# Thread: -1 = 1 Proof ! What is the mistake done to prove it ?

1. ## -1 = 1 Proof ! What is the mistake done to prove it ?

Untitled Page check out this picture !

i want to know what was the wrong step that guides us to -1=1 , thnx

2. ## Re: -1 = 1 Proof ! What is the mistake done to prove it ?

The square root has more than one solution.

3. ## Re: -1 = 1 Proof ! What is the mistake done to prove it ?

On the third step you can't just split (-1)^(2/2) into ((-1)^2)^(1/2), because that's only done for positive numbers.

4. ## Re: -1 = 1 Proof ! What is the mistake done to prove it ?

the problem, as other have pointed out, is this step:

$\sqrt{-1}\sqrt{-1} = \sqrt{(-1)(-1)}$

-1 actually has TWO square roots, i and -i. unfortunately, there is no way to tell "which is the positive one".

when you write out what this actually means you get:

$(\pm i)(\pm i) = \pm 1$ which is far less controversial.

if one writes out these complex numbers in polar form:

$(\cos\left(\frac{\pi}{2}\right) + i\sin\left(\frac{\pi}{2}\right))^2 = \cos(\pi) + i\sin(\pi) = \cos\left(\frac{2\pi}{2}\right) + i\sin\left(\frac{2\pi}{2}\right)$

that is, the "proper" square root of 1 to take on the RHS is the negative one, which leads to -1 = -1, which is hardly surprising.

put another way, "angles" aren't uniquely defined by a number, we have to restrict the range of these numbers. this can lead to confusing results, when talking about "half-angles"

(half of 360 degrees is 180 degrees, but half of 0 is 0, and these two angles (0 degrees and 360 degrees) represent "the same direction").