Flaw in this square root proof that -1 equals 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) :

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