We want to find a fraction a/b such that
[tex]\dfrac{9}{11}\leq\dfrac{a}{b}\leq\dfrac{11}{13}[/tex]
This is true if and only if
[tex]\dfrac{9b}{11}\leq a \leq\dfrac{11b}{13}[/tex]
We can choose a value for a only if the two extremes include at least one integer, i.e. if
[tex]\dfrac{11b}{13}-\dfrac{9b}{11}\geq 1[/tex]
Solving for b, we have [tex]b>35[/tex]
So, the smallest fraction is given for [tex]b=36[/tex]. We have
[tex]\dfrac{9}{11}\leq \dfrac{a}{36} \leq \dfrac{11}{13}[/tex]
Solving for a, we have a=30.
The fraction 30/36 can be simplified to 5/6. So, we have
[tex]\dfrac{9}{11}\leq \dfrac{5}{6} \leq \dfrac{11}{13}[/tex]
and this is the smallest possible denominator.
We want to find a fraction a/b such that
\dfrac{9}{11}\leq\dfrac{a}{b}\leq\dfrac{11}{13}
11
9
≤
b
a
≤
13
11
This is true if and only if
\dfrac{9b}{11}\leq a \leq\dfrac{11b}{13}
11
9b
≤a≤
13
11b
We can choose a value for a only if the two extremes include at least one integer, i.e. if
\dfrac{11b}{13}-\dfrac{9b}{11}\geq 1
13
11b
−
11
9b
≥1
Solving for b, we have b>35b>35
So, the smallest fraction is given for b=36b=36 . We have
\dfrac{9}{11}\leq \dfrac{a}{36} \leq \dfrac{11}{13}
11
9
≤
36
a
≤
13
11
Solving for a, we have a=30.
The fraction 30/36 can be simplified to 5/6. So, we have
\dfrac{9}{11}\leq \dfrac{5}{6} \leq \dfrac{11}{13}
11
9
≤
6
5
≤
13
11
and this is the smallest possible denominator.
