Flaw detector

Algebra Level 3

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

Step 1: 3 2 = ( − 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: 3 1 2 × 3 1 2 = 3 1 2 i × 3 1 2 i 3^{\frac{1}{2}} \times 3^{\frac{1}{2}}= 3^{\frac{1}{2}}i \times 3^{\frac{1}{2}}i

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

Hence, 3 = − 3 3=-3 .

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

Clarification: i = − 1 i=\sqrt{-1} .

There's no error. The calculations are perfect. Step 5 Step 2 Step 3 Step 4 Step 1

Rishik Jain by Rishik Jain , 21, India

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1 solution

Ralph James
Apr 21, 2016

Relevant wiki: List of Common Misconceptions

a b = a ⋅ b \sqrt{ab} = \sqrt{a} \cdot \sqrt{b} except when a , b < 0 a,b < 0 .

⟹ − 3 ⋅ − 3 ≠ − 3 ⋅ − 3 \implies \sqrt{-3 \cdot -3} ≠ \sqrt{-3} \cdot \sqrt{-3} , so Step 3 is mathematically invalid.

4 Helpful 0 Interesting 0 Brilliant 0 Confused

If a ⋅ b ≥ 0 a \cdot b\,≥\,0 , then a b = ∣ a ∣ ⋅ ∣ b ∣ \sqrt{ab}\,=\,\sqrt{|a|}\,\cdot\,\sqrt{|b|} .

Now, clearly, a ⋅ b ≥ 0 iff a , b ϵ R + ∪ 0 ⋁ a , b ϵ R − ∪ 0 a \cdot b\,≥\,0 \,\,\text{iff}\,\,a,b\,\epsilon\,\mathbb{R^{+}} \cup {0} \,\,\bigvee\,\, a,b \,\epsilon\,\mathbb{R^{-}} \cup {0} . a a and b b need not to be necessarily positive integers.

Aditya Sky - 5 years, 1 month ago

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Thank you, sir. I have corrected the mistake in my solution.

Ralph James - 5 years, 1 month ago

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