras Theorem. To represent 2\sqrt{2}, visualize a right-angled triangle with a base of 11 unit and a height of 11 unit. The hypotenuse, according to the theorem, is 12+12=2\sqrt{1^2 + 1^2} = \sqrt{2}. By placing a compass at the origin and drawing an arc with the hypotenuse as the radius, the point where the arc meets the number line is the exact location of 2\sqrt{2}.

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Properties of Operations: The sum or difference of a rational number and an irrational number is always irrational. Similarly, the product or quotient of a non-zero rational number and an irrational number is irrational. However, the sum, difference, product, or quotient of two irrational numbers may or may not be irrational.

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Rationalizing the Denominator: This is the process of removing a radical (square root) from the denominator of a fraction. To rationalize 1a\frac{1}{\sqrt{a}}, we multiply both the numerator and denominator by a\sqrt{a}. For expressions like 1a+b\frac{1}{a + \sqrt{b}}, we multiply by the conjugate aβˆ’ba - \sqrt{b}.

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The Square Root Spiral: This is a geometric construction where right-angled triangles are built consecutively. Starting with a triangle of legs 11 and 11 (hypotenuse 2\sqrt{2}), a second triangle is built using the 2\sqrt{2} side as a base and a new perpendicular side of 11 unit. This creates a hypotenuse of 3\sqrt{3}. Continuing this creates a visual spiral of lengths 2,3,4,5,…\sqrt{2}, \sqrt{3}, \sqrt{4}, \sqrt{5}, \dots representing the square roots of consecutive natural numbers.

πŸ“Formulae

ab=aβ‹…b\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}

ab=ab\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}

(a+b)(aβˆ’b)=aβˆ’b(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b

(a+b)(aβˆ’b)=a2βˆ’b(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b

(a+b)2=a+2ab+b(\sqrt{a} + \sqrt{b})^2 = a + 2\sqrt{ab} + b

πŸ’‘Examples

Problem 1:

Rationalize the denominator of 12+3\frac{1}{2 + \sqrt{3}}.

Solution:

Step 1: Identify the conjugate of the denominator 2+32 + \sqrt{3}, which is 2βˆ’32 - \sqrt{3}. Step 2: Multiply both the numerator and denominator by the conjugate: 1Γ—(2βˆ’3)(2+3)(2βˆ’3)\frac{1 \times (2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})}. Step 3: Apply the identity (a+b)(aβˆ’b)=a2βˆ’b2(a+b)(a-b) = a^2 - b^2 to the denominator: (2)2βˆ’(3)2(2)^2 - (\sqrt{3})^2. Step 4: Simplify the denominator: 4βˆ’3=14 - 3 = 1. Step 5: The final expression is 2βˆ’31=2βˆ’3\frac{2 - \sqrt{3}}{1} = 2 - \sqrt{3}.

Explanation:

By multiplying the numerator and denominator by the conjugate of the denominator, we use the difference of squares identity to eliminate the square root from the denominator.

Problem 2:

Simplify the expression (5+2)2(\sqrt{5} + \sqrt{2})^2.

Solution:

Step 1: Use the algebraic identity (x+y)2=x2+2xy+y2(x + y)^2 = x^2 + 2xy + y^2. Step 2: Substitute x=5x = \sqrt{5} and y=2y = \sqrt{2} into the identity. Step 3: The expression becomes (5)2+2(5)(2)+(2)2(\sqrt{5})^2 + 2(\sqrt{5})(\sqrt{2}) + (\sqrt{2})^2. Step 4: Simplify each term: 5+210+25 + 2\sqrt{10} + 2. Step 5: Combine the rational parts: (5+2)+210=7+210(5 + 2) + 2\sqrt{10} = 7 + 2\sqrt{10}.

Explanation:

This problem uses the square of a binomial identity and the property aβ‹…b=ab\sqrt{a} \cdot \sqrt{b} = \sqrt{ab} to expand and simplify the expression.