8.3 Rational Numbers, Infinite/Telescoping Fractions & Repeating Decimals (Devirli Ondalık Sayılar)

8.3 Rational Numbers, Infinite/Telescoping Fractions & Repeating Decimals (Devirli Ondalık Sayılar)

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1. Rational Numbers (Rasyonel Sayılar) & Order of Operations (İşlem Önceliği)

A rational number takes the form $\frac{a}{b}$ where $a, b \in \mathbb{Z}$ and $b \ne 0$ ($a$ is the numerator / Pay, and $b$ is the denominator / Payda). Note that $\frac{0}{b} = 0$ for $b \ne 0$, whereas every fraction with denominator zero, including $\frac{0}{0}$, is undefined (Tanımsız) as an arithmetic value. In a limit, a substitution producing $0/0$ is called an indeterminate form (Belirsiz biçim) because further analysis is needed.

Classification of Fractions (Kesir Çeşitleri)

1. Proper Fraction (Basit Kesir): The absolute value of the numerator is strictly less than the absolute value of the denominator:

$$\lvert a \rvert < \lvert b \rvert$$

Every proper fraction lies strictly between $-1$ and $1$: $-1 < \frac{a}{b} < 1$. Note that $0 = \frac{0}{b}$ IS a proper fraction! 2. Improper Fraction (Bileşik Kesir): The absolute value of the numerator is greater than or equal to the absolute value of the denominator:

$$\lvert a \rvert \ge \lvert b \rvert \implies \frac{a}{b} \le -1 \quad \text{or} \quad \frac{a}{b} \ge 1$$

1. Mixed Fraction (Tam Sayılı Kesir): Written as $A\frac{b}{c}$ for positive integers $A, b, c$ with $b < c$:

$$A\frac{b}{c} = A + \frac{b}{c} = \frac{A \cdot c + b}{c}$$ $$-2\frac{3}{5} = -\left(2 + \frac{3}{5}\right) = -2 - \frac{3}{5} = -\frac{13}{5} \qquad \left(\text{NEVER } -2 + \frac{3}{5} = -\frac{7}{5}\text{!}\right)$$

The Main Fraction Bar (Ana Kesir Çizgisi)

In a compound fraction, the main fraction bar is the horizontal bar positioned level with the $+$ / $-$ / $=$ signs. Keep the upper expression intact and multiply by the reciprocal of the lower expression:

$$\frac{\frac{a}{b}}{c} = \frac{a}{b} \cdot \frac{1}{c} = \frac{a}{bc}, \qquad \frac{a}{\frac{b}{c}} = a \cdot \frac{c}{b} = \frac{ac}{b}$$

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2. Finite & Infinite Continued Fractions (Merdivenli ve Sonsuz Kesirler)

Bottom-Up Evaluation vs. The Balloon / Onion Method (Balon Yöntemi)

• Bottom-Up Evaluation (Pure Numerical Ladders): When a finite ladder fraction contains only numbers (such as $1 + \frac{1}{2 - \frac{1}{3}}$), start at the deepest bottom step ($2 - \frac{1}{3} = \frac{5}{3}$), invert it Immediately ($\frac{1}{5/3} = \frac{3}{5}$), and add the next integer upward ($1 + \frac{3}{5} = \frac{8}{5}$).
• The Balloon Method (Equations with an Unknown $x$): When solving a nested fraction equation such as:

$$3 + \frac{12}{2 + \frac{6}{x - 1}} = 7$$

never clear denominators from the bottom up! Instead, peel the equation from the outside in using mental balloons:

1. $3 + [\text{Box}_1] = 7 \implies \text{Box}_1 = \frac{12}{2 + \frac{6}{x-1}} = 4$.
2. $\frac{12}{[\text{Box}_2]} = 4 \implies \text{Box}_2 = 2 + \frac{6}{x-1} = 3$.
3. $2 + [\text{Box}_3] = 3 \implies \text{Box}_3 = \frac{6}{x-1} = 1 \implies x - 1 = 6 \implies x = 7$.

Infinite Periodic Continued Fractions (Sonsuz Devirli Kesirler)

If an infinite ladder repeats the same pattern indefinitely, set the entire infinite expression equal to $x$ and substitute $x$ into the first repeating layer:

$$x = a + \frac{b}{a + \frac{b}{a + \ddots}} \implies x = a + \frac{b}{x} \implies x^2 - ax - b = 0$$

When $a, b > 0$, select the positive root of the quadratic equation.

Telescoping Partial Fractions (Ardışık Çarpanlı Kesir Toplamları) & Group Cancellation

Whenever denominators are products of consecutive factors with step $d$, split each fraction via the partial-fraction identity:

$$\frac{1}{k(k + 1)} = \frac{1}{k} - \frac{1}{k + 1}, \qquad \frac{1}{k(k + d)} = \frac{1}{d}\left(\frac{1}{k} - \frac{1}{k + d}\right)$$

If successive first factors also advance by $d$, intermediate terms cancel and only the first and final reciprocal remain. If the index advances by 1 while $d>1$, there are generally $d$ surviving terms at each boundary.

Similarly, for large mixed-fraction expressions such as $\left(2025 + \frac{3}{7}\right) - \left(2023 - \frac{4}{7}\right)$, never convert $2025\frac{3}{7}$ to an improper fraction! Group integers and fractions separately: $(2025 - 2023) + \left(\frac{3}{7} + \frac{4}{7}\right) = 2 + 1 = 3$.

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3. Decimals & Repeating Decimals (Ondalık ve Devirli Ondalık Sayılar)

In Turkish notation, a comma ($,$) denotes the decimal point (e.g., $0,25 = \frac{25}{100} = \frac{1}{4}$). When dividing two terminating decimals, shift the comma right by the maximum number of decimal places in either term, padding the other term with zeros: $\frac{0,036}{0,0009} = \frac{360}{9} = 40$.

Repeating Decimal Conversion Formula (Devirli Ondalık Açılım)

For a repeating decimal $a,bc\overline{de}$ where $de$ repeats infinitely ($a,bcdedede\dots$):

$$a,bc\overline{de} = \frac{\text{Entire Number (without comma or bar)} - \text{Non-Repeating Part (without comma)}}{\underbrace{99\dots 9}_{\text{count of repeating digits after comma}} \underbrace{00\dots 0}_{\text{count of non-repeating digits after comma}}} = \frac{(abcde)_{10} - (abc)_{10}}{9900}$$

• The $\overline{9}$ Shortcut: Because $0,\overline{9} = \frac{9}{9} = 1$, a repeating bar over $9$ simply increments the immediately preceding digit by $+1$ and drops the $\overline{9}$:

$$0,\overline{9} = 1, \qquad 2,3\overline{9} = 2,4, \qquad 7,\overline{9} = 8$$

• Single-Digit Pure Repeats: $0,\overline{a} = \frac{a}{9}$, so $0,\overline{a} + 0,\overline{b} = \frac{a + b}{9}$.

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4. Ordering Rational Fractions (Kesirlerde Sıralama)

To compare positive fractions (if fractions are negative, compare their positive absolute values first and reverse all inequality signs at the very end!):

1. Equal Denominators ($b_1 = b_2$): The fraction with the larger numerator is larger ($\frac{9}{13} > \frac{7}{13}$).
2. Equal Numerators ($a_1 = a_2$): The fraction with the smaller denominator is larger ($\frac{15}{7} > \frac{15}{11}$).
3. Constant Difference Between Numerator and Denominator ($\lvert b - a \rvert = k$ is constant):
   • Proper Fractions (Basit Kesirler, $a < b$): As the numbers grow larger, the fraction approaches $1$ from below, so larger numbers yield a larger fraction:

$$\frac{3}{5} < \frac{13}{15} < \frac{93}{95} \qquad (\text{constant difference } k = 2)$$

• Improper Fractions (Bileşik Kesirler, $a > b$): Since $\frac{b+k}{b} = 1 + \frac{k}{b}$, as the numbers grow larger, the fraction approaches $1$ from above, so larger numbers yield a smaller fraction:

$$\frac{95}{93} < \frac{15}{13} < \frac{5}{3} \qquad (\text{constant difference } k = 2)$$

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Summary Reference Table

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Worked Examples

Worked Example 8.3.1: Balloon Method & Telescoping Partial Fractions

Problem: Solve for $x$ in $3 + \frac{12}{2 + \frac{6}{x - 1}} = 7$, and then evaluate the telescoping sum:

$$S = \frac{1}{x(x+2)} + \frac{1}{(x+2)(x+4)} + \dots + \frac{1}{(x+16)(x+18)}$$

Solution:

1. Peel the nested fraction from the outside in (Balloon Method):
   • Since $3 + \frac{12}{2 + \frac{6}{x-1}} = 7$, the fraction $\frac{12}{2 + \frac{6}{x-1}}$ must equal $7 - 3 = 4$.
   • Since $12$ divided by the denominator $2 + \frac{6}{x-1}$ equals $4$, that denominator must equal $\frac{12}{4} = 3$.
   • From $2 + \frac{6}{x-1} = 3$, we get $\frac{6}{x-1} = 1 \implies x - 1 = 6 \implies x = 7$.
2. Substitute $x = 7$ into $S$ and apply partial fraction decomposition with step $d = 2$:

$$S = \frac{1}{7 \cdot 9} + \frac{1}{9 \cdot 11} + \dots + \frac{1}{23 \cdot 25} = \frac{1}{2}\left[\left(\frac{1}{7} - \frac{1}{9}\right) + \left(\frac{1}{9} - \frac{1}{11}\right) + \dots + \left(\frac{1}{23} - \frac{1}{25}\right)\right]$$ $$S = \frac{1}{2}\left(\frac{1}{7} - \frac{1}{25}\right) = \frac{1}{2} \cdot \frac{25 - 7}{175} = \frac{9}{175}$$

Worked Example 8.3.2: Repeating Decimals & Ordering Negative Fractions

Problem: (a) Compute $\frac{0,4\overline{9} + 1,2\overline{6}}{0,2\overline{3}}$. (b) Order the negative fractions $x = -\frac{13}{17}$, $y = -\frac{25}{29}$, and $z = -\frac{41}{45}$ from smallest to largest.

Solution:

1. Part (a):
   • By the $\overline{9}$ shortcut, $0,4\overline{9} = 0,5 = \frac{1}{2} = \frac{45}{90}$.
   • Convert $1,2\overline{6}$ and $0,2\overline{3}$ with common denominator $90$: $1,2\overline{6} = \frac{126 - 12}{90} = \frac{114}{90}$, and $0,2\overline{3} = \frac{23 - 2}{90} = \frac{21}{90}$.
   • Evaluate the ratio: $\frac{\frac{45}{90} + \frac{114}{90}}{\frac{21}{90}} = \frac{159}{21} = \frac{53}{7}$.
2. Part (b):
   • Compare the positive absolute values first: $\lvert x \rvert = \frac{13}{17}$, $\lvert y \rvert = \frac{25}{29}$, $\lvert z \rvert = \frac{41}{45}$.
   • All three are positive proper fractions ($a < b$) with a constant difference of $4$ ($17 - 13 = 29 - 25 = 45 - 41 = 4$). Therefore, larger numbers produce a larger positive fraction: $\lvert x \rvert < \lvert y \rvert < \lvert z \rvert$.
   • Multiplying by $-1$ reverses the inequalities: $z < y < x$.

Review the Method

For positive fractions with a constant difference between numerator and denominator, proper fractions ($\lvert a \rvert < \lvert b \rvert$) increase as the numbers grow larger, while improper fractions ($\lvert a \rvert > \lvert b \rvert$) decrease as the numbers grow larger.