From Irreducible Fractions to Recurring Decimals - Part I

Published by Ganit Charcha | Category - Math Articles | 2016-09-24 03:16:23

In this article we will establish the connection between Irreducible Fractions and Recurring Decimals. We will start by defining irreducible fractions. Thorughout the entire article we consider only positive fractions.

Irreducible Fraction :  An irreducible fraction is a fraction in which the numerator and denominator are integers that have no common divisor other than $1$. In other words, a fraction $\frac{a}{b}$ is irreducible iff $a$ and $b$ are co-prime to each other, i.e., $g.c.d(a, b) = 1$. If $a < b,$ then it is called an Irreducible Proper Fraction. If $a > b$, then $a$ can be written as $a = c.b + a'$, where $c \geq 1$ is an integer and $0 < a' < b$. Therefore, $\frac{a}{b}$ can be wtitten as $\frac{a}{b} = c + \frac{a'}{b}$. Since $g.c.d(a, b) = 1$, so we have  $g.c.d(a', b) = 1$ where $a' < b$.

Reducible Fraction : Fraction $\frac{a}{b}$ is called a reducible fraction if $g.c.d(a, b) = d > 1.$ Any reducible fraction can be readily converted into an irreducible fraction by dividing both numerator and denominator by their g.c.d - $d$. Therefore, $\frac{a}{b} = \frac{\frac{a}{d}}{\frac{b}{d}}$. Since, $d = g.c.d(a, b),$ so $d$ divides both $a$ and $b$ and we write $a' = \frac{a}{d}$ and  $b' = \frac{b}{d}$. Clearly, $a'$ and $b'$ are integers such that $g.c.d(a', b') = 1$. Therefore, $\frac{a'}{b'}$ is an irreducible fraction.
For example, $\frac{3}{8}$ is an irreducible proper fraction, whereas $\frac{6}{16}$ is a reducible fraction which can be converted to irreducible fraction $\frac{3}{8}$, on dividing both numerator and denominator by $g.c.d(6, 16) = 2$. Fraction $\frac{8}{3}$ is an irreducible fraction but is not proper.

Since, reducible fractions are convertible to its equivalent irreducible form, so it is sufficient to focus mainly on the decimal representations of irreducible fractions. If  $\frac{a}{b}$ is an irreducible fraction with $g.c.d(a, b) = 1$, then the decimal representation of $\frac{a}{b}$ is comprised of the digits obtained upon successive division by the denominator $b$. If $a < b$, then the process starts by dividing $10a$ by $b$ and then the successive remainders are multiplied by $10$ and then divided by $b$. The quotients $q_{i}$'s obtained successively in this process contributes to the decimal representation and is represented as $\frac{a}{b} = 0.q_{1}q_{2}q_{3}\ldots q_{n}....$. If $a > b,$ then it can be represented as $\frac{a}{b} = c + \frac{a'}{b},$ where $\frac{a'}{b}$ is an irreducible proper fraction. If the decimal representation of  $\frac{a'}{b}$ is $0.q_{1}q_{2}q_{3}\ldots q_{n}....$, then the decimal representation of $\frac{a}{b}$ will be $c.q_{1}q_{2}q_{3}\ldots q_{n}....$. This $c$ is called the integer part of the decimal representation and is placed on the left side of the decimal point ('.'). Therefore, for irreducible proper fractions the value of $c$ will be $0$ and for non-proper irreducible fractions $c \geq 1$.

Irreducible fractions can be transformed either into a terminating decimal or a recurring (or repeating) decimal. Recurring decimals are of two types - (a) pure recurring (repeating) decimal and (b) mixed recurring (repeating) decimal.
For example, $\frac{1}{2} = 0.5$ or $\frac{3}{25} = \frac{12}{100} = 0.12$ are terminating decimals.
$\frac{1}{7} = 0.142857142857142857..... = 0.\overline{142857}$ and $\frac{1}{9} = 0.09090909.... = 0.\overline{09}$ are examples of pure recurring decimals. They are called pure recurring (repeating) since each of the digits in the decimal representation is getting repeated after a certain period. For example, each of the digits in 142857 is getting repeated in the same order in the decimal representation of $\frac{1}{7}$. That is why $\frac{1}{7}$ is shortly written in its decimal form as $0.\overline{142857}$, $\frac{1}{9}$ is shortly written in its decimal form as $0.\overline{09}$.
Finally, $\frac{1}{14} = 0.0714285714285714285..... = 0.0\overline{714285}$ and  $\frac{7}{55} = 0.1272727..... = 0.1\overline{27}$ are examples of mixed recurring decimals. They are called mixed recurring (repeating) decimals since they involve a few non-recurring digits at the first and then comes the recurring digits which keep on appearing in the same order. For example, the decimal representation of $\frac{7}{55}$ starts with 1 and then the digits of 27 are getting repeated infinitely and it is shortly written as  $0.1\overline{27}$.

In all the above example and in both types of recurring decimals pure and mixed, the repeating block is indicated by a bar and this notation is used for our conveniance so that we do not require to write the repeating block multiple times. The repeating block of digits in a recurring decimal is called the period of the recurring decimal and number of digits in the repeating block is called the length of the period. For example, since $\frac{1}{7} = 0.\overline{142857}$, so the period of decimal representation of $\frac{1}{7}$ is $142857$ and the length of the period is $6$. Similarly, the period of decimal representation of $\frac{7}{55}$ is $27$ and the length of the period is $2$.

We have seen examples of different types of decimal representation of an irreducible fraction, we will now see how a recurring decimal (pure and mixed) can be converted back into an irreducible fraction.

Case I (Pure Recurring Decimal) : Let, $0.\overline{q_{1}q_{2}\ldots q_{s}}$ be the pure recurring decimal and we need to find out its equivalent irreducible proper fraction form.
Let, $x = 0.\overline{q_{1}q_{2}\ldots q_{s}}$, then we will multiply $x$ first by $10^{s}$, where $s$ is the length of the period of the pure recurring decimal $x$. Hence, $$10^{s}x = q_{1}q_{2}\ldots q_{s-1}q_{s}.\overline{q_{1}q_{2}\ldots q_{s}} = q_{1}q_{2}\ldots q_{s} + 0.\overline{q_{1}q_{2}\ldots q_{s}} = q_{1}q_{2}\ldots q_{s} + x.$$ 
Therefore, $$x = \frac{q_{1}q_{2}\ldots q_{s}}{10^{s} - 1} = \frac{q_{1}q_{2}\ldots q_{s}}{999....9},$$ where the denominator in the last expression is consisted only of $9$'s and it contains the same number of $9$'s as that of the length $s$ of the period of the pure recurring decimal. If $d = g.c.d((q_{1}q_{2}\ldots q_{s}), (999....9))$, then $$x = \frac{\frac{q_{1}q_{2}\ldots q_{s}}{d}}{\frac{999....9}{d}},$$ gives the equivalent irreducible proper fraction form of the recurring decimal $x$.
Let us illustrate this through an example. Consider $x = 0.\overline{285714}$, then $x = \frac{285714}{10^{6} - 1} = \frac{285714}{999999} = \frac{2}{7}.$

Case II (Mixed Recurring Decimal) : Let, $0.r_{1}r_{2}\ldots r_{t}\overline{q_{1}q_{2}\ldots q_{s}}$ be the mixed recurring decimal and we need to find out its equivalent irreducible proper fraction form.
$x = 0.r_{1}r_{2}\ldots r_{t}\overline{q_{1}q_{2}\ldots q_{s}},$ then we will multiply $x$ first by $10^{t},$  $t$ is the length of the starting non-repeating part of the decimal representation. Hence, $$10^{t}x = (r_{1}r_{2}\ldots r_{t}) + 0.\overline{q_{1}q_{2}\ldots q_{s}}.$$ The above expression will then be multiplied by $10^{s},$ where $s$ is the length of the period of the mixed recurring decimal $x$. We, therefore, obtain $$10^{s + t}x = 10^{s}(r_{1}r_{2}\ldots r_{t}) + q_{1}q_{2}\ldots q_{s} +  0.\overline{q_{1}q_{2}\ldots q_{s}}$$$$10^{s + t}x = 10^{s}(r_{1}r_{2}\ldots r_{t}) + q_{1}q_{2}\ldots q_{s} +  10^{t}x - (r_{1}r_{2}\ldots r_{t})$$$$10^{t}(10^{s} - 1)x = (10^{s} - 1)(r_{1}r_{2}\ldots r_{t}) + q_{1}q_{2}\ldots q_{s}.$$ Therefore, $$x = \frac{(999....9)(r_{1}r_{2}\ldots r_{t}) + q_{1}q_{2}\ldots q_{s}}{(999....9)10^{t}},$$ where the expression $999....9$ in the above equation is consisted of $s$ number of $9$'s.
Note that, $(999...9)(r_{1}r_{2}\ldots r_{t}) \le (999...9)(10^{t} - 1)$ and also $q_{1}q_{2}\ldots q_{s} \le (10^{s} - 1) = (999...9)$. Hence, $(999....9)(r_{1}r_{2}\ldots r_{t}) + q_{1}q_{2}\ldots q_{s} \le (999...9)10^{t}$. So, if not all $q_{i}$'s and $r_{i}$'s are $9$, then the numerator is strictly less than the denominator in the above expression for $x$. if $d = g.c.d((999....9)(r_{1}r_{2}\ldots r_{t}) + q_{1}q_{2}\ldots q_{s}, (999....9)10^{t}$, then $$x =  \frac{\frac{(999....9)(r_{1}r_{2}\ldots r_{t}) + q_{1}q_{2}\ldots q_{s}}{d}}{\frac{(999....9)10^{t}}{d}},$$ gives the equivalent irreducible proper fraction form of the mixed recurring decimal $x$.
Let us illustrate this through an example. Consider $x = 0.1\overline{27}$, then $x = \frac{(99).1 + 27}{99.10} = \frac{126}{990} = \frac{\frac{126}{18}}{\frac{990}{18}} = \frac{7}{55}.$

In both the above cases, we have considered pure and mixed recurring decimals whose integer part is $0$ and we have seen how they can be converted back into an irreducible proper fraction. In case the integer part is non-zero and is greater that $1$, then the decimal form $I.r_{1}r_{2}\ldots r_{t}\overline{q_{1}q_{2}\ldots q_{s}}$ can be written as $I +0.r_{1}r_{2}\ldots r_{t}\overline{q_{1}q_{2}\ldots q_{s}}$, $(I > 1).$ The irreducible proper fraction representation of the mixed recurring decimal $0.r_{1}r_{2}\ldots r_{t}\overline{q_{1}q_{2}\ldots q_{s}}$  when will be added with $I$ will have numerator greater than the denominator and hence in this case we will end up obtaining an irreducible fraction. Same thing happens for a pure recurring decimal wil non-zero and positive integer part.

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