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

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

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Answer 1 · 2016-12-12T06:59:06

$72$ $72$

Explanation:

First, find the prime factorization of each denominator:

$8 = 2 \times 2 \times 2 = 2^{3}$ $8 = 2xx2xx2 = 2^3$
$9 = 3 \times 3 = 3^{2}$ $9 = 3xx3 = 3^2$

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

$2^{3} \times 3^{2} = 8 \times 9 = 72$ $2^3xx3^2 = 8xx9 = 72$

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

$\frac{1}{8} = \frac{1 \times 9}{8 \times 9} = \frac{9}{72}$ $1/8 = (1xx9)/(8xx9) = 9/72$
$\frac{2}{9} = \frac{2 \times 8}{9 \times 8} = \frac{16}{72}$ $2/9 = (2xx8)/(9xx8) = 16/72$

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 $\frac{1}{12}$ $1/12$ and $\frac{1}{18}$ $1/18$

$12 = 2 \times 2 \times 3 = 2^{2} \times 3^{1}$ $12 = 2xx2xx3 = 2^2xx3^1$
$18 = 2 \times 3 \times 3 = 2^{1} \times 3^{2}$ $18 = 2xx3xx3 = 2^1xx3^2$

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

$2^{2} \times 3^{2} = 4 \times 9 = 36$ $2^2 xx 3^2 = 4xx9 = 36$

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

$\frac{1}{12} = \frac{1 \times 3}{12 \times 3} = \frac{3}{36}$ $1/12 = (1xx3)/(12xx3) = 3/36$
$\frac{1}{18} = \frac{1 \times 2}{18 \times 2} = \frac{2}{36}$ $1/18 = (1xx2)/(18xx2) = 2/36$

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

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