Fun:Mathematical fallacies

Part of a convergent series on Mathematics

1+1=11

Infinity Pseudomathematics Conservapedian mathematics Fibonacci to the 1000th iteration Aumann's agreement theorem Nine Central limit theorem Darrell Huff

We are number one

Alan Sokal Thomas Hobbes Ibn al-Haytham Napoleon Bonaparte 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:

$2B=B$

$2=1$

Q.E.D.

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

(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:

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

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

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

And dividing by $\pi i$ :

$1=-1$

This may be why assfly hates complex numbers.

See also

• Fun:Pi#Biblically knowing pi

External links

• Mathematical fallacies