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Fun:Mathematical fallacies

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2+2=4


(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 correct, 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