Hadday tahay dugsiga ama jaamacadda: Su'aal xiise leh oo \( 0,99999... = 1 \) iyo ka dibba waxaa ku jira su'aasha ah haddii isla'egta soo socota ay run tahay: \( 0,99999... = 1 \) . In kasta oo ay hurdo \(0,99999... = A\) ku dhacdo qaybta bidix ee isla'egta, waxaan u bixinnaa magac: \(0,99999... = A\) . Ka dib isku dhufashada qodobka \(10\) iyo isbadalka aljabra ee fudud, waxaan helnay aragti hor leh oo yaab leh.
$$ \begin{array}{rcll} 9,99999... & = & 10\cdot A & \Leftrightarrow \\ 9 + 0,99999... & = & 10 \cdot A & \Leftrightarrow \\ 9 + A & = & 10 \cdot A & \Leftrightarrow \\ 9 & = & 9 \cdot A & \Leftrightarrow \\ 1 & = & A & \Leftrightarrow \\ 1 & = & 0,99999... & \end{array} $$
Ma ahayn wax sidaas u adag gabi ahaanba. Laakiin maxaa dhacaya haddii aad eegto lambarka soo socda $$ ...99999 $$ , taas oo jaleecada hore u muuqata waxoogaa la yaab leh, taas oo aan dhammaadkeedu u baahnayn dhinaca midig ee bidixda?
Waxaan fulinaa isla isbeddellada sida kor ku xusan oo aan helno:
$$ \begin{array}{rcll} ...99999 & = & B & \Leftrightarrow \\ ...999990 & = & 10\cdot B & \Leftrightarrow \\ B - 9 & = & 10 \cdot B & \Leftrightarrow \\ - 9 & = & 9 \cdot B & \Leftrightarrow \\ -1 & = & B & \Leftrightarrow \\ -1 & = & ...99999 & \end{array} $$
Ugu dambeyntii waxaan tixgelinaynaa lambarka \( ...99999,99999... \)
waxaadna helaysaa wax umuuqda layaab aragtida koowaad
$$ \begin{array}{rcll} ...99999,99999... & = & C & \Leftrightarrow \\ ...99999,99999... & = & 10\cdot C & \Leftrightarrow \\ C & = & 10 \cdot C & \Leftrightarrow \\ 0 & = & 9 \cdot C & \Leftrightarrow \\ 0 & = & C & \end{array} $$
Laakiin tani sidoo kale waa mid iswaafaqsan, maadaama dhinac ka ah \(A + B = 0,99999... + ...99999 = 99999,99999 = C\) iyo kan kale, $$A + B = 1 + (-1) = 0 = C$$ khuseeya.
Fiiro gaar ah: Waxaa la muujiyey in haddii qofku qeexo \(A, B\) iyo \(C\) oo uu u yeelo qiime macquul ah, markaa qiimayaashu waa \(1, -1\) iyo \(0\) .