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DigiNut
You kids get off my lawn!

Registered: Dec 2002
Location: Toronto, Self-proclaimed Centre of the Universe
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| quote: | Originally posted by NeoPhono
If you like some "fun" proofs that also are a little wacky, try this classic.
| quote: | Step 1: Let a=b.
Step 2: Then a^2 = ab
Step 3: a^2 + a^2 = a^2 + ab
Step 4: 2a^2 = a^2 + ab
Step 5: 2a^2 - 2ab = a^2 + ab - 2ab
Step 6: 2a^2 - 2ab = a^2 - ab
Step 7: reduced to 2(a^2 - ab) = 1(a^2 - ab)
Setp 8: cancelling (a^2 - ab) on both sides leaves 2 = 1 |
There's also a similar proof using complex numbers, but I thought this one would be more "kid friendly." |
Haha, I've seen that one before. Of course, when you cut out all the eye candy, it's more like:
10 - 10 = 5 - 5
10(1 - 1) = 5(1 - 1)
10 = 5
Ah, the lovely things you can do with the number zero. The only problem is, it's mathematically invalid to "factor out" a zero, so the whole proof is invalid, just like the God proof. 
I think I've seen the one with complex numbers too, doesn't it exploit the periodic nature of the complex exponential and attempt to prove an invalid construct using multiples of 2j*pi as the argument?
/offtopic 
As for proving the existence of God, the only really viable way to do it would be proof by contradiction (i.e. assume the contrary and prove it is impossible). However, no form of proof for God could ever be constructed this way, because God is a catchall explanation, and proof by contradiction requires an alternative hypothesis (form of disproof).
___________________
My party schedule:
2009-02-21 - DJ Attention @ I'm So Popular
2009-06-18 - DJ Annoying @ People Need To Know Where I'll Be
2012-11-32 - DJ Insufferable ɸ Or At Least the Stalkers I Complain About
2048-06-66 - Spastic & Whocares ¶ Although I'm Actually Flattered
9999-45-81 - Tweaker Gimp ☼ I Probably Won't Even Go To This But I Have To Make Sure I Fill Up All The Available Space Here
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Jan-05-2004 23:04
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NeoPhono
Übermensch

Registered: Sep 2003
Location: In Orbit
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| quote: | | I think I've seen the one with complex numbers too, doesn't it exploit the periodic nature of the complex exponential and attempt to prove an invalid construct using multiples of 2j*pi as the argument? |
Actually, the one I know of is much simpler than that. In fact, it could be argued that even though imaginary numbers are involved, it has nothing to do with them.
Here is is. (sorry it's so ugly, couldn't figure out a square root sign, so I used a negative 2 exponent instead)
Step 1: -1/1 = 1/-1
Step 2: Taking the square root of both sides: (-1/1)^-2 = (1/-1)^-2
Step 3: Simplifying: (-1)^-2 / (1)^-2 = (1)^-2 / (-1)^-2
Step 4: In other words, i/1 = 1/i.
Step 5: Therefore, i / 2 = 1 / (2i),
Step 6: i/2 + 3/(2i) = 1/(2i) + 3/(2i),
Step 7: i (i/2 + 3/(2i) ) = i ( 1/(2i) + 3/(2i) ),
Step 8: (i^2)/2 + (3i)/(2i) = i/(2i) + (3i)/(2i)
Step 9: (-1)/2 + 3/2 = 1/2 + 3/2,
Step 10: 1=2
I actually thought this one was easier to find the false step than the first. Probably because I was looking too hard in the "easier" one.
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Jan-06-2004 05:06
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DigiNut
You kids get off my lawn!

Registered: Dec 2002
Location: Toronto, Self-proclaimed Centre of the Universe
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| quote: | Originally posted by NeoPhono
Actually, the one I know of is much simpler than that. In fact, it could be argued that even though imaginary numbers are involved, it has nothing to do with them.
Here is is. (sorry it's so ugly, couldn't figure out a square root sign, so I used a negative 2 exponent instead)
Step 1: -1/1 = 1/-1
Step 2: Taking the square root of both sides: (-1/1)^-2 = (1/-1)^-2
Step 3: Simplifying: (-1)^-2 / (1)^-2 = (1)^-2 / (-1)^-2
Step 4: In other words, i/1 = 1/i.
Step 5: Therefore, i / 2 = 1 / (2i),
Step 6: i/2 + 3/(2i) = 1/(2i) + 3/(2i),
Step 7: i (i/2 + 3/(2i) ) = i ( 1/(2i) + 3/(2i) ),
Step 8: (i^2)/2 + (3i)/(2i) = i/(2i) + (3i)/(2i)
Step 9: (-1)/2 + 3/2 = 1/2 + 3/2,
Step 10: 1=2
I actually thought this one was easier to find the false step than the first. Probably because I was looking too hard in the "easier" one. |
Bahahaha, that proof is ridiculous! Whoever wrote it seems to have forgotten that sqrt(x) = x^(1/2), not x^(-2). So (-1)^-2 is most definitely NOT i, it's actually 1. Basically everything from step 2 onward makes no sense. 
___________________
My party schedule:
2009-02-21 - DJ Attention @ I'm So Popular
2009-06-18 - DJ Annoying @ People Need To Know Where I'll Be
2012-11-32 - DJ Insufferable ɸ Or At Least the Stalkers I Complain About
2048-06-66 - Spastic & Whocares ¶ Although I'm Actually Flattered
9999-45-81 - Tweaker Gimp ☼ I Probably Won't Even Go To This But I Have To Make Sure I Fill Up All The Available Space Here
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Jan-06-2004 13:10
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