A proof that " 2 = 0 "

Logic Level 2

Here is a proof that "2 = 0". Can you find out the mistake?

Step 1 : $\text{2 = 1 +1}$

Step 2 : $\text {2 = 1 +}\sqrt{1}$

Step 3 : $\text{2 = 1 +}\sqrt{(-1)(-1)}$

Step 4 : $\text {2 = 1 +}\sqrt{-1} \sqrt{-1}$

Step 5 : $\text {2 = 1 + }(i)(i)$

Step 6 : $\text {2 = 1 +} i^2$

Step 7 : $\text {2 = 1 + (-1)}$

$\large \color{#3D99F6}\therefore \text {2 = 0}$

Step 5 Step 4 Step 3 Step 1 Step 6 Step 2 Step 7

2 solutions

Zach Abueg
Oct 10, 2017

$\sqrt{ab} = \sqrt{a}\sqrt{b}$ is only valid when $a$ and $b$ are nonnegative, real numbers.

I don't think that's true. It is valid when at least one of $a$ and $b$ are nonnegative numbers.

For example, $\sqrt{-24} = \sqrt{24}\sqrt{-1} = (2\sqrt{6})i$ is true.

Siva Budaraju - 3 years, 8 months ago

If you deny that piece of fact, then tell where-else is the mistake in this proof?

Syed Hamza Khalid - 3 years, 6 months ago

I did not deny that. I just clarified it.

Siva Budaraju - 3 years, 6 months ago

Syed Hamza Khalid
Oct 15, 2017

The product rule, which is used in step 4 states that:

$\large \color{#20A900}\text{If a and b are greater or equal to 0, then}\sqrt{ab} = \sqrt{a}\sqrt{b}$

In this case it is lower than 0, therefore Step 4 is wrong.