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avatar+2353 

Funny fallacious 'proofs' (ordered by difficulty level);

 

'Proof' that a dog has nine legs;

A dog has nine legs;

No dog has five legs

A dog had four more legs than no dog

A dog had nine legs.

 

'Proof' that 2 = 1

 

'Proof' that 1 =  0

$$(n+1)^2 = n^2 + 2n + 1\\
(n+1)^2 -(2n+1) = n^2\\
(n+1)^2 - (2n+1) - n(2n+1) = n^2 - n(2n+1)\\
(n+1)^2 - (n+1)(2n+1) = n^2 - n(2n+1)\\
(n+1)^2 - (n+1)(2n+1)+\frac{(2n+1)^2}{4} = n^2-n(2n+1)++\frac{(2n+1)^2}{4}\\
((n+1)-\frac{2n+1}{2})^2 = (n-\frac{2n+1}{2})^2\\
(n+1)-\frac{2n+1}{2} = n - \frac{2n+1}{2}\\
n+1 = n\\
1 = 0\\$$

 

 

'Proof' that 1 is the largest number

 

 

 

 

'Proof' that 1 = -1

$$1 = \sqrt{1} = \sqrt{-1*-1} = \sqrt{-1}*\sqrt{-1} = i*i = i^2 = -1$$

 

'Proof' that all cars have the same color

Proof by induction

If there is one car, all cars have the same color.

Fix an arbitrary n. Suppose that for any set of n cars they have the same color. Consider any set of n+1 cars. Then car 1 through n have the same color. We also know that cars 2 through n+1 have the same color by the inductive hypothesis. Hence all cars must have the same color. 

 

'Proof' that 0 = 1

Using integration by parts we have

$$\int\frac{1}{x} = x*\frac{1}{x} - \int x d\frac{1}{x}\\
\int\frac{1}{x} = 1 - \int x(-\frac{1}{x^2})dx\\
\int\frac{1}{x} = 1 + \int \frac{1}{x}dx\\
0 = 1$$

 

'Proof' that $$e^x = 1 \mbox{ }\forall x$$

$$e^x = exp(i2\pi*\frac{x}{i2\pi}) = exp(i2\pi)^\frac{x}{i2\pi} = 1^{\frac{x}{i2\pi}} = 1$$

 

 

 

Have I just disproven your world? 

reinout-g  Jun 18, 2014

Best Answer 

 #2
avatar+91432 
+3

A great post reinout-g.    

Melody  Jun 18, 2014
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6+0 Answers

 #1
avatar+405 
+3

I cant even find a reaction picture for this....

Ornstein  Jun 18, 2014
 #2
avatar+91432 
+3
Best Answer

A great post reinout-g.    

Melody  Jun 18, 2014
 #3
avatar+11757 
0

looks like the devil is really old !lol!

rosala  Jun 18, 2014
 #4
avatar+91432 
0

I have just added this address to the sticky topic "Puzzles"

Melody  Jun 19, 2014
 #5
avatar+2353 
0

Aah, I guess if someone's interested (s)he could try and see where the fault lies in these.

 

Well thought of Melody 

reinout-g  Jun 19, 2014
 #6
avatar+91432 
0

Indeed he could!

Melody  Jun 19, 2014

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