\n| C<\/td>\n | \\[3,\\left( 1 \\right) = \\frac{{28}}{9}\\]<\/td>\n | \\[3,\\left( 1 \\right) = 3 + 0,\\left( 1 \\right) = 3 + \\frac{{\\left( {0,3} \\right)}}{3} = 3 + \\frac{{{\\textstyle{1 \\over 3}}}}{3} = 3 + \\frac{1}{9} = \\frac{{28}}{9}\\]<\/td>\n<\/tr>\n |
\n| D<\/td>\n | \\[2,8\\left( 3 \\right) = \\frac{{17}}{6}\\]<\/td>\n | \\[2,8\\left( 3 \\right) = \\frac{{28,\\left( 3 \\right)}}{{10}} = \\frac{{28 + \\left( {0,3} \\right)}}{{10}} = \\frac{{28 + {\\textstyle{1 \\over 3}}}}{{10}} = \\frac{{28}}{{\\mathop {10}\\limits_{\\left( { \\times 3} \\right)} }} + \\frac{1}{{30}} = \\frac{{85}}{{\\mathop {30}\\limits_{\\left( { \\div 5} \\right)} }} = \\frac{{17}}{6}\\]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n \u00a0<\/p>\n Apesar de acima estar realizada de outra maneira, a determina\u00e7\u00e3o da fra\u00e7\u00e3o irredut\u00edvel de cada uma das d\u00edzimas infinitas peri\u00f3dicas pode ser feita pelo m\u00e9todo habitual.\u00a0<\/p>\n Exemplifica-se seguidamente para o ponto B.<\/p>\n \\[ – 0,\\left( {27} \\right) = – \\frac{3}{{11}}\\]<\/p>\n Designando a d\u00edzima por \\(x\\), vem: \\(x = 0,\\left( {27} \\right)\\).
Logo, multiplicando por \\(100\\) os dois membros da igualdade anterior, temos: \\(100x = 27,\\left( {27} \\right)\\).
Subtraindo, membro a membro, as duas equa\u00e7\u00f5es anteriores, temos:
\\[\\begin{array}{*{20}{r}}{}&{100x}& = &{27,\\left( {27} \\right)}\\\\ – &x& = &{0,\\left( {27} \\right)}\\\\\\hline{}&{99x}& = &{27\\quad\\;\\;\\;\\;\\,}\\end{array}\\]
Donde, \\(x = \\frac{{27}}{{99}} = \\frac{{3}}{{11}}\\).
Portanto, \\( – 0,\\left( {27} \\right) = – \\frac{3}{{11}}\\).<\/p>\n \u00a0<\/p>\n Apenas vamos representar o primeiro dos n\u00fameros na reta num\u00e9rica.<\/p>\n \u00a0<\/p>\n \r\n\r\n |