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Number Systems - Operations on Real Numbers

Grade 9CBSE

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

🔑Concepts

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Operations on Rational and Irrational Numbers: The sum, difference, product, and quotient of a rational number and an irrational number is always irrational. For example, 2+32 + \sqrt{3} and 525\sqrt{2} are both irrational. However, if we add, subtract, multiply or divide two irrational numbers, the result may be rational or irrational.

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Geometric Representation of Square Roots: To represent x\sqrt{x} for any positive real number xx on a number line, we visualize a geometric construction. We mark a distance AB=xAB = x units and BC=1BC = 1 unit on a straight line. By finding the midpoint OO of ACAC and drawing a semicircle with ACAC as diameter, a perpendicular line drawn from point BB to the semicircle at point DD will have a length BD=xBD = \sqrt{x}.

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Radical Identities: Real numbers under square roots follow specific algebraic identities similar to polynomial identities. These include splitting roots across multiplication and division, such as ab=ab\sqrt{ab} = \sqrt{a}\sqrt{b} and ab=ab\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}. These identities are essential for simplifying complex radical expressions.

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Rationalizing the Denominator: This process involves eliminating the radical from the denominator of a fraction. If the denominator is of the form a+b\sqrt{a} + b, we multiply both the numerator and the denominator by its conjugate a−b\sqrt{a} - b. This uses the identity (x+y)(x−y)=x2−y2(x+y)(x-y) = x^2 - y^2 to turn the irrational denominator into a rational one.

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Laws of Exponents for Real Numbers: When dealing with real numbers as bases and rational numbers as exponents, the standard laws of indices apply. For a positive real number aa and rational numbers pp and qq, the product law states ap⋅aq=ap+qa^p \cdot a^q = a^{p+q} and the power of a power law states (ap)q=apq(a^p)^q = a^{pq}.

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Definition of nthn^{th} Roots: For a real number a>0a > 0 and a positive integer nn, the nthn^{th} root of aa is denoted by an=b\sqrt[n]{a} = b such that bn=ab^n = a. Visually, this represents finding a value bb which, when multiplied by itself nn times, equals aa. In exponential form, this is written as a1na^{\frac{1}{n}}.

📐Formulae

ab=ab\sqrt{ab} = \sqrt{a}\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

(a+b)(a−b)=a2−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

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

amn=(an)m=amna^{\frac{m}{n}} = (\sqrt[n]{a})^m = \sqrt[n]{a^m}

💡Examples

Problem 1:

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

Solution:

Step 1: Identify the conjugate of the denominator 3−5\sqrt{3} - \sqrt{5}, which is 3+5\sqrt{3} + \sqrt{5}.\Step 2: Multiply both numerator and denominator by the conjugate:\53−5×3+53+5\frac{5}{\sqrt{3} - \sqrt{5}} \times \frac{\sqrt{3} + \sqrt{5}}{\sqrt{3} + \sqrt{5}}\Step 3: Simplify the numerator and apply the identity (a−b)(a+b)=a2−b2(a-b)(a+b) = a^2 - b^2 in the denominator:\5(3+5)(3)2−(5)2=5(3+5)3−5\frac{5(\sqrt{3} + \sqrt{5})}{(\sqrt{3})^2 - (\sqrt{5})^2} = \frac{5(\sqrt{3} + \sqrt{5})}{3 - 5}\Step 4: Perform the final subtraction in the denominator:\5(3+5)−2=−52(3+5)\frac{5(\sqrt{3} + \sqrt{5})}{-2} = -\frac{5}{2}(\sqrt{3} + \sqrt{5})

Explanation:

To rationalize the denominator, we use the conjugate to create a difference of squares, which effectively removes the square root symbols from the bottom of the fraction.

Problem 2:

Simplify: (125)−13(125)^{-\frac{1}{3}}

Solution:

Step 1: Express 125 as a power of 5. We know that 5×5×5=1255 \times 5 \times 5 = 125, so 125=53125 = 5^3.\Step 2: Substitute this into the expression:\(53)−13(5^3)^{-\frac{1}{3}}\Step 3: Use the law of exponents (ap)q=apq(a^p)^q = a^{pq}:\53×(−13)=5−15^{3 \times (-\frac{1}{3})} = 5^{-1}\Step 4: Convert the negative exponent to a fraction using a−n=1ana^{-n} = \frac{1}{a^n}:\5−1=155^{-1} = \frac{1}{5}

Explanation:

This problem uses the laws of exponents to simplify a base with a fractional negative power. By prime factorizing the base, we can easily cancel out the denominator of the fractional exponent.