Fun:Mathematical fallacies
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1+1=11 |
(Incorrect) proof that 1 = 2[edit]
assume:
multiply both sides by :
subtract from both sides:
factor both sides:
divide both sides by :
as A and B are equal, substitute all s with s:
continuing:
A good example of why dividing by zero is a bad move.
(Incorrect) proof that 0 = -1 (or 1 = 2 if you prefer)[edit]
substitute :
Integrate by parts, ^{[1]} assume :
but so:
we substract both sides by :
then:
(Incorrect) proof that 1 = -1[edit]
assume:
rewrite -1 two different ways:
take the square root of both sides:
using laws of square roots, rewrite both sides:
multiply both sides by and reduce:
the square root of a number squared equals the number itself, so:
(Incorrect) proof that an elephant and a mosquito have the same mass[edit]
Let = mass of elephant in kg
Let = mass of mosquito in kg
Let = their combined mass in kg
Then:
multiplying the two latter equations:
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a^2-ay=x^2-xy}
adding Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(\frac{y}{2}\right)^2} to both sides:
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a^2-ay+\left(\frac{y}{2}\right)^2=x^2-xy+(\frac{y}{2})^2}
which can be rewritten:
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(a-\frac{y}{2}\right)^2=\left(x-\frac{y}{2}\right)^2}
from which derives:
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a-\frac{y}{2}=x-\frac{y}{2}}
and finally:
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x\in\R}
, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt{x^2}=|x|}
. So, the last line should not be Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a-\frac{y}{2}=x-\frac{y}{2}}
, but Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left|a-\frac{y}{2}\right|=\left|x-\frac{y}{2}\right|}
. In essence, the "proof" is claiming that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (-x)^2=(+x)^2}
implies Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -x=+x}
.
Another proof[edit]
Consider the function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)=x} , with domain the positive reals. Write
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=\underbrace{1+\cdots+1}_{x\text{ times}}}
Then multiplying through by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} we obtain
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^2=\underbrace{x+\cdots+x}_{x\text{ times}}}
Differentiating yields
Since by assumption we may divide through by , whence .
(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]
And dividing by :
This may be why assfly hates complex numbers.
See also[edit]
External links[edit]
References[edit]
- ↑ 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
- or in an even shorter form, if we let and the differentials , then it is in the form in which it is most often seen: