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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