1 + 1 = 0

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!

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