Fun:Mathematical fallacies

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[edit] (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.

[edit] (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 and 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 ʃtan x dx:
\int\tan x\;dx - \int\tan x\;dx = - 1 + \int\tan x\;dx - \int\tan x\;dx
then:
0 = - 1

[edit] (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}


[edit] (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:
a + x = y
a = y - x
a - y = -x
multiplying the two latter equations:
a^2 - ay = x^2 - xy
adding (\frac{y}{2})^2 to both sides:
a^2 - ay + (\frac{y}{2})^2 = x^2 - xy + (\frac{y}{2})^2
which can be rewritten:
(a - \frac{y}{2})^2 = (x - \frac{y}{2})^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∈ℝ,  \sqrt{x^2}=|x| . So, the last line should not be  a - \frac{y}{2} = x - \frac{y}{2}, but  |a - \frac{y}{2}| = |x - \frac{y}{2}|. In essence, the "proof" is claiming that  (-x)^2=(+x)^2 implies -x=+x.

[edit] Another proof

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

x = \underbrace{1 + \cdots + 1}_{x \; \textrm{times}}.

Then multiplying through by x we obtain

x^2 = \underbrace{x + \cdots + x}_{x \; \textrm{ times}}

Differentiating yields

2x = \underbrace{1 + \cdots + 1}_{x \; \textrm{ times}} = x.

Since by assumption x > 0 we may divide through by x, whence 2=1.

[edit] (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 correct, in the sense that the conclusion follows from the premise. It's just that the premise is wrong.)

[edit] Another (incorrect) proof that 1 = -1

\log\left((-i)^2\right) = \log\left((-i)^2\right)
\log\left((-i)^2\right) = 2\log(-i)
\log(-1) = 2\left(\frac{-i\pi}{2}\right)
i\pi = -i\pi

And dividing by iπ:

1 = -1

This may be why assfly hates complex numbers.

[edit] References

  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 and dv = g′(x) dx, then it is in the form in which it is most often seen:
    \int u\,dv=uv-\int v\,du.
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