# 0,99999... = 1?1214

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.

$$\begin{array}{rcll} 9,99999... & = & 10\cdot A & \Leftrightarrow \\ 9 + 0,99999... & = & 10 \cdot A & \Leftrightarrow \\ 9 + A & = & 10 \cdot A & \Leftrightarrow \\ 9 & = & 9 \cdot A & \Leftrightarrow \\ 1 & = & A & \Leftrightarrow \\ 1 & = & 0,99999... & \end{array}$$

Kwakungekho nzima kangako. Kepha kwenzekani uma ubheka le nombolo elandelayo $$...99999$$ , okuthi lapho uthi nhlá ibonakale isimanga, lapho okungapheli kungadluleli kwesokudla kepha ngakwesobunxele?

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: Kuyaboniswa ukuthi uma umuntu echaza $$A, B$$ ne- $$C$$ bese ebabela inani elifanele kubo, khona-ke amanani angama- $$1, -1$$ Kanye no- $$0$$ .

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