155 lines
6 KiB
Markdown
155 lines
6 KiB
Markdown
---
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tags:
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- Mathematics
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- Prealgebra
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- fractions
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- divisors
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---
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## Reducing fractions to their lowest terms
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>
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> A fraction is said to be *reduced to its lowest terms* if the [greatest common divisor](Factors%20and%20divisors.md#greatest-common-divisor) of the numerator and the denominator is $1$.
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>
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> Whenever we reduce a fraction, the resultant fraction will always be [equivalent](Equivalent%20fractions.md) to the fraction we started with.
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Thus the fraction $\frac{2}{3}$ is reduced to its lowest terms because the greatest common divisor is 1. Neither the numerator or the denominator can be reduced to any lower terms. In contrast, the fraction $\frac{4}{6}$ is not reduced to its lowest terms because the greatest common divisor of both 4 and 6 is 2, not 1.
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### 1. Reducing with repeated application of divisors
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The following demonstrates the process of reducing a fraction to its lowest terms in a series of steps:
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$$
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\\frac{18}{24} = \frac{18/2}{24/2} = \frac{9}{12} = \frac{9/3}{12/3} = \frac{3}{4}
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$$
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\_Once we get to $\frac{3}{4}$ the greatest common divisor is 1, therefore $\frac{18}{24}$ has been reduced to its lowest terms \_.
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### 2. Reducing in one step with the highest common divisor
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In the previous example the reduction took two steps: first we divided by two and then we divided by three. There is a more efficient way: find the [highest common divisor](Factors%20and%20divisors.md#greatest-common-divisor) of the numerator and denominator and then use this as the basis for the reduction. With this method, the reduction can be completed in a single step.
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The greatest common divisor of 18 and 24 is 6, thus:
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$$
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\\frac{18}{24} = \frac{18/6}{24/6} = \frac{3}{4}
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$$
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Note how our earlier two divisors 2 and 3 are [factors](Factors%20and%20divisors.md#factors) of 6, showing the consistency between the two methods.
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### 3. Reducing with factors and cancellation
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The two methods above are not very systematic and are quite heuristic. The third approach is more systematic and relies on the [interchangeability of factors and divisors](Factors%20and%20divisors.md).
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Instead of thinking asking what is the greatest common divisor of 18 and 24 we could ask: which single number can we multiply by to get 18 and 24? Obviously both numbers are in the six times table. This is therefore to say that 6 is a [factor](Factors%20and%20divisors.md#factors) of both: we can multiply some number by 6 to arrive at both 18 and 24. The numbers are 3 and 4 respectively:
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$$
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\\begin{split}
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3 \cdot 6 = 18 \\
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4 \cdot 6 = 24
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\\end{split}
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$$
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Here, 3 and 4 are the multiplicands of the factor 6. As $\frac{3}{4}$ doesn't have a lower common factor, it is therefore defined in its lowest terms.
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Once we have reached this point, we no longer need the common factor 6, we can therefore cancel it out, leaving the multiplicands as the reduced fraction:
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$$
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\\begin{split}
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3 \cancel{\cdot6= 18}\\
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4 \cancel{\cdot6= 24}
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\\end{split}
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$$
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### 4. Reducing with prime factorisation
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This is still a bit long-winded however particularly when finding the factors of larger numbers because we have to go through the factors of both numbers to find the largest held in common.
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A better method is to utilise [prime factorization](Prime%20factorization.md) combined with the canceling technique.
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First we find the prime factors of both the numerator and denominator:
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This gives us:
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$$
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\\frac{18}{24} = \frac{2 \cdot 3 \cdot 3}{2 \cdot 2 \cdot 2 \cdot 3}
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$$
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We then cancel out the factors held in common between the numerator and denominator:
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$$
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\\frac{\cancel{2} \cdot \cancel{3} \cdot 3}{\cancel{2} \cdot 2 \cdot 2 \cdot \cancel{3}}
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$$
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This gives us:
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$$
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\\frac{3}{2 \cdot 2}
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$$
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We then simplify the fraction as normal to its lowest term (conducting any multiplications required by what is left from the prime factorization):
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$$
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\\frac{3}{4}
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$$
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## Reducing fractions that contain variables
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Superficially this looks to be more difficult but in fact we can apply the same prime factorization method to get the result.
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### Demonstration
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*Reduce the following fraction to its lowest terms: $$\frac{25a^3b}{40a^2b^3}$$*
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The prime factors of the numerator and denominator:
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$$
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\\begin{split}
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25 = {5, 5} \\
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40 = {2,2,2,5}
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\\end{split}
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$$
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Now we apply canceling but we include the variable parts, treating them exactly the same as the coefficients. We break them out of their exponents however.
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$$\frac{25a^3b}{40a^2b^3} =\frac{5 \cdot 5 \cdot a \cdot a \cdot a \cdot b}{2 \cdot 2 \cdot 2 \cdot 5 \cdot a \cdot a \cdot b \cdot b \cdot b }$$
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Canceled:
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$$\frac{\cancel{5} \cdot 5 \cdot \cancel{a} \cdot \cancel{a} \cdot a \cdot \cancel{b}}{2 \cdot 2 \cdot 2 \cdot \cancel{5} \cdot \cancel{a} \cdot \cancel{a} \cdot \cancel{b} \cdot b \cdot b }$$
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Which gives us:
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$$
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\\frac{5 \cdot a}{2 \cdot 2 \cdot 2 \cdot b \cdot b} = \frac{5a}{8b^2}
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$$
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## Reducing fractions that contain negative values
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*Reduce the following fraction to its lowest terms: $$\frac{14y^5}{-35y^3}$$*
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* This fraction is an instance of a [fraction with unlike terms](Handling%20negative%20fractions.md#fractions-with-unlike-terms).
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* Apply [Prime factorization](Prime%20factorization.md):
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* Cancel the coefficients and variable parts
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$$
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\\frac{14y^5}{-35y^3}=\frac{5 \cdot 7 \cdot 2 \cdot y \cdot y \cdot y \cdot y \cdot y}{-5 \cdot 7 \cdot y \cdot y \cdot y} = - \frac{2y^2}{5}
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$$
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*Reduce the following fraction to its lowest terms:
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$$\frac{- 12xy^2}{ - 18xy^2}$$*
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* This fraction is an instance of a [fraction with like terms](Handling%20negative%20fractions.md#fractions-with-like-terms).
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* Apply [Prime factorization](Prime%20factorization.md):
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* Cancel the coefficients and variable parts
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$$
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* \\frac{12xy^2}{18xy^2}=\frac{3 \cdot 2 \cdot 2 \cdot x \cdot y \cdot y}{3 \cdot 7 \cdot 3 \cdot 2 \cdot x \cdot x \cdot y} = - \frac{2y}{3x}
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$$
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