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# Find three different irrational numbers between the rational numbers $$\frac{5}{7}$$and $$\frac{9}{11}$$.

To find irrational numbers, firstly we shall divide 5 by 7 and 9 by 11,
So, for $$\frac{9}{11}$$,
we get, $$\require{enclose} \begin{array}{rll} 0.8181 && \hbox{(Explanations)} \\[-3pt] 11 \enclose{longdiv}{90}\kern-.2ex \\[-3pt] \underline{88\phantom{00}} && \hbox{(8 \times 11 = 88)} \\[-3pt] \phantom{0}20 && \hbox{(90 - 88 = 2)} \\[-3pt] \underline{11\phantom{00}} && \hbox{(1 \times 11 = 11)} \\[-3pt] \phantom{0}90 && \hbox{(20 - 11 = 9)} \\[-3pt] \underline{\phantom{0}88} && \hbox{(8 \times 11 = 88)} \\[-3pt] \phantom{00}2 \end{array}$$ Now, for $$\frac{5}{7}$$,
we get,$$\require{enclose} \begin{array}{rll} 0.714285 && \hbox{(Explanations)} \\[-3pt] 7 \enclose{longdiv}{50}\kern-.2ex \\[-3pt] \underline{49\phantom{00}} && \hbox{(7 \times 7 = 49)} \\[-3pt] \phantom{0}10 && \hbox{(50 - 49 = 1)} \\[-3pt] \underline{7\phantom{00}} && \hbox{(1 \times 7 = 7)} \\[-3pt] \phantom{0}30 && \hbox{(10 - 7 = 3)} \\[-3pt] \underline{28\phantom{00}} && \hbox{(4 \times 7 = 28)} \\[-3pt] \phantom{0}20 && \hbox{(30 - 28 = 2)} \\[-3pt] \underline{14\phantom{00}} && \hbox{(2 \times 7 = 14)} \\[-3pt] \phantom{0}60 && \hbox{(20 - 14 = 6)} \\[-3pt] \underline{56\phantom{00}} && \hbox{(8 \times 7 = 56)} \\[-3pt] \phantom{0}40 && \hbox{(60 - 56 = 4)} \\[-3pt] \underline{\phantom{0}35} && \hbox{(5 \times 7 = 35)} \\[-3pt] \phantom{00}5 \end{array}$$ Thus, $$\frac{5}{7} = 0\overline{.714285} = 0.71428571428...$$

$$\frac{9}{11} = 0\overline{.818181} = 0.8181818.....$$

Therefore, the three rational numbers between these will be

0.74074007400074000074...

0.7750775007750007750000...

0.80800800080000...