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# Fun:Mathematical fallacies

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 Part of aconvergent series onMathematics 1+1=11 v - t - e

## (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, [1] 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:
${\displaystyle {\frac {1}{-1}}={\frac {-1}{1}}}$
take the square root of both sides:
${\displaystyle {\sqrt {\frac {1}{-1}}}={\sqrt {\frac {-1}{1}}}}$
using laws of square roots, rewrite both sides:
${\displaystyle {\frac {\sqrt {1}}{\sqrt {-1}}}={\frac {\sqrt {-1}}{\sqrt {1}}}}$
multiply both sides by ${\displaystyle {\sqrt {1}}{\sqrt {-1}}}$ and reduce:
${\displaystyle {\sqrt {1}}{\sqrt {1}}={\sqrt {-1}}{\sqrt {-1}}}$
the square root of a number squared equals the number itself, so:
${\displaystyle \mathbf {1} =\mathbf {-1} }$

## (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{aligned}&a+x=y\\&a=y-x\\&a-y=-x\end{aligned}}}

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 ${\displaystyle x\in \mathbb {R} }$ , ${\displaystyle {\sqrt {x^{2}}}=|x|}$ . 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

${\displaystyle 2x=\underbrace {1+\cdots +1} _{x{\text{ times}}}=x}$

Since by assumption ${\displaystyle x>0}$ we may divide through by ${\displaystyle x}$ , whence ${\displaystyle 2=1}$ .

## (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

{\displaystyle {\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 ${\displaystyle \pi i}$ :

${\displaystyle 1=-1}$

This may be why assfly hates complex numbers.