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I saw this video (link at bottom), with a supposed "proof" that $3=0$. It goes as follows:

Let $x$ be a solution of $$x^2+x+1=0 \tag1$$
Since $x\neq0$, we can divide both sides by $x$: $$\frac{x^2+x+1}{x}=\frac0x\implies x+1+\frac1x=0 \tag2$$
From $(1)$, $$x^2+x+1=0\implies x+1=-x^2$$
Substitute $x+1=-x^2$ into $(2)$ $$\begin{align*} -x^2+\frac1x&=0 \tag3\\ \frac1x&=x^2\\ 1&=x^3\implies x=1 \tag4 \end{align*}$$ Substitute $x=1$ into $(1)$ $$\begin{align*} 1^2+1+1&=0\\ 3&=0 \end{align*}$$

The explanation given in the video is

Substituting $x+1=-x^2$ into $(2)$ creates the extraneous solution $x=1$ which is not a solution to the original equation $(1)$, $x^2+x+1=0$.
Equations $(1)$ and $(2)$ have solutions $\frac{-1\pm i\sqrt3}{2}$, but after the substitution, equation $(3)$ has these two solutions and $1$.

Basically, it is saying that the issue is substituting $x+1=-x^2$, but I'm not sure if this is actually the problem. How can a substitution cause an issue if everything before the substitution is correct?

After reading the comments, I realised many of them say that the real issue is $(4)$, because $1=x^3$ could also mean that $x=\frac{-1\pm i\sqrt3}{2}$. Not considering these solutions is the issue with the "proof". One also needs to check these solutions before making conclusions, and "pick" whichever one is correct.

So, my question is, what is the issue with the above "proof" that $3=0$?

Video: "Prove" 3 = 0. Can You Spot The Mistake? https://www.youtube.com/watch?v=SGUZ-8u1OxM.

Just_A_User
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    an nth degree polynomial has atmost n distinct roots. – Aniruddha Deb Aug 17 '20 at 06:58
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    Why is this being downvoted? It has everything you want in a question, except a conclusion that is wrong, but the question is not silly. +1, in fact I think this should be encouraged. – Sarvesh Ravichandran Iyer Aug 17 '20 at 07:00
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    x=1 is not the root of $x^2+x+1$. it is the real rout of $x^3-1=(x-1)(x^2+x+1)$. X=1 must satisfy $x-1=0$ not $x^2+x+1=0$. – sirous Aug 17 '20 at 07:12
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    As others have noted, the problem is that $x^3 = 1$ does not imply $x=1$, at least in the complex numbers. If you work in the real numbers, then $x^3 = 1$ does imply $x = 1$. In that case, this is a correct proof by contradiction that $x^2 + x +1 = 0$ has no (real) solutions. – Jair Taylor Aug 17 '20 at 07:24
  • @JamesA Somebody have the heavy hand to put a downvote :-))) +1 for me. – Sebastiano Aug 17 '20 at 07:46
  • @Sebastiano: I didn't downvote, but the question was asked a few days ago. –  Aug 17 '20 at 08:16
  • I've thought the same as @sirous: the cubic's solutions includes the two complex conjugate solutions of (1) plus x=1. Going from (1) to (2) to construct a new expression that equates to 0, plus the re-arrangement of (1) and substitution into (2) to get cubic (3) is all valid. If you work out what the complex conjugate roots z and z* have to be if (x - 1)(x - z)(x - z*) = x^3 - 1 = 0, you'll see the roots of (4) are those of (1) plus x=1. You can put any numerical value into (1) to test its candicacy as a root. Thus the result "3=0" is no "proof" but just shows that x=1 fails the root test. – Jimbo Sep 06 '20 at 07:47

4 Answers4

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The problem is $x^3=1$ does not imply that $x=1$. The equation $x^3-1=0$ has three possible roots and the root $x=1$ is an additionally generated root.

Anand
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    $x^2 + x +1 = 0 \Rightarrow -x^2 +1/x = x^3 - 1 = 0$ but the converse is not true. Hence why the root $x=1$ is additionally generated. @James A – Enforce Aug 17 '20 at 07:22
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Substituting a member of an equation into itself can introduce alien solutions.

E.g. $$x=x^2\implies x^2=x^2.$$

You can do that, provided you keep the initial equation as well.

Safe operations are:

  • adding a term to both members;

  • multiplying both members by a nonzero factor;

  • applying an invertible transformation to both members.

Anything else (e.g. squaring both members) must be done with care.

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Substitution is able to cause an extraneous root because it's an irreversible step. That is, it is clear that if $x^2 + x + 1 = 0$, then we have $x + 1 + 1/x = 0$, $x+1 = -x^2$, and by substitution, $$ -x^2 + 1/x = 0. $$ However, the reverse is not true: if $-x^2 + 1/x = 0$, then it does not necessarily hold that $-x^2 = x+1$, from which it would follow that $x^2 + x + 1 = 0$.

Indeed, we see that this is how the solution $x = 1$ fits in: it satisfies $-x^2 + 1/x = 0$, but not $-x^2 = x+1$.

Another perspective: the substitution can be summarized with the following multiplication: $$ x^2 + x + 1 = 0 \implies\\ (-1 + 1/x)(x^2 + x + 1) = 0 \implies\\ -(x^2 + x + 1) + \frac 1x(x^2 + x + 1) = 0 \implies\\ -x^2 + 1/x = 0. $$ Multiplying $x^2 + x + 1$ by another factor has given the polynomial another root.

Ben Grossmann
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Let $x\ne0$. Then

$$x+1=-x^2\\\iff\\x+1=-\frac1x$$ is true. But

$$x+1=-x^2\land x+1=-\frac1x\color{red}\iff-x^2=-\frac1x$$ is not* ! The logical consequence is only left to right.

As shown on the plot, the curves of $-x^2$ and $-\dfrac1x$ do intersect, but not with $x+1$. By equating the two above RHS, you lose information and introduce non-solutions.

enter image description here

*If you think of it, it would be like saying

$$a=b\implies a=c\land b=c$$ whatever $c$.

  • +1 The Graph shows that this is an ex falso quodlibet. Both equations don't have a solution, hence they are false, hence anything follows. But ex falso quodlibets are not reversible. – rexkogitans Aug 18 '20 at 07:22