187 lines
5.9 KiB
Markdown
187 lines
5.9 KiB
Markdown
---
|
|
tags:
|
|
- prealgebra
|
|
- fractions
|
|
---
|
|
|
|
# Reducing fractions to their lowest terms
|
|
|
|
> 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 $1$.
|
|
|
|
> Whenever we reduce a fraction, the resultant fraction will always be
|
|
> [equivalent](Equivalent%20fractions.md) to the fraction we started with.
|
|
|
|
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.
|
|
|
|
### 1. Reducing with repeated application of divisors
|
|
|
|
The following demonstrates the process of reducing a fraction to its lowest
|
|
terms in a series of steps:
|
|
|
|
$$
|
|
\\frac{18}{24} = \frac{18/2}{24/2} = \frac{9}{12} = \frac{9/3}{12/3} = \frac{3}{4}
|
|
$$
|
|
|
|
\_Once we get to $\frac{3}{4}$ the greatest common divisor is 1, therefore
|
|
$\frac{18}{24}$ has been reduced to its lowest terms \_.
|
|
|
|
### 2. Reducing in one step with the highest common divisor
|
|
|
|
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.
|
|
|
|
The greatest common divisor of 18 and 24 is 6, thus:
|
|
|
|
$$
|
|
\frac{18}{24} = \frac{18/6}{24/6} = \frac{3}{4}
|
|
$$
|
|
|
|
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.
|
|
|
|
### 3. Reducing with factors and cancellation
|
|
|
|
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).
|
|
|
|
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:
|
|
|
|
$$
|
|
\\begin{split}
|
|
3 \cdot 6 = 18 \\
|
|
4 \cdot 6 = 24
|
|
\\end{split}
|
|
$$
|
|
|
|
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.
|
|
|
|
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:
|
|
|
|
$$
|
|
\\begin{split}
|
|
3 \cancel{\cdot6= 18}\\
|
|
4 \cancel{\cdot6= 24}
|
|
\\end{split}
|
|
$$
|
|
|
|
### 4. Reducing with prime factorisation
|
|
|
|
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.
|
|
|
|
A better method is to utilise [prime factorization](Prime%20factorization.md)
|
|
combined with the canceling technique.
|
|
|
|
First we find the prime factors of both the numerator and denominator:
|
|
|
|
|
|
This gives us:
|
|
|
|
$$
|
|
\\frac{18}{24} = \frac{2 \cdot 3 \cdot 3}{2 \cdot 2 \cdot 2 \cdot 3}
|
|
$$
|
|
|
|
We then cancel out the factors held in common between the numerator and
|
|
denominator:
|
|
|
|
$$
|
|
\\frac{\cancel{2} \cdot \cancel{3} \cdot 3}{\cancel{2} \cdot 2 \cdot 2 \cdot \cancel{3}}
|
|
$$
|
|
|
|
This gives us:
|
|
|
|
$$
|
|
\\frac{3}{2 \cdot 2}
|
|
$$
|
|
|
|
We then simplify the fraction as normal to its lowest term (conducting any
|
|
multiplications required by what is left from the prime factorization):
|
|
|
|
$$
|
|
\\frac{3}{4}
|
|
$$
|
|
|
|
## Reducing fractions that contain variables
|
|
|
|
Superficially this looks to be more difficult but in fact we can apply the same
|
|
prime factorization method to get the result.
|
|
|
|
### Demonstration
|
|
|
|
_Reduce the following fraction to its lowest terms: $$\frac{25a^3b}{40a^2b^3}$$_
|
|
|
|
The prime factors of the numerator and denominator:
|
|
|
|
$$
|
|
\\begin{split}
|
|
25 = {5, 5} \\
|
|
40 = {2,2,2,5}
|
|
\\end{split}
|
|
$$
|
|
|
|
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.
|
|
|
|
$$\frac{25a^3b}{40a^2b^3} =\frac{5 \cdot 5 \cdot a \cdot a \cdot a \cdot b}{2 \cdot 2 \cdot 2 \cdot 5 \cdot a \cdot a \cdot b \cdot b \cdot b }$$
|
|
|
|
Canceled:
|
|
|
|
$$\frac{\cancel{5} \cdot 5 \cdot \cancel{a} \cdot \cancel{a} \cdot a \cdot \cancel{b}}{2 \cdot 2 \cdot 2 \cdot \cancel{5} \cdot \cancel{a} \cdot \cancel{a} \cdot \cancel{b} \cdot b \cdot b }$$
|
|
|
|
Which gives us:
|
|
|
|
$$
|
|
\\frac{5 \cdot a}{2 \cdot 2 \cdot 2 \cdot b \cdot b} = \frac{5a}{8b^2}
|
|
$$
|
|
|
|
## Reducing fractions that contain negative values
|
|
|
|
_Reduce the following fraction to its lowest terms: $$\frac{14y^5}{-35y^3}$$_
|
|
|
|
- This fraction is an instance of a
|
|
[fraction with unlike terms](Handling%20negative%20fractions.md#fractions-with-unlike-terms).
|
|
|
|
- Apply [Prime factorization](Prime%20factorization.md):
|
|
|
|
|
|
|
|
- Cancel the coefficients and variable parts
|
|
|
|
$$
|
|
\\frac{14y^5}{-35y^3}=\frac{5 \cdot 7 \cdot 2 \cdot y \cdot y \cdot y \cdot y \cdot y}{-5 \cdot 7 \cdot y \cdot y \cdot y} = - \frac{2y^2}{5}
|
|
$$
|
|
|
|
_Reduce the following fraction to its lowest terms:
|
|
$$\frac{- 12xy^2}{ - 18xy^2}$$_
|
|
|
|
- This fraction is an instance of a
|
|
[fraction with like terms](Handling%20negative%20fractions.md#fractions-with-like-terms).
|
|
|
|
- Apply [Prime factorization](Prime%20factorization.md):
|
|
|
|
|
|
|
|
- Cancel the coefficients and variable parts
|
|
|
|
$$
|
|
|
|
* \\frac{12xy^2}{18xy^2}=\frac{3 \cdot 2 \cdot 2 \cdot x \cdot y \cdot y}{3 \cdot 7 \cdot 3 \cdot 2 \cdot x \cdot x \cdot y} = - \frac{2y}{3x}
|
|
$$
|