Whether at school or at university: An interesting question that \( 0,99999... = 1 \) and then includes the question of whether the following equation is true: \( 0,99999... = 1 \) . Although infinity \(0,99999... = A\) in the left part of the equation, we give it a name: \(0,99999... = A\) . After multiplication by the factor \(10\) and simple algebraic transformations, we get a first astonishing insight.
$$ \begin{array}{rcll} 9,99999... & = & 10\cdot A & \Leftrightarrow \\ 9 + 0,99999... & = & 10 \cdot A & \Leftrightarrow \\ 9 + A & = & 10 \cdot A & \Leftrightarrow \\ 9 & = & 9 \cdot A & \Leftrightarrow \\ 1 & = & A & \Leftrightarrow \\ 1 & = & 0,99999... & \end{array} $$
But what happens if you look at the following number $$ ...99999 $$, which at first sight seems a bit strange, in which infinity does not extend to the right but to the left?
We carry out the same transformations as above and receive:
$$ \begin{array}{rcll} ...99999 & = & B & \Leftrightarrow \\ ...999990 & = & 10\cdot B & \Leftrightarrow \\ B - 9 & = & 10 \cdot B & \Leftrightarrow \\ - 9 & = & 9 \cdot B & \Leftrightarrow \\ -1 & = & B & \Leftrightarrow \\ -1 & = & ...99999 & \end{array} $$
Finally, we consider the number \( ...99999,99999... \)
and receive the at first sight amazing result
$$ \begin{array}{rcll} ...99999,99999... & = & C & \Leftrightarrow \\ ...99999,99999... & = & 10\cdot C & \Leftrightarrow \\ C & = & 10 \cdot C & \Leftrightarrow \\ 0 & = & 9 \cdot C & \Leftrightarrow \\ 0 & = & C & \end{array} $$
But even this is quite consistent, because on the one hand \(A + B = 0.99999... + ...99999 = 99999.99999 = C\) and on the other hand $$A + B = 1 + (-1) = 0 = C$$.
Note: It is shown that if one defines \(A, B\) and \(C\) and assigns a reasonable value to them, then the values are \(1, -1\) and \(0\) .