Mistake proving that $3 = 0$

Question

I was watching this video in which they proved that $3 = 0$ and the objective is to find the mistake in the proof. They did the following:

• Let $x$ be a solution for the equation: $x^2 + x + 1 = 0$ $(1)$.

• Because $x \neq 0$ we can devide both sides by $x$: $x + 1 + \frac{1}{x} = 0$ $(2)$

• From eq. $(1)$ we conclude that $x +1 = -x^2$ and if we substitude that in eq. $(2)$ we get: $-x^2 + \frac{1}{x} = 0$ $(3)$

• Simplifying eq $(3)$ we get: $x^2 = \frac{1}{x} \iff x^3 = 1$, so we get that $x = 1$ is a solution.

• If we substitude the solution $x = 1$ back into eq. $(1)$ we get $1^2 + 1 + 1 = 0 \iff 3 = 0$

I'll leave the answer and the rest of my question as a spoiler if you first want to try this yourself:

He explains that the mistake is when we substitute $x + 1=-x^2$ into eq. $(2)$, because when we do so we are actually adding the solution $x = 1$. The thing is that in the video he only says that and that got my wondering, why is that so?

Answer 1

Consider the equation $x=x^2$. Its solutions are easy to find: they are $0$ and $1$, right?!

However, if $x$ is a solution of that equation, then $x=x^2$, and therefore $x=(x^2)^2=x^4$. But the equation $x=x^4$ has other solutions, besides $0$ and $1$, namely $-\frac12+\frac{\sqrt3}2i$.

The problem lies in the if-then clause of the first sentence of the previous paragraph: we are saying (correctly) that if $x$ is a solution of the given equation, then $x$ is also a solution of another equation. But there is no reason for you to think that the new equation only has the solutions of the original one!

That's what happens in your example: if $x$ is a solution of $x^2+x+1=0$, then $x$ is a solution of $x^2=\frac1x$, but you have no reason to suppose that every solution of $x^2=\frac1x$ is also a solution of $x^2+x+1=0$.

Answer 2

Call $f(x)=x^2 + x + 1$, then his equation is $f(x)=0$.

From there, he derives the equation $(2)$, $\frac{f(x)}{x}=0$.

To reach equation $(3)$, he substracts the second equation to the first one, obtaining the equation $\left(\frac{1}{x}-1\right)f(x)=0$.

Thus, the equation $(3)$ will be true when either $f(x)=0$, which are the solutions of the original problem, or when $\frac{1}{x}-1=0$, that is, $x=1$. That´s what he means when he says he is adding the solution $x=1$.

Answer 3

The reasonning only shows that if $x$ satisfies $x^2 + x+1=0$ then it satisfies $x^3=1$. The fact is that you cannot "go back" to prove the reverse, since at step 3 you reuse the hypothesis that $x^2+x+1=0$.

Answer 4

You made a mistake When you said that if $$x^3=1$$

then $$x=1$$

In fact,

$$x^3=1\iff x^3-1=0\iff (x-1)(x^2+x+1)=0$$

$$\implies x=1 \text{ xor } x^2+x+1=0$$

Answer 5

Here is a really stupid version of proof that noone would fall for:

Start with equation (1). Multiply both sides with $x - 1$ and simplify to get: $x^3 - 1 = 0$. Conclude that $x = 1$ is a solution and proceed as in the original.

The question is now: how is the stupid version equivalent to the original? In other words:

How can substitution be equivalent to multiplication?

The other answers don't address this issue and I am also not so sure myself. I'd like to see answer to this, if I find one I will edit it in to this answer.

Answer 6

In an exercise where your task is solving equations, equations are not theorems: equations are conditions. The theorem is, if we want to see it in set-theoretical terms, that the set of the values satisfying the first condition is equal to the set of values satisfying the last one. It is certainly true (and your proof is as good as any) that $\{x\in\Bbb R\,:\, x^2+x+1=0\}=\{x\in\Bbb R\,:\, 3=0\}$. Does that prove that $3=0$? No, of course. It just proves that $\{x\in\Bbb R\,:\, x^2+x+1=0\}=\varnothing$. Notice, as a side remark, that at some point you use the fact that $\{x\in\Bbb R\,:\, x^3=1\}=\{x\in\Bbb R\,:\, x=1\}$ and that this, while crucial in your proof that $\{x\in\Bbb R\,:\, x^2+x+1=0\}=\{x\in\Bbb R\,:\, 3=0\}$, is no longer true if you substitute $x\in\Bbb R$ with $x\in\Bbb C$.