-1 = -1

1/-1 = -1
sqrt(1/-1) = sqrt(-1)
1/sqrt(-1) = sqrt(-1)
1 = sqrt(-1) * sqrt(-1)
1 = -1

it's MAGIC!
:P

But can you divide by zero!

only the infinite monkeys can divide by zero!

and I quote:
"Ever had one of those moments where you were in dire need of a banana? This was one of those times. With rage, they made the final move of the game.

... π/0

Division by Zero. The world had not seen such horror since World War II, when the Americans used it on Hiroshima. Kasparov stood up, stunned, then immediately resigned. "

excerpt from the article Kasparov vs the infinite monkeys

0.999... = 1

fluff = ffulf

martian is nub.


sqrt(1/-1) = sqrt(-1)
1/sqrt(-1) != sqrt(-1)


sqrt(1 / (-1)) = i
sqrt(-1) = i
1 / sqrt(-1) = - i

I had a kasperov computerized chess set. In UNI we used to do a fluff ton of coke off the lid of said chess set so "chess" or "kasperov" became code for "let's do a fluffton of blow"

just thought I'd share

but toma: sqrt(a/b) = sqrt(a)/sqrt(b)
:P
1/-1 = -1
thus if you take sqrt of both sides they should still be equal!
really...
:P

BACK TO MATH CLASS FOR YOU!

that's not what google said!

The real problem with your argument is that sqrt() is not a injective function.

For example
lets declare f(x) = 0;

0 = 0
f(-1) = f(1)
-1 = 1

actually no.

-1 = -1 correct

1/-1 = -1 correct
sqrt(1/-1) = sqrt(-1) correct
1/sqrt(-1) = sqrt(-1) correct
the thing is that y=a^x does not have a unique solution

when I do this:

1 = sqrt(-1) * sqrt(-1) (incorrect in the sense that each of those sqrt(-1) from before represent different solutions of sqrt(-1))

hence you get the statement:
1 = -1 which shouldn't really follow from the above because I'm not using sqrt() in the same way for each.

for example

9 = 9
(-3)^2 = (3)^2
sqrt((-3)^2) = sqrt(3^2)
thus -3 = 3
same as above:P

nerds!

pot.. meet kettle!

Is this math ever going to come in handy with any of you? Outside of this argument? Cause I know the fluff at wawa won't take .99$ on the dollar, even after I told her its the same thing.

Your Momma Can't Dance, And Your Daddy Just Whacks His Pole.
Your Hobby Involves Squinting And A Telescope, And Trying To Find A Black Hole.
Bet When Evening Comes Around,
And It's Time To Hit The Town,
You Just Sit Around Thinking All About Wanking On Your Gawd Dang Pole.

The action of taking a square root is NOT well-defined, thus it cannot be used in proofs.

Its like saying that 47=95 because 0=0 and 47*0=95*0
It just makes no sense.

um, don't square roots always provide 2 answers anyway? a positive and a negative value?

my math is real rusty, so you can yell at me for about an inch worth of forum for asking silly questions...

Martian: http://en.wikipedia.org/wiki/Injective_function

Oh god, enough math to make my head hurt. XD

actually no, your example is something different.
sqrt is a specific power (1/2) and it is very well and precisely definsed in mathematics as are all powers and you can definately use them in proofs if you apply them correctly and completely(which is not what I did)

Certain operations alter the domain of possible solutions and hence once you've done that you get a solution set for the new equation that won't work for the old one.
for example if you try to solve
x^2 = 1
x = 1, x = -1
but if you square both sides
x^4 = 1
to which the solutions are:
x = 1, x= -1, x=i, x= -i
but those aren't the solutions to the original equation but you have "created solutions" by raising both sides to a new power.

you can also get situations like this (without complex numbers)
by solving equations in the form (sqrt(ax^2+bx+c))= dx+e
(x is the variable, a,b,c,d,e are real numbers)
or even sqrt(ax+b) = cx

So to summarize the issue isn't with the definition of the functions, it's that we aren't being rigerous enough in the illustrated solutions and hence we end up with the wrong answer.

Division by 0 is undefined for the reason you stated.
as is lim as x->0 of sin(1/x)
:P

.999... = 1

1 = 2

It's an axiom, so don't worry about proof.

Martian - for taking the square root to be a well defined action, one must always come up with the same result if they do it twice. Taking the square root gives two different answers, which therefore makes it NOT well-defined.

When I said 'well defined' I was not referring to commonfolk speak, but the actual math term 'well defined'. See http://mathworld.wolfram.com/Well-Defined.html

The expression 'square root of negative one' is not well defined because it gets assigned the value 'i', but it can also be assigned the value '-i'.

well actually I'm not correctly applying the definition of sqrt.
but yes, the way I am applying it you would be correct.
:P
The moment you cross into complex analysis you can't use powers the same way either..