A common method for adding fractions is to find the lowest common multiple of the denominators. This is called the LCD algorithm, Lowest Common Denominator algorithm. Even though we have the smallest possible denominator that allows us to combine the fractions, sometimes the sum will still reduce. I have always wondered why.

Consider \(\frac{1}{3}+\frac{1}{6}\). The **LCD** = 6, so
the sum becomes \(\frac{2}{6}+\frac{1}{6}=\frac{3}{6}\) or
\(\frac{1}{2}\). This is reducible.

Now consider \(\frac{1}{3}+\frac{1}{2}\). The **LCD** = 6,
so the sum becomes \(\frac{2}{6}+\frac{3}{6}=\frac{5}{6}\). This is
clearly **not** reducible.

I finally found an article explaining the theory behind when a fraction sum is reducible or irreducible. Please try our exploration tools before checking out number theory details.

View Example

Consider the table below comparing 5^{ths} to 10^{ths}.
The columns are the numerators for the denominator 5 and the rows are the
numerators for the denominator 10. Notice the sums are only reducible sometimes.

10 vs 5 | 1 | 2 | 3 | 4 |
---|---|---|---|---|

1 | 3/10 | 1/2 | 7/10 | 9/10 |

3 | 1/2 | 7/10 | 9/10 | 11/10 |

7 | 9/10 | 11/10 | 13/10 | 3/2 |

9 | 11/10 | 13/10 | 3/2 | 17/10 |

Are you sure that you are ready?

Did you really Explore some denominators first?

It's way more fun to explore first.

But if you insist, here's why.

References:

Biscuits of Number Theory by Arthur T. Benjamin, Ezra Brown, 2006, MAA URL

Originally published as: Reducing the Sum of Two Fractions Shultz, Harris S. and Ray C. Shiflett. Mathematics Teacher. vol. 98, no. 7 (March 2005): pp.486-490. URL

If the denominators have a commom prime power in the prime factorizations, then some sums may reduce, otherwise, the sums are always irreducible. For denominators \(12=2^2\cdot3\) and \(15=3\cdot5\) there exist numerators for which the sum will reduce since they have a common prime power of \(3^1\), and, for example, \(\frac{1}{12}+\frac{1}{15}=\frac{9}{60}=\frac{3}{20}\). For denominators \(8=2^3\) and \(12=2^2\cdot3\) there are no numerators for which the sum reduces, since they do not have a common prime power. Here I am assuming that the numerators are relatively prime to their denominators, i.e., they have a gcd of 1.

Even better, if the common prime power is based on 2, then the sums are
always reducible. Otherwise, there exist numerators for which the sum
will reduce. The problem then becomes finding those numerators. **This is
where the fun begins.**