Problems from Sorting fractions

Development:

We compare the fractions one by one.

• ${\displaystyle \frac{4}{3}}={\displaystyle \frac{12}{9}}$ and then it ${\displaystyle \frac{4}{3}}$ is bigger than ${\displaystyle \frac{7}{9}}$.
• ${\displaystyle \frac{4}{3}}={\displaystyle \frac{32}{24}}$ and ${\displaystyle \frac{11}{8}}={\displaystyle \frac{33}{24}}$ then ${\displaystyle \frac{4}{3}}$ is less than ${\displaystyle \frac{11}{8}}$.
• ${\displaystyle \frac{4}{3}}={\displaystyle \frac{8}{6}}$ and then ${\displaystyle \frac{4}{3}}$ is bigger than ${\displaystyle \frac{5}{6}}$.
• $1={\displaystyle \frac{3}{3}}$ and then ${\displaystyle \frac{4}{3}}$ is bigger than $1$.

Then, the biggest is ${\displaystyle \frac{11}{8}}$ and the second biggest is ${\displaystyle \frac{4}{3}}$. We compare three remaining ones. ${\displaystyle \frac{5}{6}}={\displaystyle \frac{15}{18}}$ and ${\displaystyle \frac{7}{9}}={\displaystyle \frac{14}{18}}$, then ${\displaystyle \frac{5}{6}}$ is bigger than ${\displaystyle \frac{7}{9}}.$ $1={\displaystyle \frac{9}{9}}$ and then ${\displaystyle \frac{7}{9}}$ is less than $1$.

In conclusion, sorted from greatest to smallest we have the fractions ${\displaystyle \frac{7}{9}},{\displaystyle \frac{5}{6}},1,{\displaystyle \frac{4}{3}}$ and ${\displaystyle \frac{11}{8}}.$

Solution:

Sorted from smallest to greatest we have the fractions ${\displaystyle \frac{7}{9}},{\displaystyle \frac{5}{6}},1,{\displaystyle \frac{4}{3}}$ and ${\displaystyle \frac{11}{8}}.$

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