73 lines
2.2 KiB
Markdown
73 lines
2.2 KiB
Markdown
---
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tags:
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- prealgebra
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- fractions
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- arithmetic
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---
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# Multiplying fractions
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To find the product of two fractions $\frac{a}{b}$ and $\frac{c}{d}$ multiply
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their numerators and denominators and then reduce:
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$$\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}$$
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### Example
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$$
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\frac{1}{3} \cdot \frac{2}{5} = \frac{1 \cdot 2}{3 \cdot 5} = \frac{2}{15}
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$$
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## Prime factorisation in place
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The example above did not require a reduction, so here is a more complex
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example:
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$$
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\frac{14}{15} \cdot \frac{30}{140} = \frac{420}{2100}
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$$
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It would be laborious to reduce such a large product using factor trees or the
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repeated application of divisors, as defined in
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[reducing fractions](Reducing_fractions.md). We can use a more efficient
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method. This method can be applied at the point at which we conduct the
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multiplication rather than afterwards once we have the product. We express the
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the initial multiplicands as prime factors:
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$$
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\frac{14}{15} \cdot \frac{30}{140} = \frac{(2 \cdot 7) \cdot (2 \cdot 3 \cdot 5) }{(3 \cdot 5) \cdot (2 \cdot 2 \cdot 7 \cdot 5)}
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$$
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We now have the product in factorised form before we have applied the
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multiplication so we can go ahead and cancel:
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$$
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\frac{\cancel{2}, \cancel{7}, \cancel{2}, \cancel{3}, \cancel{5}}{\cancel{3}, \cancel{5}, \cancel{2}, \cancel{2}, \cancel{7}, 5} = \frac{1}{5}
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$$
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**Note that in the above case, there was only a single 5 left as a denominator
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and no value left as a numerator. This is equivalent to there just being "one
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five" so we write $\frac{1}{5}$**
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## Example of multiplying fractions with negative fractions containing variables
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Calculate: $$- \frac{6x}{55y} \cdot - \frac{110y^2}{105x^2}$$
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First multiply in place:
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$$
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\frac{(3 \cdot 2 \cdot x) \cdot (5 \cdot 2 \cdot 11 \cdot y \cdot y)}{(5 \cdot 11 \cdot y) \cdot (7 \cdot 5 \cdot 3 \cdot x \cdot x)}
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$$
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Then cancel:
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$$
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\frac{(\cancel{3} \cdot 2 \cdot \cancel{x}) \cdot (\cancel{5} \cdot 2 \cdot \cancel{11} \cdot \cancel{y} \cdot y)}{(\cancel{5} \cdot \cancel{11} \cdot \cancel{y}) \cdot (7 \cdot 5 \cdot \cancel{3} \cdot \cancel{x} \cdot x)} =
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\frac{2 \cdot 2 \cdot y}{7 \cdot 5 \cdot x}
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$$
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Then reduce:
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$$
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\frac{2 \cdot 2 \cdot y}{7 \cdot 5 \cdot x} = \frac{4y}{35x}
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$$
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