# Fun:Mathematical fallacies

## (Incorrect) proof that 1 = 2

assume:

$\displaystyle A=B$

multiply both sides by $\displaystyle A$  :

$\displaystyle AA=AB$

subtract $\displaystyle B^2$ from both sides:

$\displaystyle A^2-B^2=AB-B^2$

factor both sides:

$\displaystyle (A-B)(A+B)=B(A-B)$

divide both sides by $\displaystyle A-B$  :

$\displaystyle A+B=B$

as A and B are equal, substitute all $\displaystyle A$ s with $\displaystyle B$ s:

$\displaystyle B+B=B$

continuing:

$\displaystyle 2B=B$
$\displaystyle 2=1$

A good example of why dividing by zero is a bad move.

## (Incorrect) proof that 0 = -1 (or 1 = 2 if you prefer)

$\displaystyle \int\tan(x)dx=\int\tan(x)dx$

substitute $\displaystyle \tan(x)$  :

$\displaystyle \int\tan(x)dx=\int\sin(x)\sec(x)dx$

Integrate by parts,  assume $\displaystyle u=\sec(x),dv=\sin(x)dx$  :

$\displaystyle \int\tan(x)dx=-\sec(x)\cos(x)+\int\cos(x)\tan(x)\sec(x)dx$

but $\displaystyle \cos(x)\sec(x)=1$ so:

$\displaystyle \int\tan(x)dx=-1+\int\tan(x)dx$

we substract both sides by $\displaystyle \int\tan(x)dx$  :

$\displaystyle \int\tan(x)dx-\int\tan(x)dx =-1+\int\tan(x)dx-\int\tan(x)dx$

then:

$\displaystyle 0=-1$

## (Incorrect) proof that 1 = -1

assume:
$\displaystyle \mathbf{-1}=\mathbf{-1}$
rewrite -1 two different ways:
$\displaystyle \frac1{-1}=\frac{-1}{1}$
take the square root of both sides:
$\displaystyle \sqrt{\frac1{-1}}=\sqrt{\frac{-1}{1}}$
using laws of square roots, rewrite both sides:
$\displaystyle \frac{\sqrt1}{\sqrt{-1}}=\frac{\sqrt{-1}}{\sqrt1}$
multiply both sides by $\displaystyle \sqrt1\sqrt{-1}$ and reduce:
$\displaystyle \sqrt1\sqrt1=\sqrt{-1}\sqrt{-1}$
the square root of a number squared equals the number itself, so:
$\displaystyle \mathbf1=\mathbf{-1}$

## (Incorrect) proof that an elephant and a mosquito have the same mass

Let $\displaystyle a$ = mass of elephant in kg
Let $\displaystyle x$ = mass of mosquito in kg
Let $\displaystyle y$ = their combined mass in kg
Then:

\displaystyle \begin{align}&a+x=y\\&a=y-x\\&a-y=-x\end{align}

multiplying the two latter equations:

$\displaystyle a^2-ay=x^2-xy$

adding $\displaystyle \left(\frac{y}{2}\right)^2$ to both sides:

$\displaystyle a^2-ay+\left(\frac{y}{2}\right)^2=x^2-xy+(\frac{y}{2})^2$

which can be rewritten:

$\displaystyle \left(a-\frac{y}{2}\right)^2=\left(x-\frac{y}{2}\right)^2$

from which derives:

$\displaystyle a-\frac{y}{2}=x-\frac{y}{2}$

and finally:

$\displaystyle a=x$

that is, mass of elephant = mass of mosquito.

The fallacy lies in the second to last step, when you take the square root of both sides. For all $\displaystyle x\in\R$ , $\displaystyle \sqrt{x^2}=|x|$ . So, the last line should not be $\displaystyle a-\frac{y}{2}=x-\frac{y}{2}$ , but $\displaystyle \left|a-\frac{y}{2}\right|=\left|x-\frac{y}{2}\right|$ . In essence, the "proof" is claiming that $\displaystyle (-x)^2=(+x)^2$ implies $\displaystyle -x=+x$ .

## Another proof

Consider the function $\displaystyle f(x)=x$ , with domain the positive reals. Write

$\displaystyle x=\underbrace{1+\cdots+1}_{x\text{ times}}$

Then multiplying through by $\displaystyle x$ we obtain

$\displaystyle x^2=\underbrace{x+\cdots+x}_{x\text{ times}}$

Differentiating yields

$2x=\underbrace {1+\cdots +1} _{x{\text{ times}}}=x$ Since by assumption $x>0$ we may divide through by $x$ , whence $2=1$ .

## (Incorrect) proof that I am the Pope

This is a classic by the mathematician G. H. Hardy.

The Pope and I are two. [That is, two people.]

By the previous theorem, 2 = 1.

Therefore, the Pope and I are one.

(Technically, this proof is valid, in the sense that the conclusion follows from the premise. It's just that the premise is wrong.)

## Another (incorrect) proof that 1 = -1

{\begin{aligned}&\log \left((-i)^{2}\right)=\log \left((-i)^{2}\right)\\&\log \left((-i)^{2}\right)=2\log(-i)\\&\log(-1)=2\left({\frac {-\pi i}{2}}\right)\\&\pi i=-\pi i\end{aligned}} And dividing by $\pi i$ :

$1=-1$ This may be why assfly hates complex numbers.