-1 = 1

1. Nov 1, 2004

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"?

2. Nov 1, 2004

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.

3. Nov 1, 2004

mathlete

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

4. Nov 1, 2004

Tom McCurdy

$$\sqrt{-1}=i$$

imaginay numbers allow for negitive sqroots

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

5. Nov 1, 2004

6. Nov 1, 2004

7. Nov 2, 2004

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

8. Nov 2, 2004

Gokul43201

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

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