A cikin wannan tsari, daidaito 1 = 0.999... da 1 = 1.000... suna nunawa, bi da bi, gaskiyar cewa 1 yana cikin duka [0, 1] da [1, 2], don haka mutum na iya zaɓar ko dai wani yanki lokacin da yake nemo lambobin sa. Don tabbatar da cewa wannan alamar ba ta amfani da alamar "=" ba, mutum yana buƙatar hanyar sake gina ainihin lambobi na musamman ga kowane lambobi na goma. Ana iya yin wannan tare da iyaka, amma sauran gine-gine suna ci gaba da batun tsari. Zaɓin kai tsaye ɗaya shine ka'idar da aka saita, wanda ya ba da tabbacin cewa an ba da jerin tsararru, tsararru wanda tsawonsa ya zama ƙananan ƙananan, ƙayyadaddun sun ƙunshi ainihin ainihin lambobi a cikin haɗarsu. Don haka b0.1b2b3... an ayyana shi a matsayin ainihin lambar da ke ƙunshe a cikin duka [0, 1b] da [0, 1b1], don haka mutum zai iya zaɓar ko dai wani ɓangaren ƙayyadaddun da ke ƙunshe a cikin waɗannan lambobi [0, 1b] da [0, 1b2b] da kuma a kan kowane
lambar ta musamman