Question

use the benchmark \\(\frac{1}{2}\\) to compare \\(\frac{2}{9}\\) and \\(\frac{9}{10}\\). first, compare each fraction to \\(\frac{1}{2}\\). \\(\frac{2}{9}\\) is? \\(\frac{1}{2}\\), \\(\frac{9}{10}\\) is? \\(\frac{1}{2}\\)

Explanation:

Step1: Compare $\frac{2}{9}$ and $\frac{1}{2}$

To compare $\frac{2}{9}$ and $\frac{1}{2}$, we can find a common denominator. The common denominator of 9 and 2 is 18.
$\frac{2}{9}=\frac{2\times2}{9\times2}=\frac{4}{18}$
$\frac{1}{2}=\frac{1\times9}{2\times9}=\frac{9}{18}$
Since $4 < 9$, we have $\frac{4}{18}<\frac{9}{18}$, so $\frac{2}{9}<\frac{1}{2}$.

Step2: Compare $\frac{9}{10}$ and $\frac{1}{2}$

To compare $\frac{9}{10}$ and $\frac{1}{2}$, we find a common denominator. The common denominator of 10 and 2 is 10.
$\frac{1}{2}=\frac{1\times5}{2\times5}=\frac{5}{10}$
Since $9 > 5$, we have $\frac{9}{10}>\frac{5}{10}$, so $\frac{9}{10}>\frac{1}{2}$.

Answer:

$\frac{2}{9}$ is less than $\frac{1}{2}$, $\frac{9}{10}$ is greater than $\frac{1}{2}$ (and thus $\frac{2}{9}<\frac{9}{10}$)