I could swear that I've seen some math threads overrun by these, but I can't find them anymore. Anyway, why are these proofs wrong?:
a = b
a^2 = ab
a^2-b^2 = ab-b^2
(a+b)(a-b) = b(a-b)
a+b = b
2b = b
2 = 1


x = (Pi+3)/2
2x = Pi+3
2x(Pi-3) = (Pi+3)(Pi-3)
2Pix-6x = Pi^2-9
9-6x = Pi^2-2Pix
9-6x+x^2 = Pi^2-2Pix+x^2
(3-x)^2 = (Pi-x)^2
3-x = Pi-x
Pi = 3


-1 = -1
-1/1 = -1/1
-1/1 = 1/-1
sqrt(-1/1) = sqrt(1/-1)
i/1 = 1/i
i = 1/i
i * i = 1
-1 = 1

(a+b)(a-b) = b(a-b)
a+b = b
This is a divide by zero


I suspect the second one has a divide by zero error as well, but I'm too lazy to find it.



sqrt(-1/1) = sqrt(1/-1)
i/1 = 1/i

sqrt(a/b) = sqrt(a)/sqrt(b) only for non negative b

1/i = -i anyway so that should set of alarm bells for #3

x = (Pi+3)/2
2x = Pi+3
2x(Pi-3) = (Pi+3)(Pi-3)
2Pix-6x = Pi^2-9
9-6x = Pi^2-2Pix
9-6x+x^2 = Pi^2-2Pix+x^2
(3-x)^2 = (Pi-x)^2 .................... [1]
3-x = Pi-x ................................ [2]
Pi = 3
(red bits mine)

[2] is just one solution to eq'n [1]. Here are the others:

-(3-x) = (pi-x)
(3-x) = -(pi-x)
-(3-x) = -(pi-x)

Clearly, the solution you wrote down ([2] above) is incorrect. The correct solution is either:
-(3-x) = (pi-x)
or:
(3-x) = -(pi-x)
which both simplify to:
x = (pi+3)/2
as per the first line.

I saw this one the other day:

Time = Money
Women = Time*Money
Women = Money^2
Money = sqrtEvil
Money^2 = Evil
Women = Evil

pizza/me+my friends=number of slices i want-1=BS
((popular girl in high school+30 years)-job)couch=fat
(rate)time=distance(wich also=)$of a hooker

pizza/me+my friends=number of slices i want-1=BS
((popular girl in high school+30 years)-job)couch=fat
(rate)time=distance(wich also=)$of a hooker
You're stealing material from Dmitri Martin!!!:mad:

That'll show him, Dmititri Martin stole material form me. :mad:

First I should call your attention to this thread, (http://forums.civfanatics.com/showthread.php?t=308225) particularly posts 23, and 27.

And for a new one:
n^2 = n*n, where n != 0
n^2 = n + n +...+ n (n times).
2n = 1 + 1 +...+ 1 (n times) -- take derivatives.
2n = n
2 = 1 -- divide by n (n != 0)

You didn't take the derivative of the (n times)

How about this one?

e^1 = e^((j*pi)/(j*pi))
=(e^(j*pi))^(1/(j*pi))
using Euler's Theorem
=(cos (j*pi) +j*sin(j*pi))^(1/(j*pi))
cos->1, sin->0
=(1-j*0)^(1/(j*pi))
=1^(1/(j*pi))
=1
e=1

Yep .

I could swear that I've seen some math threads overrun by these, but I can't find them anymore. Anyway, why are these proofs wrong?:[FONT=Verdana]
a = b
a^2 = ab
a^2-b^2 = ab-b^2
(a+b)(a-b) = b(a-b)
a+b = b
2b = b
2 = 1

Answer Should be a^2-b^2=ab-b^2 Which just gets you back to where you were before.

There's no misstep there, Cubsfan. That's a Perfectly okay algebraic manipulation.

why is this in the humar forum?

Why is this in the Humor Forum?
This is on topic though:
PROOF THAT GIRLS ARE EVIL:
http://dansemacabre.files.wordpress.com/2007/03/proof_girls_are_evil.jpg

How about this one?

e^1 = e^((j*pi)/(j*pi))
=(e^(j*pi))^(1/(j*pi))
using Euler's Theorem
=(cos (j*pi) +j*sin(j*pi))^(1/(j*pi))
cos->1, sin->0
=(1-j*0)^(1/(j*pi))
=1^(1/(j*pi))
=1
e=1

yea but stuff like this is not funny

How about this one?

e^1 = e^((j*pi)/(j*pi))
=(e^(j*pi))^(1/(j*pi))
using Euler's Theorem
=(cos (j*pi) +j*sin(j*pi))^(1/(j*pi))
cos->1, sin->0
=(1-j*0)^(1/(j*pi))
=1^(1/(j*pi))
=1
e=1

Well you didn't use Euler's formula correctly (should read simply cos(pi) +..., furthermore that would evalualte to -1) but the 'error' is most likely the 2nd equals sign. That's only allowed with real exponents - although I'm not quite sure what the reason is. Probably to do with the fact that you have many expressions all evaluating to the same number.

this thread should be bumped:mad:

I could swear that I've seen some math threads overrun by these, but I can't find them anymore. Anyway, why are these proofs wrong?:
a = b
a^2 = ab
a^2-b^2 = ab-b^2
(a+b)(a-b) = b(a-b)
a+b = b
2b = b
2 = 1


x = (Pi+3)/2
2x = Pi+3
2x(Pi-3) = (Pi+3)(Pi-3)
2Pix-6x = Pi^2-9
9-6x = Pi^2-2Pix
9-6x+x^2 = Pi^2-2Pix+x^2
(3-x)^2 = (Pi-x)^2
3-x = Pi-x
Pi = 3


-1 = -1
-1/1 = -1/1
-1/1 = 1/-1
sqrt(-1/1) = sqrt(1/-1)
i/1 = 1/i
i = 1/i
i * i = 1
-1 = 1

i^2 is -1, not 1.

That's not the misstep though