Prove me wrong.
0 = 1
Ian Anderson
Mason Diaz
> 0(0-10)=0
> => 0 = 0 or 10
> => 0=10
Gabriel Taylor
10 doesn't work as a value for A but 1 does.
A(A-1)=0
A=0 and A=1 both are valid.
A is constant though, not a variable.
0=A=1
Cameron James
> A=0 or 1
> A=0 and 1
> Derp.
Hudson Parker
>A=0 and A=1 both are valid.
No, they are not
Oliver Fisher
yes they are
Adam Nguyen
No, only one of them is, since A is a constant. Unless, of course, 0 = 1, but proving 0 = 1 by assuming 0 = 1 is logically fallacious
David Gutierrez
Where's the assumption? A is defined then it's shown that A evaluates to 0 as well as 1.
Okay so I know sqrt(4)=+/-2 doesn't imply -2=+2 but it is at least true that (-2)^2=sqrt(4)=(+2)^2. Same magnitude different sign.
So with A maybe 0=/=1 but 0^2=A=1^2 doesn't work either. Different magnitudes.
Try B:
B=sqrt(2+sqrt(2+...))
B^2=2+B
B^2-B-2=0
(B-2)(B+1)=0
B=2 or -1
Is it actually possible to correctly evaluate these things?
Sebastian Campbell
I know this is bait but you're dividing 0 by 0 and you can't do that shithead
Andrew Howard
>Where's the assumption? A is defined then it's shown that A evaluates to 0 as well as 1.
No, it evaluates to one of the two. You do not know at that point to which unless you can rule one out
>Is it actually possible to correctly evaluate these things?
Yes, limits
You need limits anyway if you want to cleanly define infinite sums (or in this case infinitely stacked square roots)