$2=1$ 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}
 A: Why not show all numbers are equal to 1: 
For any $z\in\mathbb R$,
$$
\sum_{n=-\infty}^{\infty}z^{n}=z\sum_{n=-\infty}^{\infty}z^{n-1}=z\sum_{n=-\infty}^{\infty}z^{n}.
$$
So 
$$
\sum_{n=-\infty}^{\infty}z^{n}=z\sum_{n=-\infty}^{\infty}z^{n}\Rightarrow 1=z.
$$
A: Another enjoyable "paradox": We first denote
$$S:=\sum_{n\in\mathbb N}\dfrac{(-1)^{n+1}}{n}$$
The fact that $0\neq S\in\mathbb R$ can be established using elementary tools.
We then write: 
$S=\frac{1}{1}-\frac {1}{2}+\frac{1}{3}-...+...-...$
$2S = 2(\frac{1}{1}-\frac {1}{2}+\frac{1}{3}-...+...-...)=\frac{2}{1}-\frac {2}{2}+\frac{2}{3}-\frac{2}{4}+\frac{2}{5}-\frac{2}{6}+\frac{2}{7}-...=$ 
$=\color{red}{\frac{2}{1}}\color{red}{-\frac {2}{2}}\color{green}{+\frac{2}{3}}\color{blue}{-\frac{2}{4}}+\frac{2}{5}\color{green}{-\frac{2}{6}}+\frac{2}{7}-...=\color{red}{\frac{1}{1}}\color{blue}{-\frac{1}{2}}\color{green}{+\frac{1}{3}}-...=S$

And at last:
$$2S = S \Longrightarrow 2=1$$
A: Here is a simple one:
$$
x=y\\
x^2=xy\\
2x^2=x^2+xy\\
2x^2-2xy=x^2-xy\\
2(x^2-xy)=1(x^2-xy)\\
2=1
$$
The error is quite obviously division by zero (from the 5th to 6th step).
A: $x = \underbrace{1 + 1 + 1 + \ldots + 1}_{x \textrm{ times}} = \underbrace{\frac{\mathrm{d}}{\mathrm{d}x}\left(x\right) + \frac{\mathrm{d}}{\mathrm{d}x}\left(x\right) + \frac{\mathrm{d}}{\mathrm{d}x}\left(x\right) + \ldots + \frac{\mathrm{d}}{\mathrm{d}x}\left(x\right)}_{x \textrm{ times}} = \frac{\mathrm{d}}{\mathrm{d}x}\underbrace{\left(x + x + x + \ldots + x\right)}_{x \textrm{ times}} = \frac{\mathrm{d}}{\mathrm{d}x}\left(x^2\right) = 2x$
A: Does this count or is it too obvious where things go wrong, considering there is only one step?
Define $f_n(x) = n \cdot 1_{x \le \frac 1n}$
Clearly for every $x$, $$\lim_{n \to \infty} f_n(x) = 0$$
Therefore 
$$\lim_{n \to \infty} \int_0^1 f_n(x)\  dx = \int_0^1 0 \ dx = 0$$
But $\int_0^1 f_n(x) \ dx = 1$ for every $n$; hence it is proved that 
$$1 = \lim_{n \to \infty} 1 = 0$$
A: Here's one I just made up. $\log_{b} b^x = x$. And $\log_{b} 1 = 0$.
Let $b = 1, x = 1$, and $b^x = 1$.  Then $0 = \log_b 1 = \log_b b^x = x = 1$
A: In the same vein as your example, let's integrate $\frac1x$ by parts.
Let $I = \int\frac1x\ \textrm dx$, and set $u = \frac1x, \textrm dv = \textrm dx$. Then:
$$
\begin{align}
I = \int u\ \textrm dv &= uv - \int v\ \textrm du \\
&= \frac1x\cdot x - \int x\left(\frac{-1}{x^2}\right) \textrm dx \\
&= 1 + \int\frac1x\ \textrm dx \\
&= 1 + I
\end{align}
$$
Therefore $0 = 1$, so clearly $1 = 2$.
A: Here is one of my favorites.

Proof that $1=0$
Let's consider for real $x$ the function $f(x)=xe^{-x^2}$. Note, the following integral representation of $f$ is valid (substitute: $u=x^2/y$).
  \begin{align*}
\int_{0}^{1}\frac{x^3}{y^2}e^{-x^2/y}\,dy
=\left[xe^{-x^2/y}\right]_0^1
=xe^{-x^2}
\end{align*}
We obtain for all $x$ the following relationship
\begin{align*}
e^{-x^2}(1-2x^2)&=\frac{d}{dx}\left(xe^{-x^2}\right)\\
&=\frac{d}{dx}\int_0^1\frac{x^3}{y^2}e^{-x^2/y}\,dy\\
&=\int_0^1\frac{\partial}{\partial x}\left(\frac{x^3}{y^2}e^{-x^2/y}\right)\,dy\\
&=\int_0^1e^{-x^2/y}\left(\frac{3x^2}{y^2}-\frac{2x^4}{y^3}\right)\,dy
\end{align*}
and observe by setting $x=0$ the left-hand side is one while the right-hand side is zero.
  \begin{align*}
\text{LHS: }\qquad e^0(1-0)&=1\\
\text{RHS: }\qquad \int_0^1 0\,dy&=0
\end{align*}

Note: This example can be found in Counterexamples in Analysis by B.R. Gelbaum and J.H.M. Holmsted.
A: $(1 - x)(1 + x + x^2 + ..... )=$
$(1 + x + x^2 .... )(-x - x^2 - x^3 -......)  = 1 + (x -x) + (x^2 - x^2)... = 1$ so
$1 + x + x^2 + .... = \frac{1}{1 - x}$
Let x = -1.
$ 1 - 1 + 1 - 1 + 1 - 1 .... = \frac{1}{1 -(-1)} = \frac 12$
but clearly $1 - 1 + 1 - 1 +... = (1-1) + (1-1) +... = 0$.
So $0 = \frac 12$ (and also 1, and -1).
A: Although it's not quite what you're looking for, the Banach-Tarski paradox shows that, in a certain sense, $1$ does equal $2$:
Given a solid ball in $\mathbb R^3$, there is a way to decompose the ball into $5$ disjoint sets, move them by rigid motions, and obtain two solid balls of the same radius.
The catch is that these are non-measurable sets (and, of course, you need the Axiom of Choice).
A: Let $x=(x_{ij})$ be the infinite matrix (where omitted entries are $0$), $$x = \begin{pmatrix}1  \\ -1 & 1 \\ &-1 & 1 \\ &&-1 & 1\\ &&&\ddots&
\end{pmatrix}$$
i.e. $x_{ij} = \Bbb 1_{i=j} - \Bbb 1_{i=j+1}$. Here $i$ is the row, $j$ is the column.
Then $$∑_{ij} x_{ij}=∑_i\left(∑_jx_{ij}\right) = ∑_i0 = 0$$
While also
$$∑_{ij} x_{ij}=∑_j\left(∑_ix_{ij}\right) = ∑_j\Bbb 1_{j=1} = 1$$
So $1=2$.

For the interested, this is a violation of Fubini.
A: I always liked the one that "proves" $1+2+3+4+...=1/12$
https://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF
A: 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.
A: $$0 = (1-1)+(1-1)+ … = 1 -(1-1)-(1-1)-… = 1 \implies 2=1$$ 
A: Let $U_n$ be a probability measure on $[0,1]$ such that when restricted to $\{0,\frac{1}{n},…,\frac{n-1}{n},1\}$, is the uniform measure on that set. i.e.
$$ U_n\left( A \right) := \left| \left\{ k∈{0,…,n} : \frac{k}{n} ∈ A\right\}\right|$$
Of course, as you send $n→∞$, $U_n$ tends to the (continuous) uniform measure on $[0,1], U_{[0,1]}$,
$$U_n→ U_{[0,1]}$$
Note that if $Q=\Bbb Q∩ [0,1]$, $Q$ is measurable and $U_{[0,1]}$-null, so
$$1 = U_n(Q) → U_{[0,1]}(Q) = 0$$
Hence  $2=1$.
