Don't Drink and Derive: A Simple Proof that 1 = 2

Let $\begin{array}{r} (1) & a, b, c \in R ∖ {0} . \end{array}$

Now let us define a simple equation $\begin{array}{r} (2) & a = b + c \end{array}$

and play around with it a little bit.

\begin{aligned} a & = b + c & | \cdot a \\ a^{2} & = a b + a c & | - b^{2}, - c^{2} \\ a^{2} - b^{2} - c^{2} & = a b + a c - b^{2} - c^{2} & | - 2 b c \\ a^{2} - b^{2} - c^{2} - 2 b c & = a b + a c - b^{2} - c^{2} - 2 b c \\ a^{2} - b^{2} - c^{2} - 2 b c & = a b - b^{2} - b c + a c - b c - c^{2} \\ a^{2} - b^{2} - c^{2} - 2 b c & = b (a - b - c) + c (a - b - c) \end{aligned}

We then add the terms $a b$ and $a c$ to the left side of the equation and then subtract them again and rearrange the terms of the equation:

\begin{aligned} a^{2} - b^{2} - c^{2} - 2 b c + a b - a b + a c - a c & = b (a - b - c) + c (a - b - c) \\ a^{2} + a b + a c - a b - b^{2} - b c - a c - b c - c^{2} & = b (a - b - c) + c (a - b - c) \\ (a^{2} + a b + a c) - (a b + b^{2} + b c) - (a c + b c + c^{2}) & = b (a - b - c) + c (a - b - c) \\ a (a + b + c) - b (a + b + c) - c (a + b + c) & = b (a - b - c) + c (a - b - c) \\ (a + b + c) (a - b - c) & = b (a - b - c) + c (a - b - c) & | \div (a - b - c) \end{aligned}

Now, the equation simplifies to:

\begin{aligned} a + b + c & = b + c \\ a + (b + c) & = (b + c) \end{aligned}

If we use our original relation $a = b + c$ from Eq. $(2)$ and insert $a$ for the term $b + c$ we finally get:

\begin{aligned} a + a & = a \\ 2 a & = a & | \div a \\ 2 & = 1 \end{aligned}

So, starting with Eq. $(2)$ we could prove that actually $1 = 2$ . Or is there something wrong up there?

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