# 0,99999... = 1?1214

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 $$0$$ .

Emva