I saw something on YouTube today that was interesting. To relieve you of having to watch the extremely slow-moving video, I will just provide the proof right here.

First assume that

`i = √(-1)`

Next, assume that

`√(ab) = √(a) * √(b)`

Now, we will derive the proof:

`1 + 1 = 1 + √(1)`

= 1 + √[(-1)(-1)]

= 1 + √(-1)√(-1) = 1 + (√(-1))²

= 1 + i²

= 1 + (-1)

= 1 - 1

= 0

Does that not blow your mind? Now, obviously this is crazy. So let’s take a look at how the proof is flawed.

The assumption that `√(ab) = √(a) * √(b)`

is only true if at least one of `a`

and `b`

are positive numbers (including zero). If both `a`

and `b`

are negative then `√(ab) = -√(a) * √(b)`

. Hence,

`1 + 1 = 1 + √(1)`

= 1 + √[(-1)(-1)]

= 1 - √(-1)`√`

`(-1) = 1 - (`

`√`

`(-1))²`

= 1 - i²

= 1 - (-1)

= 2

And the world makes sense again. Now I feel like my feet are back on the ground!