Fun False Proofs

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rwald
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Fun False Proofs

Post by rwald » Sun Jun 13, 2004 11:47 am

One thing that always amused me were those proofs that appeared to be true, but which had some patently untrue conclusion, such as 2 = 1. I'm curious how many such proofs people here know. I'll start off with one of the classics:

a + b = c
4a + 4b = 4c
5c = 5a + 5b
4a + 4b + 5c = 5a + 5b + 4c
4a + 4b - 4c = 5a + 5b - 5c
4(a + b - c) = 5(a + b - c)
4 = 5

The reason this one is false is fairly simple, but not all such proofs are so easy. Does anyone know some better proofs of false statements?

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Tez
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Post by Tez » Sun Jun 13, 2004 11:11 pm

sqrt[(x-y)]=sqrt[-(y-x)]=sqrt[-1]*sqrt[(y-x)]=i*sqrt[(y-x)]=i*sqrt[-(x-y)]=i*sqrt[-1]*sqrt[(x-y)]=i*i*sqrt[(x-y)]=-sqrt[(x-y)]

Yahweh
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Post by Yahweh » Mon Jun 14, 2004 1:47 am

The Laws of Physics control atoms control your brains controls YOU!
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rwald
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Post by rwald » Mon Jun 14, 2004 2:18 am

Ah yes, I remember that one. Don't forget the car example...

Also, here's another more mathematically-oriented proof:

Code: Select all

x^2 = x + ... + x
        x times
d/dx(x^2) = d/dx(x + ... + x)
                   x times
2x = 1 + ... + 1
       x times
2x = x
2 = 1
For the record, I don't actually know anything. Not even this.

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Zombified
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Post by Zombified » Mon Jun 14, 2004 4:02 am

Euler's formula:

exp(iπ) + 1 = 0

exp(iπ) = -1

(exp(iπ))^2 = (-1)^2

exp(2iπ) = 1

log exp(2iπ) = log 1

2iπ = 0

π = 0

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Harry
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Post by Harry » Thu Jun 17, 2004 8:58 am

Zombified wrote:Euler's formula:

exp(iπ) + 1 = 0

exp(iπ) = -1

(exp(iπ))^2 = (-1)^2

exp(2iπ) = 1

log exp(2iπ) = log 1

2iπ = 0

π = 0
Just want to say that the symbol for pi looks bloody awful in my browser :)

My favourite false proof is the one that you've spent months working on. You've checked it, rechecked it and are absolutely sure that it's correct. Then some smart assed undergrad points out an obvious counter example. Seen way too many of those!
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Brian the Snail
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Post by Brian the Snail » Thu Jun 17, 2004 9:41 am

Tez wrote:sqrt[(x-y)]=sqrt[-(y-x)]=sqrt[-1]*sqrt[(y-x)]=i*sqrt[(y-x)]=i*sqrt[-(x-y)]=i*sqrt[-1]*sqrt[(x-y)]=i*i*sqrt[(x-y)]=-sqrt[(x-y)]
Or, sqrt[(x-y)]=sqrt[i^4]*sqrt[(x-y)]=i^2*sqrt[(x-y)]=-sqrt[(x-y)]