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

Let \(\begin{align} a,b,c \in \mathbb{R} \backslash \{ 0\}. \end{align}\)

Now let us define a simple equation \(\begin{align} a=b+c \label{eq:start} \end{align}\)

and play around with it a little bit.

\[\begin{align*} a &= b+c & & \vert\ \cdot a\\ a^2 &= ab + ac & & \vert\ -b^2, -c^2\\ a^2-b^2-c^2 &= ab + ac - b^2 -c^2 & & \vert\ -2bc\\ a^2-b^2-c^2 -2bc &= ab + ac - b^2 -c^2 -2bc & & \\ a^2-b^2-c^2 -2bc &= ab - b^2 - bc + ac - bc - c^2 & & \\ a^2-b^2-c^2 -2bc &= b(a - b - c) + c(a - b - c) & & \\ \end{align*}\]

We then add the terms \(ab\) and \(ac\) to the left side of the equation and then subtract them again and rearrange the terms of the equation:

\[\begin{align*} a^2-b^2-c^2 -2bc + ab -ab +ac -ac &= b(a - b - c) + c(a - b - c) & & \\ a^2 + ab +ac -ab -b^2 -bc -ac -bc -c^2 &= b(a - b - c) + c(a - b - c) & & \\ (a^2 + ab +ac) - (ab +b^2 +bc) - (ac +bc +c^2) &= b(a - b - c) + c(a - b - c) & & \\ a(a + b +c) - b(a +b + c) - c(a +b +c) &= b(a - b - c) + c(a - b - c) & & \\ (a + b +c) (a-b-c) &= b(a - b - c) + c(a - b - c) & & \vert\ \div (a-b-c)\\ \end{align*}\]

Now, the equation simplifies to:

\[\begin{align*} a+b+c &= b+c \\ a+ (b+c) &= (b+c) \\ \end{align*}\]

If we use our original relation \(a=b+c\) from Eq. \eqref{eq:start} and insert \(a\) for the term \(b+c\) we finally get:

\[\begin{align*} a+ a &= a & & \\ 2a &= a & & \vert\ \div a \\ 2 &= 1 \end{align*}\]

So, starting with Eq. \eqref{eq:start} we could prove that actually \(1=2\). Or is there something wrong up there?