-1 = 1

[T@P]

heres a little problem that at a first glance is real:

\frac{1}{-1} = \frac{-1}{1}

so
\sqrt{\frac{1}{-1}} = \sqrt{\frac{-1}{1}}

by splitting it the square root into two parts...

\frac{i}{1} = \frac{1}{i}
and i^2 = 1

-1 = 1

wonder if there are any more similar "proofs"?

[AKG]

You can't split the square root into two parts. There are plenty of similar "proofs". You can search the web for them, and there are a number of them on this site alone.

[mathlete]

You can not take the square root of a negative number.

[Tom McCurdy]

You can not take the square root of a negative number.

\sqrt{-1}=i

imaginay numbers allow for negitive sqroots

he just violated a law in the way he split up his negitive signs.

[Tom McCurdy]

http://mathworld.wolfram.com/ImaginaryNumber.html
imagianary numbers

[Tom McCurdy]

http://en.wikipedia.org/wiki/Imaginary_numbers

more explanation on how you can take sqrt of negitive numbers

[shmoe]

A thinly veiled version of the same, though the fallacy is perhaps more transparent:

Euler's formula tells us:

e^{i\theta}=\cos(\theta)+i\sin(\theta)

So we see that:

e^{-i\pi}=e^{i\pi}

taking roots gives:

(e^{-i\pi})^{1/2}=(e^{i\pi})^{1/2}
e^{-i\frac{\pi}{2}}=e^{i\frac{\pi}{2}}

Using Euler's formula again and we get:

-i=i

or -1=1

[Gokul43201]

Here's another (though this one cheats in a different way) :

1 = \sqrt{1} = \sqrt{(-1)(-1)} = \sqrt{-1} \sqrt{-1} = i^2 = -1