Let $a$ and $b$ be positive whole numbers such that $\frac{4.5}{11}<\frac{a}{b}<\frac{5}{11}$. Find the fraction $\frac{a}{b}$ for which the sum $a+b$ is as small as possible. Justify your answer.

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By multiplying numerators and denominators by 7 , we can rewrite the inequalities as follows:

$$
\frac{7 \cdot 4.5}{7 \cdot 11}<\frac{a}{b}<\frac{7 \cdot 5}{7 \cdot 11} \Rightarrow \frac{31.5}{77}<\frac{a}{b}<\frac{35}{77}
$$

We now see that the fraction $\frac{a}{b}=\frac{33}{77}$ works: $\frac{31.5}{77}<\frac{33}{77}<\frac{35}{77}$. The fraction reduces to $\frac{a}{b}=\frac{3}{7}$, and its sum $a+b$ is $3+7=10$.

But is this the minimal possible sum $a+b$ ? To show that no other fraction with a smaller $a$ works, we multiply the inequalities by $11 b$ and solve for $b$ :

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