Mathematical Riddles

[dwaxe]

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

 

[Perfection]

(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

 

[ParadigmShifter]

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

 

[Mise]

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.

 

[Onionsoilder]

I saw this one the other day:

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

 

[dragonforce]

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

 

[Aramazd]

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!!!

 

[Perfection]

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

 

[Souron]

First I should call your attention to this thread, 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)

 

[Perfection]

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

 

[taper]

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

 

[Souron]

Yep .

 

[cubsfan6506]

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

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

 

[Perfection]

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

 

[dragonforce]

why is this in the humar forum?

 

[mechaerik]

Why is this in the Humor Forum?
This is on topic though:
PROOF THAT GIRLS ARE EVIL:

 

[dragonforce]

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

 

[Riffraff]

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.

 

[dragonforce]

this thread should be bumped

 

[Huayna Capac357]

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.