# 0,99999... = 1?1214

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 $$0$$ .

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