# Flaw detector

Algebra Level 3

Here's how we can prove $$3=-3$$.

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

Step 2: $$\sqrt{3 \times 3}=\sqrt{-3 \times -3}$$

Step 3: $$\sqrt{3} \times \sqrt{3}=\sqrt{-3} \times \sqrt{-3}$$

Step 4: $$3^{\frac{1}{2}} \times 3^{\frac{1}{2}}= 3^{\frac{1}{2}}i \times 3^{\frac{1}{2}}i$$

Step 5: $$3^{\frac{1}{2}+\frac{1}{2}}=3^{\frac{1}{2}+\frac{1}{2}} \times i^2$$

Hence, $$3=-3$$.

But we know that $$3 \ne -3$$. In which step has the error been committed?

Clarification: $$i=\sqrt{-1}$$.

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