Complex, yet Faulty

Algebra Level 2

What is the wrong step in the following proof that 1 = − 1 1 = -1 ?

  1. Let w w be a complex number such that ( w + 1 ) 3 = ( w − 1 ) 3 (w + 1)^3 = (w - 1)^3 .

  2. Solving this equation gives w = ± i 3 3 w = \pm \frac{i \sqrt{3}}{3} .

  3. Since ( w + 1 ) 3 = ( w − 1 ) 3 (w + 1)^3 = (w - 1)^3 for our previously mentioned values of w w , cube rooting both sides gives w + 1 = w − 1 w + 1 = w - 1 .

  4. Subtracting w w from both sides gives 1 = − 1 1 = -1 .

2 1 3 The proof is completely correct. 4 Two or more steps are incorrect.

Steven Yuan by Steven Yuan , 20, USA

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2 solutions

Steven Yuan
Mar 24, 2015

We are assuming the principal root for both sides of the equation in step 3. However, the principal cube root of ( w − 1 ) 3 (w - 1)^3 is not w − 1 w - 1 . It is actually ( w − 1 ) e 2 π i / 3 (w - 1)e^{2 \pi i / 3} . In this case, the two sides are still equal, which can be checked using Wolfram Alpha. Thus, the incorrect step is 3 \boxed{3} .

Note that step 4 is not incorrect because it logically follows from the previous step, even though the previous step is incorrect.

19 Helpful 2 Interesting 0 Brilliant 2 Confused

Honestly, I think this question is simpler if solved by simple process of elimination.

  • Step 1 is just a statement.

  • Step 2 is just solving an equation and the answer can be verified independently.

  • Step 4 is just subtraction.

  • The proof cannot possibly be correct.

  • Since 3 of 4 steps are invalid there are not "two or more" erroneous steps.

Therefore, Step 3 \boxed{\text{Step 3}} \text{ } must be the incorrect one.

Alex Delhumeau - 6 years ago

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Sure, you can do it that way. But it doesn't explain why step 3 is incorrect, which is the key in the problem.

Steven Yuan - 6 years ago

Although the answer is technically correct, I think the problem is posed somewhat confusingly. Why is Step 2 necessary to the "proof"?

Jace Harker - 5 years, 2 months ago

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The reason step #2 was shown is to verify that there are complex numbers satisfying step #1. They should've probably defined w=i*sqrt(3)/3, then did some algebra to get to (w+1)^3=(w-1)^3, and finally take the cube root incorrectly!

Math Nerd 1729 - 4 months, 1 week ago

Oleg Yovanovich

I think you can't equate (w+1)^3 = (w-1)^3, since when this equation will be simplified you get the left side equal to the negative value of the same on the right side, which is not possible. So, there is a fallacy from the very start of this proof, i.e. the first step is wrong.

Oleg Yovanovich - 1 year, 7 months ago
Tim Thielke
Aug 31, 2016

The key is that (w+1)^3 = (w-1)^3 = 0. You can't just equate any two complex numbers that when raised to the same power reduce to 0.

So expressing the cube root of 0 as (w-1) or as (w+1) is invalid. It's similar to dividing by zero.

3 Helpful 1 Interesting 0 Brilliant 0 Confused

From where do you conclude: "The key is that (w+1)^3 = (w-1)^3 = 0."? (I mean the "= 0" part)

Özgür Kesim - 3 years, 9 months ago

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