Think on this

Given: a=b a^2=ab a^2-b^2=ab-b^2 (a+b)(a-b)=b(a-b) (a+b)=b a+a=a 2a=a 2=1

at one point u will hit that 2 is actually 1 O_o

you divided by 0.

you divided by 0.

a=b => a-b=0

:D

incidently, though, 0.9 recuring is 1.

1/9+8/9=9/9=1
1/9=0.1...
8/9=0.8...
...  means recurring
 therefor 9/9= 0.1...+0.8...=0.9...
therefor 9/9=1=0.9...
QED

B) I know Im bad-ass

Haykuro wrote: Given: a=b a^2=ab a^2-b^2=ab-b^2 (a+b)(a-b)=b(a-b) (a+b)=b a+a=a 2a=a 2=1

at one point u will hit that 2 is actually 1 O_o

if a=b then a-b = 0, so (a+b)(a-b) = 0.

BobbyB wrote: [quote]Haykuro wrote: Given: a=b a^2=ab a^2-b^2=ab-b^2 (a+b)(a-b)=b(a-b) (a+b)=b a+a=a 2a=a 2=1

at one point u will hit that 2 is actually 1 O_o

if a=b then a-b = 0, so (a+b)(a-b) = 0.[/quote]

dude you're too late.

Curses…

I'll get you yet, wolfmankurd…