# Fun:Mathematical fallacies

 Part of aconvergent series onMathematics 2+2=4 v - t - e

## (Incorrect) proof that 1 = 2

assume:

$A=B$

multiply both sides by $A$ :

$AA=AB$

subtract $B^2$ from both sides:

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

factor both sides:

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

divide both sides by $A-B$ :

$A+B=B$

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

$B+B=B$

continuing:

$2B=B$
$2=1$

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

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

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

substitute $\tan(x)$ :

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

Integrate by parts, [1] assume $u=\sec(x),dv=\sin(x)dx$ :

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

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

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

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

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

then:

$0=-1$

## (Incorrect) proof that 1 = -1

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

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

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

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

multiplying the two latter equations:

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

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

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

which can be rewritten:

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

from which derives:

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

and finally:

$a=x$

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 $x\in\R$ , $\sqrt{x^2}=|x|$ . So, the last line should not be $a-\frac{y}{2}=x-\frac{y}{2}$ , but $\left|a-\frac{y}{2}\right|=\left|x-\frac{y}{2}\right|$ . In essence, the "proof" is claiming that $(-x)^2=(+x)^2$ implies $-x=+x$.

## Another proof

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

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

Then multiplying through by $x$ we obtain

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

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{align} &\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{align}

And dividing by $\pi i$ :

$1=-1$

This may be why assfly hates complex numbers.

$\int f(x)g'(x)dx =f(x)g(x)-\int f'(x)g(x)dx$
or in an even shorter form, if we let $u=f(x),v=g(x)$ and the differentials $du=f'(x)dx,dv=g'(x)dx$ , then it is in the form in which it is most often seen:
$\int u\,dv=uv-\int v\,du$