Proving 2=0

Algebra Level 3

I will now try to prove that $2=0$ . These are the steps:

Step 1: $2=1+1$

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

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

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

Step 5: $2=1+(i)(i)$

Step 6: $2=1+i^2$

Step 7: $2=1+(-1)=0$

Therefore we proved that 2=0 . This is obviously an absurd contradiction. Which of the above steps is incorrect?

Step 4 Step 5 Step 3 Step 2

1 solution

Sathvik Acharya
May 21, 2017

We know the fact that $\sqrt{ab}=\sqrt{a}\sqrt{b}$ . But this property comes with the condition that $\sqrt{ab}=\sqrt{a}\sqrt{b}$ is true only if $a, b\geq0$ . But in Step 4 we are violating this rule. Therefore, neglecting these conditions can lead to such absurd results.

Proving 2=0

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