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express-the-folowing-in-the0-6-bar-form-of-p-q-where-p-and-q-are-integers-and-q-is-not-0-0-6-

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Question Number 139 by shaleen last updated on 25/Jan/15
express the folowing in the0.6(bar)= form of p/q ;where p and q are integers and q is not =0 0.6^�
$$\boldsymbol{{express}}\:\boldsymbol{{the}}\:\boldsymbol{{folowing}}\:\boldsymbol{{in}}\:\boldsymbol{{the}}\mathrm{0}.\mathrm{6}\left(\boldsymbol{{bar}}\right)= \\ $$$$\boldsymbol{{form}}\:\boldsymbol{{of}}\:\boldsymbol{{p}}/\boldsymbol{{q}}\:;{where}\:{p}\:{and}\:{q} \\ $$$${are}\:{integers}\:{and}\:{q}\:{is}\:{not}\:=\mathrm{0} \\ $$$$\mathrm{0}.\bar {\mathrm{6}} \\ $$
Answered by vkulkarni last updated on 10/Dec/14
0.6^(−) =0.6666666.... =(6/(10))+(6/(100))+(6/(1000))+(6/(10^4 ))+.... This is an infinite G.P. with first term, a=(6/(10)) and r=(1/(10)), hence 0.6^− =(a/(1−r))=(( (6/(10)))/(1−(1/(10))))=(6/9)=(2/3)
$$\mathrm{0}.\overline {\mathrm{6}}=\mathrm{0}.\mathrm{6666666}…. \\ $$$$=\frac{\mathrm{6}}{\mathrm{10}}+\frac{\mathrm{6}}{\mathrm{100}}+\frac{\mathrm{6}}{\mathrm{1000}}+\frac{\mathrm{6}}{\mathrm{10}^{\mathrm{4}} }+…. \\ $$$$\mathrm{This}\:\mathrm{is}\:\mathrm{an}\:\mathrm{infinite}\:\mathrm{G}.\mathrm{P}.\:\mathrm{with}\:\mathrm{first} \\ $$$$\mathrm{term},\:{a}=\frac{\mathrm{6}}{\mathrm{10}}\:\mathrm{and}\:{r}=\frac{\mathrm{1}}{\mathrm{10}},\:\mathrm{hence} \\ $$$$\mathrm{0}.\overset{−} {\mathrm{6}}=\frac{{a}}{\mathrm{1}−{r}}=\frac{\:\frac{\mathrm{6}}{\mathrm{10}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{10}}}=\frac{\mathrm{6}}{\mathrm{9}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$

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