4000 + 2000 = 0 ... ? ? ?

[ron1245]

intresting but...!!!!!!!!!!!!!!!!!!!!!!!!!1111

[InFiNiTyPwN]

This can actually be proven wrong...but anything can = anything...nubs

[Blade™]

let a = 4000,

b = 2000 and

c = 6000

a + b = c

Multiply both sides by a + b

(a + b) (a + b ) = c (a + b)

a*a + ab + ba + b*b = ca + cb

a*a + ab - ca = cb - b*b - ba

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

so a = - b

a + b = 0

Hence 4000 + 2000 = 0.

Ok, where is the mistake. I mean, where is the wrong step. This can't be.

[shilo]

this is lame....
not funny makes me cry!!!

[Livegull]

it's not funny, maybe next time

[kpojha]

if a=4000;b=2000 & c=6000;
you can't cancel out (a+b-c) from both side of the equation. bcoz (a+b-c)=0.
a( a + b -c) = -b(b + a - c) ,
You r doing wrong in this line, This is beyond the low of mathematics, If you will do so a*0 = -b*0, cancel out 0 from both side you will get a=-b, This is not possible boss?

[clansx]

Haha, that was a funny one

[kaurin]

I remember my math teacher doing something like this back in high school. It was entertaining to say the least.

[minisoji]

very interesting jokes.

Not lend money to mathematician

[samik]

Its an interesting maths problem