Flaw detector

Algebra Level 3

Here's how we can prove 3=33=-3.

Step 1: 32=(3)2\sqrt{3^2}=\sqrt{(-3)^2}

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

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

Step 4: 312×312=312i×312i3^{\frac{1}{2}} \times 3^{\frac{1}{2}}= 3^{\frac{1}{2}}i \times 3^{\frac{1}{2}}i

Step 5: 312+12=312+12×i23^{\frac{1}{2}+\frac{1}{2}}=3^{\frac{1}{2}+\frac{1}{2}} \times i^2

Hence, 3=33=-3.

But we know that 333 \ne -3. In which step has the error been committed?

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


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