# 1 = 2

Posted onMany people have seen “proofs” that 1 = 2. However most of them have an obvious fallacy – a division by 0.

A classic such example is:

\begin{eqnarray*}

a &=& b, \\

a^2 &=& ab, \\

a^2 – b^2 &=& ab – b^2, \\

(a+b)(a-b) &=& b(a-b), \\

(a+b) &=& b, \\

a+a &=& a, \\

2a &=& a, \\

2 &=& 1.

\end{eqnarray*}

This is a decent “trick” to show people, but as I said above, the mistake simply comes from the division by 0 on line 5. However, here is a different “proof”, can you see the mistake?

Consider the equation, \(2=x^{x^{x^{\ldots}}}\), with an infinite number of \(x\)’s. If we add some brackets to the equation we get, \(2=x^{\left(x^{x^{\ldots}}\right)}\) and then we can substitute the first equation in to get: \[2=x^2.\] So \(x=\sqrt{2}\), and hence \[2=\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots}}}\].

Then if we repeat the exact same above process with the equation \(4=x^{x^{x^{\ldots}}}\) we see that \(4=x^4\) and so again \(x=\sqrt{2}\). So this time, \[4=\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\ldots}}}.\]

And so \(2=4\) and thus \(1=2\)!

Where is the mistake?

Is it real or just for fun?