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8. Find three different irrational numbers between the rational numbers 5/7 and 9/11.
Answer :

To find irrational numbers, firstly we shall divide 5 by 7 and 9 by 11,
So, for 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 5/11,
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, \(5/7 = 0\overline{.714285} = 0.71428571428...\)
\(9/11 = 0\overline{.818181} = 0.8181818.....\)

Therefore, the three rational numbers between these will be

0.74074007400074000074...

0.7750775007750007750000...

0.80800800080000...