$2=1$ Paradoxes repository

I really like to use paradoxes in my math classes, in order to awaken the interest of my students. Concretely, these last weeks I am proposing paradoxes that achieve the conclusion that 2=1. After one week, I explain the solution in the blackboard and I propose a new one. For example, I posted the following one some months ago: What is wrong with the sum of these two series? I would like to increase my repertoire of fake-proofs. I would be glad to read your proposals and discuss them! My students are 18 years old, so don't be too cruel :) Here is my own contribution:

\begin{equation} y(x) = \tan x \end{equation} \begin{equation} y^{\prime} = \frac{1}{\cos^{2} x} \end{equation} \begin{equation} y^{\prime \prime} = \frac{2 \sin x}{\cos^{3} x} \end{equation} This can be rewritten as: \begin{equation} y^{\prime \prime} = \frac{2 \sin x}{\cos^{3} x} = \frac{2 \sin x}{\cos x \cdot \cos^{2} x} = 2 \tan x \cdot \frac{1}{\cos^{2} x} = 2yy^{\prime} = \left( y^{2} \right)^{\prime} \end{equation} Integrating both sides of the equation $y^{\prime \prime} = \left( y^{2} \right)^{\prime}$: \begin{equation} y^{\prime} = y^{2} \end{equation} And therefore \begin{equation} \frac{1}{\cos^{2} x} = \tan^{2} x \end{equation} Now, evalueting this equation at $x = \pi / 4$ \begin{equation} \frac{1}{(\sqrt{2}/2)^{2}} = 1^{2} \end{equation} \begin{equation} 2 = 1 \end{equation}

Solution 1:

One of my favorites, and very simple to understand for most algebra students:

$2 = 1+1$

$2 = 1+\sqrt{1}$

$2 = 1+\sqrt{(-1)(-1)}$

$2 =^* 1+\sqrt{-1}\sqrt{-1}$

$2 = 1+i*i$

$2 = 1+i^2$

$2 = 1+(-1)$

$2 = 0$

$^*$ The wrong step

Divide both sides by 2 and add 1 and you would get $2=1$, as you desired.

To be thorough, the mistake occurs in the fourth line where the square root is split. In reality, the rule is:

$\sqrt{ab}=\sqrt{a}\sqrt{b}$ when either $a\geq0$ or $b\geq0$

$\sqrt{ab}=-\sqrt{a}\sqrt{b}$ when $a<0$ and $b<0$

So you would have an equality if you follow that rule, but many students aren't going to catch the error.

Solution 2:

$$0 = (1-1)+(1-1)+ … = 1 -(1-1)-(1-1)-… = 1 \implies 2=1$$