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The World of Numbers - Real Numbers: Decimals and Cyclic Patterns

Grade 9CBSE

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

🔑Concepts

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The collection of all rational and irrational numbers together forms the set of Real Numbers, denoted by R\mathbb{R}.

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A rational number pq\frac{p}{q} has a decimal expansion that is either terminating (e.g., 0.1250.125) or non-terminating recurring/cyclic (e.g., 0.333...=0.3‾0.333... = 0.\overline{3}).

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A decimal expansion is terminating if and only if the prime factorization of the denominator qq (in its simplest form) is of the form 2n5m2^n 5^m, where nn and mm are non-negative integers.

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Irrational numbers have decimal expansions that are non-terminating and non-recurring (e.g., π=3.14159...\pi = 3.14159... or 2=1.41421...\sqrt{2} = 1.41421...).

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The 'period' is the repeating block of digits in a cyclic decimal, and 'periodicity' is the number of digits in that repeating block. For example, in 0.142857‾0.\overline{142857}, the periodicity is 66.

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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.

📐Formulae

Rational Number=pq, where p,q∈Z and q≠0\text{Rational Number} = \frac{p}{q}, \text{ where } p, q \in \mathbb{Z} \text{ and } q \neq 0

For terminating decimals: q=2n×5m for n,m≥0\text{For terminating decimals: } q = 2^n \times 5^m \text{ for } n, m \geq 0

General conversion: x=0.a‾  ⟹  x=a9\text{General conversion: } x = 0.\overline{a} \implies x = \frac{a}{9}

General conversion: x=0.ab‾  ⟹  x=ab99\text{General conversion: } x = 0.\overline{ab} \implies x = \frac{ab}{99}

💡Examples

Problem 1:

Show that 0.2353535...=0.235‾0.2353535... = 0.2\overline{35} can be expressed in the form pq\frac{p}{q}.

Solution:

Let x=0.2353535...x = 0.2353535... (Equation 1). Since two digits are repeating, we multiply xx by 100100: 100x=23.53535...100x = 23.53535... (Equation 2). Subtracting Equation 1 from Equation 2: 100.0x=23.53535...−1.0x=0.23535...99.0x=23.30000...\begin{array}{r} 100.0x = 23.53535... \\ - 1.0x = 0.23535... \\ \hline 99.0x = 23.30000... \end{array} This gives 99x=23.399x = 23.3, which means x=23.399=233990x = \frac{23.3}{99} = \frac{233}{990}.

Explanation:

To convert a recurring decimal, we multiply by a power of 1010 corresponding to the number of repeating digits to align the cyclic patterns, then subtract to eliminate the decimal part.

Problem 2:

Without actual division, determine if the rational number 13125\frac{13}{125} has a terminating or non-terminating repeating decimal expansion.

Solution:

The denominator is q=125q = 125. The prime factorization of 125125 is: 125=5×5×5=53125 = 5 \times 5 \times 5 = 5^3 We can write this in the form 2n×5m2^n \times 5^m as: 125=20×53125 = 2^0 \times 5^3 Since the denominator is of the form 2n5m2^n 5^m, the decimal expansion is terminating.

Explanation:

By checking the prime factors of the denominator, we can predict the nature of the decimal expansion without performing long division. If only 22 and 55 are factors, it terminates.

Problem 3:

Find an irrational number between 17\frac{1}{7} and 27\frac{2}{7}.

Solution:

First, find the decimal values: 17=0.142857‾...\frac{1}{7} = 0.\overline{142857}... 27=0.285714‾...\frac{2}{7} = 0.\overline{285714}... To find an irrational number between them, we choose a non-terminating non-recurring pattern starting after 0.140.14 and before 0.280.28. One such number is: 0.15015001500015...0.15015001500015...

Explanation:

An irrational number must be non-terminating and non-recurring. By creating a pattern where the number of zeros between the digits '1515' increases, we ensure it never repeats.