Noma ngabe usesikoleni noma eyunivesithi: Umbuzo othakazelisayo \( 0,99999... = 1 \) bese ufaka umbuzo wokuthi lesi sibalo siyiqiniso yini: \( 0,99999... = 1 \) . Yize \(0,99999... = A\) okungapheli engxenyeni yesobunxele se-equation, sikunikeza igama: \(0,99999... = A\) . Ngemuva kokuphindaphindwa ngesici \(10\) nokuguqulwa okulula kwe-algebra, sithola ukuqonda kokuqala okumangazayo.
$$ \begin{array}{rcll} 9,99999... & = & 10\cdot A & \Leftrightarrow \\ 9 + 0,99999... & = & 10 \cdot A & \Leftrightarrow \\ 9 + A & = & 10 \cdot A & \Leftrightarrow \\ 9 & = & 9 \cdot A & \Leftrightarrow \\ 1 & = & A & \Leftrightarrow \\ 1 & = & 0,99999... & \end{array} $$
Lokho kwakungenzima kangako. Kodwa kwenzekani uma ucabanga ngenombolo elandelayo, ebonakala ingajwayelekile ekuqaleni: $$ ...99999 $$ , lapho i-infinity ingena khona ngakwesokudla, kodwa kwesobunxele?
Senza ukuguqulwa okufanayo njengoba ngenhla futhi sithola:
$$ \begin{array}{rcll} ...99999 & = & B & \Leftrightarrow \\ ...999990 & = & 10\cdot B & \Leftrightarrow \\ B - 9 & = & 10 \cdot B & \Leftrightarrow \\ - 9 & = & 9 \cdot B & \Leftrightarrow \\ -1 & = & B & \Leftrightarrow \\ -1 & = & ...99999 & \end{array} $$
Ekugcineni sibheka inombolo \( ...99999,99999... \)
futhi uthola okubukeka kumangalisa lapho uqala ukukubona
$$ \begin{array}{rcll} ...99999,99999... & = & C & \Leftrightarrow \\ ...99999,99999... & = & 10\cdot C & \Leftrightarrow \\ C & = & 10 \cdot C & \Leftrightarrow \\ 0 & = & 9 \cdot C & \Leftrightarrow \\ 0 & = & C & \end{array} $$
Kepha lokhu futhi kuyahambisana impela, ngoba ngakolunye uhlangothi \(A + B = 0,99999... + ...99999 = 99999,99999 = C\) bese ngakolunye, $ $ $$A + B = 1 + (-1) = 0 = C$$ iyasebenza.
Qaphela: Kuboniswa ukuthi amanani \(1, -1\) kanye \(0\) achazwa lapho \(A, B\) kanye \(C\) kuchazwa khona futhi kunikezwa inani elifanele.