Let's Prove that 3 = 0 3=0

Algebra Level 4

I found this proof on a Chinese Q&A forum:

  1. x 2 + x + 1 = 0 x^2+x+1=0
  2. x 2 = − x − 1 x^2=-x-1
  3. x = − 1 − 1 x x=-1-\frac{1}{x}
  4. x 2 + ( − 1 − 1 x ) + 1 = 0 x^2+\big(-1-\frac{1}{x}\big)+1=0\qquad (substituting the equation in step 3 into the equation in step 1)
  5. x 2 − 1 x = 0 x^2-\frac{1}{x}=0
  6. x 2 = 1 x x^2=\frac{1}{x}
  7. x 3 = 1 x^3=1
  8. x = 1 x=1
  9. 1 2 + 1 + 1 = 0 1^2+1+1=0\qquad (substituting the equation in step 8 into the equation in step 1)
  10. 3 = 0. 3=0.

In which of these steps did I first make a mistake by using flawed logic? Or did we just prove that mathematics is inconsistent ?

If you enjoyed this problem and know some calculus, you may want to consider trying this problem .

An extraneous solution was introduced in step 3 3 Dividing both sides by x x at step 3 3 is prohibited An extraneous solution was introduced in step 4 4 Multiplying both sides by x x at step 6 6 is prohibited x 3 = 1 x^3=1 has other real solutions No real solution x x exists for x 2 + x + 1 = 0 x^2+x+1=0 3 = 0 3=0 can be true under certain axioms (e.g. the Tortuellian axioms) We just proved mathematics is inconsistent

Nick Turtle by Nick Turtle , 21, USA

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1 solution

Nick Turtle
Jan 28, 2018

The original equation x 2 + x + 1 = 0 x^2+x+1=0 had only two complex solutions: − 1 ± i 3 2 \frac{-1±i\sqrt{3}}{2} . However, while substituting in step 4 4 , an extraneous solution x = 1 x=1 was introduced.

At step 3 3 , no extraneous solutions were introduced. The equation x = − 1 − 1 x x=-1-\frac{1}{x} still has the same two complex solutions as the original. In addition, since x ≠ 0 x≠0 , the division of x x from both sides is allowed.

At step 6 6 , multiplying both sides by x x is allowed. (The multiplication of both sides of an equation by the same number is never prohibited.)

x 3 = 1 x^3=1 has only one real solution x = 1 x=1 , but its complex solutions coincide with that of x 2 + x + 1 = 0 x^2+x+1=0 as x 3 − 1 = ( x − 1 ) ( x 2 + x + 1 ) x^3-1=(x-1)(x^2+x+1) .

By definition, 3 ≠ 0 3≠0 for any consistent set of axioms, and the "proof" would still "work" under normal axioms. In addition, there's no such thing as the Tortuellian axioms.

15 Helpful 0 Interesting 0 Brilliant 0 Confused

I agree that (4) introduces a foreign solution, since this line can be obtained by multiplying with the factor, x − 1 x \dfrac{x-1}{x} . What is also wrong, is 7 ⇒ \Rightarrow 8 … unless we are working with the constraint x ∈ R x\in\mathbb{R} .

R Mathe - 3 years, 1 month ago

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