# Which step is wrong in this proof

Proof: Consider the quadratic equation $x^2+x+1=0$. Then, we can see that $x^2=−x−1$. Assuming that $x$ is not zero (which it clearly isn't, from the equation) we can divide by $x$ to give $$x=−1−\frac{1}{x}$$ Substitute this back into the $x$ term in the middle of the original equation, so $$x^2+(−1−\frac{1}{x})+1=0$$ This reduces to $$x^2=\frac{1}{x}$$ So, $x^3=1$, so $x=1$ is the solution. Substituting back into the equation for $x$ gives $1^2+1+1=0$

Therefore, $3=0$.

What happened?

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Essentially you eliminated the $x=1$ solution from $x^3 = 1$ and then wondered where it went... – Benjamin Dickman Nov 10 '13 at 1:35

Up to $x^3=1$ everything is fine. This allows you to conclude that $x\in \{z\in \Bbb C\colon z^3=1\}$. Since $\{z\in \Bbb C\colon z^3=1\}=\left\{\dfrac{-1 + i\sqrt 3}{2}, \dfrac{-1 - i\sqrt 3}{2},1\right\}$, then $x$ is one of the elements of this set.

You made a reasoning consisting of logical consequences, not one of logical equivalences. That's why you can tell that $x\in \{z\in \Bbb C\colon z^3=1\}$, but you can't say which one is it.

See this for a similar issue.

An even more simpler version of your mistake is this: suppose $x^2=1$, then $x=1$.
You can convince yourself that this is wrong and that you did the same thing in your question.

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in your answer there are three real roots. – daulomb Nov 10 '13 at 2:54
@user40615 Thanks, tbao10 already fixed it. – Git Gud Nov 10 '13 at 9:02
@tbao10 Thanks for fixing the typo. – Git Gud Nov 10 '13 at 9:02

What you have proved is that there is no real number $x$ such that $x^2+x+1=0$.

On the other hand, the two complex solutions of $x^2+x+1=0$ do indeed satisfy $x^3=1$.

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if $x^2+x+1=0$ then $(x-1)(x^2+x+1)=0$ thus $x^3-1=0$, thus: $x^3=1$.

if ... then .... is not a logical equivalence but only a logical implication.

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$x^2+x+1=0$ is a quadratic. This means it has two answers.

$x^3=1$ is a cubic. This means it has three answers. Therefore if you solve for $x$, then one of the answers you get won't fit the original equation.

This "extraneous" solution just happens to be $x=1$.

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