6
$\begingroup$

I recently saw this: ($x\in\mathbb{C}$)

  1. $$x^2+x+1=0,x\neq0,\pm1$$
  2. $$x+1+\frac{1}{x}=0$$ Plugging (2) into (1) we get $x^2-\frac{1}{x}=0$ so $x=1$

Why does this happen? I know that eq. (1) has solutions $x=\frac{-1\pm\sqrt{-3}}{2}$, and that these still are solutions for eq. (2), but where is the extra solution $x=1$ coming from? (And why is that step not valid? Is it like squaring both sides?)

$\endgroup$
  • 1
    $\begingroup$ Where did you see this? What is $x$? A real number? A complex number? $\endgroup$ – Lord Shark the Unknown Mar 15 '18 at 10:13
  • $\begingroup$ Plugging $(2)$ into $(1)$ gets $x^2-\frac{1}{x}=0$. $\endgroup$ – Rócherz Mar 15 '18 at 10:15
  • $\begingroup$ @Rócherz My mistake, I've updated the question. $\endgroup$ – cansomeonehelpmeout Mar 15 '18 at 10:16
  • $\begingroup$ Another way to look at this is considering the 'chain' of implication. You'll find that the statements don't prove each other in the way implicit in the 'proof'! $\endgroup$ – Oly Mar 15 '18 at 12:00
12
$\begingroup$

In effect, you have taken the equation $$x^2+x+1=0,$$ divided by $x$ to get $$x+1+\frac1x=0$$ and then multiplied by $x-1$ to get $$x^2-\frac1x=0.$$ The extraneous root $1$ is the solution of $x-1=0$, and $x-1$ is the factor you multiplied by.

$\endgroup$
  • $\begingroup$ So letting $x+1=-x^2=-\frac{1}{x}$ is the same as multiplying by $(x-1)$? $\endgroup$ – cansomeonehelpmeout Mar 15 '18 at 10:21
  • 4
    $\begingroup$ If your second equation is $f(x)=0$, your first is $xf(x)=0$ and so your third is $(x-1)f(x)=0$. @cansomeonehelpmeout $\endgroup$ – Lord Shark the Unknown Mar 15 '18 at 10:23
  • $\begingroup$ That makes sense, thank you! $\endgroup$ – cansomeonehelpmeout Mar 15 '18 at 10:25
5
$\begingroup$

Another way to look at this.

  1. $$x^2 + x + 1 = 0, x \neq 0, \pm 1$$

  2. $$x + 1 + \frac{1}{x} = 0$$

  3. $$x^2 - \frac{1}{x} = 0$$

  4. $$x = 1$$

We're asking for 'the solutions' to 1. What does that mean? It means, find the set of numbers, $X \subset \mathbb{C}$ with

$$x \in X \iff 1.$$

We know that

  1. $\implies$ 2.

And clearly 2. $\implies$ 1.

So 1. $\iff$ 2.

We know that, together,

  1. and 2. $\implies$ 3.

BUT, it is not the case that

  1. $\implies$ 1.

This is an assumption that is implicit in the 'proof' given.

We also know that

  1. $\implies$ 3.

BUT, again, it is not the case that

  1. $\implies$ 4.

With these implications made explicit, it should be clear why this reasoning is unsound. We are effectively making the claim that 4. $\implies$ 1!

The particular fallacy is called Affirming the Consequent because we 'affirm' the right hand of an implication, and fallaciously deduce from that that the left hand must be true.

This kind of fallacy is particularly common when 'equational' reasoning is used without making the implications explicit.

$\endgroup$
1
$\begingroup$

If $$x^2-\dfrac{1}{x}=0,$$ then $$\frac{x^3-1}{x}=0;$$ but note that $$\frac{x^3-1}{x} =\frac{(x-1)(x^2+x+1)}{x}=0,$$ so you just basically multiplied both sides of $(1)$ by $(x-1)$.

$\endgroup$
1
$\begingroup$

When you perform equation (1) - equation (2), you are multiplying $\left(1-\frac1x\right)$ to both sides of equation (1):

$$\begin{align*} x^2+x+1 &= 0\\ (x^2+x+1)\left(1-\frac1x\right) &= 0\\ x^2 - \frac1x &= 0 \end{align*}$$

So you introduced the root $\frac1x = 1$, i.e. $x=1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.