Nokuba usesikolweni okanye eyunivesithi: Umbuzo onomdla \( 0,99999... = 1 \) kwaye ubandakanye umbuzo wokuba ingaba le nxaki iyinyani: \( 0,99999... = 1 \) . Nangona ubuthongo obungapheliyo kwinxalenye yasekhohlo ye-equation, \(0,99999... = A\) igama: \(0,99999... = A\) . Emva kokuphindaphinda ngento \(10\) kunye notshintsho olulula lwealgebra, sifumana umbono wokuqala omangalisayo.
$$ \begin{array}{rcll} 9,99999... & = & 10\cdot A & \Leftrightarrow \\ 9 + 0,99999... & = & 10 \cdot A & \Leftrightarrow \\ 9 + A & = & 10 \cdot A & \Leftrightarrow \\ 9 & = & 9 \cdot A & \Leftrightarrow \\ 1 & = & A & \Leftrightarrow \\ 1 & = & 0,99999... & \end{array} $$
Kwakungekho nzima kwaphela. Kodwa kwenzeka ntoni xa ujonga eli nani lilandelayo $$ ...99999 $$ , ethi ekuqalekeni ibonakale ingaqhelekanga, apho ubude obungapheliyo budlulela ngasekunene kodwa ngasekhohlo?
Senza iinguqu ezifanayo njengasentla kwaye samkela:
$$ \begin{array}{rcll} ...99999 & = & B & \Leftrightarrow \\ ...999990 & = & 10\cdot B & \Leftrightarrow \\ B - 9 & = & 10 \cdot B & \Leftrightarrow \\ - 9 & = & 9 \cdot B & \Leftrightarrow \\ -1 & = & B & \Leftrightarrow \\ -1 & = & ...99999 & \end{array} $$
Okokugqibela sijonga inani \( ...99999,99999... \)
kwaye ufumana oko kubonakala kumangalisa ekuboneni kokuqala
$$ \begin{array}{rcll} ...99999,99999... & = & C & \Leftrightarrow \\ ...99999,99999... & = & 10\cdot C & \Leftrightarrow \\ C & = & 10 \cdot C & \Leftrightarrow \\ 0 & = & 9 \cdot C & \Leftrightarrow \\ 0 & = & C & \end{array} $$
Kodwa oku kuyahambelana, kuba kwelinye icala \(A + B = 0,99999... + ...99999 = 99999,99999 = C\) kwelinye icala $$A + B = 1 + (-1) = 0 = C$$ iyasebenza.
Inqaku: Kuyaboniswa ukuba umntu uchaza \(A, B\) kunye \(C\) kwaye abele ixabiso elifanelekileyo kubo, ke amaxabiso \(1, -1\) kunye \(0\) .