Irreducible Fractions | Elementary

A fraction is irreducible when the numerator and the denominator share only |1| as a common factor. Thus, the fraction has been “reduced” to its simplest form.

For example, the fraction |\dfrac{12}{15}| can be simplified again.

In this example, |4| and |5| only share |1| as a common factor. The irreducible fraction is |\dfrac{4}{5}| since |4| divided by |1| equals |4| and |5| divided by |1| equals |5|. Thus, the fraction has been reduced to its simplest form.

A simplified fraction must always be equivalent to the initial fraction.

1. I find a common factor for the numerator and the denominator.

2. I divide the numerator by the number.

3. I divide the denominator by the number.

4. I repeat steps |1|, |2| and |3| until no common factor exists other than |1.|

Reduce the fraction |\dfrac{36}{48}| to its lowest terms.

I find a common factor for the numerator and the denominator. I can use the divisibility rules to help.To learn more about the rules, see the Divisibility Criteria concept sheet. Since |36| and |48| are even numbers, they are both divisible by |2.|
I divide the numerator by that number.
I divide the denominator by that number.
I repeat steps |1|, |2| and |3| until no common factor exists other than |1.| Since |18| and |24| are both divisible by |3|, divide the numerator and the denominator by |3| to get |\dfrac{6}{8}|. Since |6| and |8| are both divisible by |2|, divide the numerator and the denominator by |2| to get |\dfrac{3}{4}|. Since |3| and |4| only have |1| as a common factor, the fraction is irreducible.

The fraction  |\dfrac{3}{4}| is the irreducible equivalent of |\dfrac{36}{48}.|

1. I find the GCF of the numerator and the denominator.

2. I divide the numerator by the GCF.

3. I divide the denominator by the GCF.

Reduce the fraction |\dfrac{24}{32}| to its lowest terms.

Find the GCF of the numerator and the denominator. To learn how, see the Greatest Common Factor (GCF) concept sheet. The greatest common factor of |24| and of |32| is |8.|
I divide the numerator by the GCF.
I divide the denominator by the GCF. Since the numerator and the denominator only share |1| as a common factor, the fraction is irreducible.

The fraction |\dfrac{3}{4}| is the irreducible equivalent of |\dfrac{24}{32}.|