Information icon.svg Our policy on articles on living people is under review. Your comments and inputs are welcome on the relevant project talkpage.

Nominations and campaigning for the RationalWiki 2020 Moderator Election is underway and will end on November 23.

Fun:Mathematical fallacies

From RationalWiki
(Redirected from Mathematical fallacies)
Jump to: navigation, search
Part of a
convergent series on

Mathematics
Icon math.svg
1+1=11


(Incorrect) proof that 1 = 2[edit]

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 As with Bs:

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)[edit]

\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[edit]

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[edit]

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[edit]

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[edit]

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[edit]

\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.

See also[edit]

References[edit]

  1. For a more complete discussion of this tactic, see Wikipedia. Here is a quick explanation of what is being done here:
    In the traditional calculus curriculum, this rule is often stated using indefinite integrals in the form
    \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