# Half the posts over here are trying 2 prove that 1=2 or 0=-1 or

half the posts over here are trying 2 prove that 1=2 or 0=-1 or sumthing like that lol so i thought ill try sumthing of that sort too...its silly...but it fascinated me when i was in class 8.

Let a = b where a, b are any 2 real numbers
a^2 = ab
a^2-b^2=ab-b^2
(a-b)(a+b)=b(a-b)
a+b=b
But a = b therefore b = 2b for any real number b

enjoy! dont curse me for wasting ur time when u find the mistae!

a+b=b;
a=b;
so b+b=b
so b=a=0?

rachmaninoff

"(a-b)(a+b)=b(a-b)"
a-b=0, so it is meaningless to divide both sides by (a-b)

Let a = b where a, b are any 2 real numbers

a2 = ab
Subtract both sides with b2 ==> a2-b2 = ab-b2
Simplify ==> (a-b)(a+b)=b(a-b)
divide with (a-b) ==> a+b = b
a=b gives ==> 2b = b

And the error is as said very simple.

toocool_sashi said:
(1) (a-b)(a+b)=b(a-b)
(2) a+b=b
to get eq. (2) from eq. (1)

you have to divide both sides by (a-b)
since a=b
therefore, (a-b)=0...

in addition, you have divided two sides by zero...

a number divided by zero is undefined in math...

got ya!