Tool to reduce a fraction in lowest term. A Fraction in Lowest Terms (Irreducible Fraction) is a reduced fraction in shich the numerator and the denominator are coprime (they do not share common factors)

Irreducible Fractions - dCode

Tag(s) : Arithmetics, Symbolic Computation, Mathematics

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Tool to reduce a fraction in lowest term. A Fraction in Lowest Terms (Irreducible Fraction) is a reduced fraction in shich the numerator and the denominator are coprime (they do not share common factors)

To simplify a fraction \( a / b \) or \( frac{a}{b} \) composed of a numerator \( a \) and a denominator \( b \), you have to find the greatest common divisor (GCD) of the numbers \( a \) and \( b \). The irreducible fraction is obtained by dividing the numerator and the denominator by the obtained PGCD.

Consider the fraction \( 12/10 \), with \( 12 \) the numerator and \( 10 \) the denominator. You can calculate that \( GCD(12,10) = 2 \) and divide both the numerator \( 12/2 = 6 \) and the denominator \( 10/2 = 5 \), so the corresponding irreducible fraction is \( 6/5 \)

dCode uses formal calculations in order to keep variables and find the irreducible form.

If the number has a **limited decimal development** then it only needs to be multiplied by the right power of 10, then simplify the fraction and solve the equation.

Consider the number \( 0.14 = 0.14/1 \), you can multiplied by \( 10/10 \) until having no comma: \( 0.14/1 = 1.4/10 = 14/100 \) then you can simplify \( 14/100 = 7/50 \)

If the number has a **non finite decimal expansion**, then it is necessary to locate the repeating portion of the number after the repeating decimal point.

Consider the number \( 0.166666666 ... \) where the \( 6 \) is repeated

We call \( x \) the number, and \( n \) the size (number of digits) of the smallest repeated portion. To obtain a fraction, you can multiply \( x \) by \( 10^n \) and then subtract \( x \).

\( x = 0.1666666 ... \), the smallest repeated portion is \( 6 \), which has a single digit so that \( n = 1 \). You then compute \( 10^1 \ times x = 1.6666666 ... \) and \( 10x-x \).

$$ 10x-x = 9x = 1.666666 ... - 0.1666666 ... = 1.5 \\ \iff 9x = 1.5 \\ \Rightarrow x = 1.5 / 9 = 15/90 = 1/6 $$

So \( 1/6 = 0.1666666 ... \)

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