How can you the least find denominator for 1/8 and 2/9?

Dec 12, 2016

$72$

Explanation:

First, find the prime factorization of each denominator:

$8 = 2 \times 2 \times 2 = {2}^{3}$
$9 = 3 \times 3 = {3}^{2}$

Next, find the product of the greatest powers of each prime that occurs:

${2}^{3} \times {3}^{2} = 8 \times 9 = 72$

In this case, the least common denominator is $72$.

$\frac{1}{8} = \frac{1 \times 9}{8 \times 9} = \frac{9}{72}$
$\frac{2}{9} = \frac{2 \times 8}{9 \times 8} = \frac{16}{72}$

In the above case, we get the same result by just multiplying the two denominators. For an example where that is not the case, consider $\frac{1}{12}$ and $\frac{1}{18}$

$12 = 2 \times 2 \times 3 = {2}^{2} \times {3}^{1}$
$18 = 2 \times 3 \times 3 = {2}^{1} \times {3}^{2}$

The only primes which appear are $2$ and $3$. The greatest power of $2$ is ${2}^{2}$. The greatest power of $3$ is ${3}^{2}$. Multiplying them, we get

${2}^{2} \times {3}^{2} = 4 \times 9 = 36$

So the least common denominator between $\frac{1}{12}$ and $\frac{1}{18}$ is $36$.

$\frac{1}{12} = \frac{1 \times 3}{12 \times 3} = \frac{3}{36}$
$\frac{1}{18} = \frac{1 \times 2}{18 \times 2} = \frac{2}{36}$