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The World of Numbers - Conclusion: The Never-Ending Journey

Grade 9CBSE

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

🔑Concepts

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The Real Number System represents the union of all rational and irrational numbers. Every real number is represented by a unique point on the number line, and every point on the number line represents a unique real number.

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A number is rational if its decimal expansion is either terminating (e.g., 0.250.25) or non-terminating recurring (e.g., 0.333...0.333... or 0.3‾0.\overline{3}).

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A number is irrational if its decimal expansion is non-terminating and non-recurring (e.g., 2≈1.4142135...\sqrt{2} \approx 1.4142135... or π≈3.1415926...\pi \approx 3.1415926...).

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Density Property: There are infinitely many rational and irrational numbers between any two given real numbers.

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Rationalization: The process of eliminating a radical or imaginary number from the denominator of an algebraic fraction. For an expression like 1a+b\frac{1}{a + \sqrt{b}}, we multiply the numerator and denominator by the conjugate a−ba - \sqrt{b}.

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Operations on Real Numbers: The sum, difference, product, or quotient of a non-zero rational number and an irrational number is always irrational. However, the sum or product of two irrational numbers may be rational or irrational.

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Laws of Exponents: These rules extend to real number bases and rational exponents, providing a systematic way to simplify complex numerical expressions.

📐Formulae

ab=a⋅b\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}

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

(a+b)(a−b)=a−b(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b

1a+b=a−ba−b2\frac{1}{\sqrt{a} + b} = \frac{\sqrt{a} - b}{a - b^2}

ap⋅aq=ap+qa^p \cdot a^q = a^{p+q}

(ap)q=apq(a^p)^q = a^{pq}

apaq=ap−q\frac{a^p}{a^q} = a^{p-q}

ap⋅bp=(ab)pa^p \cdot b^p = (ab)^p

a0=1,a≠0a^0 = 1, a \neq 0

💡Examples

Problem 1:

Rationalize the denominator of 53−5\frac{5}{\sqrt{3} - \sqrt{5}}.

Solution:

53−5×3+53+5=5(3+5)(3)2−(5)2=5(3+5)3−5=−52(3+5)\frac{5}{\sqrt{3} - \sqrt{5}} \times \frac{\sqrt{3} + \sqrt{5}}{\sqrt{3} + \sqrt{5}} = \frac{5(\sqrt{3} + \sqrt{5})}{(\sqrt{3})^2 - (\sqrt{5})^2} = \frac{5(\sqrt{3} + \sqrt{5})}{3 - 5} = -\frac{5}{2}(\sqrt{3} + \sqrt{5})

Explanation:

To rationalize the denominator, we multiply the numerator and denominator by the conjugate of the denominator, which is (3+5)(\sqrt{3} + \sqrt{5}). We then apply the identity (a−b)(a+b)=a2−b2(a-b)(a+b) = a^2 - b^2.

Problem 2:

Simplify: 6412×(6412+1)64^{\frac{1}{2}} \times (64^{\frac{1}{2}} + 1).

Solution:

6412=64=864^{\frac{1}{2}} = \sqrt{64} = 8 8×(8+1)=8×9=728 \times (8 + 1) = 8 \times 9 = 72

Explanation:

First, find the square root of 6464 (which is 641/264^{1/2}). Then, substitute the value into the expression and follow the order of operations.

Problem 3:

Subtract 32+533\sqrt{2} + 5\sqrt{3} from 82−238\sqrt{2} - 2\sqrt{3}.

Solution:

(82−23)−(32+53)52−73\begin{array}{r} (8\sqrt{2} - 2\sqrt{3}) \\ -(3\sqrt{2} + 5\sqrt{3}) \\ \hline 5\sqrt{2} - 7\sqrt{3} \end{array}

Explanation:

Subtract like terms (terms with the same radical). 82−32=528\sqrt{2} - 3\sqrt{2} = 5\sqrt{2} and −23−53=−73-2\sqrt{3} - 5\sqrt{3} = -7\sqrt{3}.