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Consider the quadratic equation $x^2+x+1=0$.

$x=0$ is not a solution to this equation, so it is safe to divide both sides by $x$.

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

Rearrange:

$x+1=-\frac{1}{x}$

And substitute into the original equation.

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

Multiply both sides by $x$.

$x^3-1=0$

Trivially, a solution of this equation is $x=1$ (as $1^3-1=0$).

Substitute into the original equation:

$1^2+1+1=0$

$\therefore3=0$

What's gone wrong here?

Edit for attribution: I found the puzzle on the Mind Your Decisions blog.

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12 Answers 12

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Let us say initial equation is $E=0$

You divided this equation by $x$, thus you got $\frac{E}{x} = 0$.

Then, you subtracted this from original equation, thus you got $E - \frac{E}{x} = 0$.

In the next step, you multiplied the result by $x$, so you got $xE - E = 0$, which is the same as $(x-1)E=0$.

So, basically what you have done is multiplied the initial equation with $x-1$, and hence you got $1$ as the root for the final equation.

You then assumed both the equations to be the same, and hence first equation will share all the roots of the second equation, which is WRONG. This is simply because since both equations are different, they can have a different root, which means you cannot substitute $1$ in initial equation.

Moreover, its obvious to note that multiplying a new factor to initial equation, has increased the number of roots in final equation, which in itself guarantees that both equations are different.

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    $\begingroup$ It isn't immediately clear that the OP's description (Rearrange, then substitute) is equivalent to your description (subtract this from the original equation). They are equivalent, of course. But seeing it takes deeper mathematical reasoning than the other steps. $\endgroup$ Mar 30, 2023 at 16:13
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    $\begingroup$ I love that E abstraction! Basically when we have (𝑥−1)𝐸 = 0, it means that at least one of the following (but not necessarily all of them) is true: (𝑥−1) = 0, 𝐸 = 0 or both. Hence, assuming that (𝑥−1) = 0 is incorrect. $\endgroup$
    – GalAbra
    Mar 31, 2023 at 9:55
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    $\begingroup$ @GalAbra Agreed. I've known about this paradox for a few years and have only really understood it as "it inserts an extra factor somehow" while handwaving the details. This answer is new to me so I'm definitely glad I thought to post it on this stack. $\endgroup$ Apr 2, 2023 at 23:26
  • $\begingroup$ @ApexPolenta also, you can see the same fallacy by directly multiplying the original equation by x. In the new equation, x=0 is a solution. Substituting back to the original equation, you get 1=0. $\endgroup$
    – justhalf
    Apr 5, 2023 at 6:52
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The problem is substituting $x=1$ into the original equation. In the first part, we show that:

If $x$ is a root of $x^2+x+1$ then it is also a root of $x^3-1$.

Then in the second part we assert:

Since $x=1$ is a root of $x^3-1$ it must also be a root of $x^2+x+1$.

This is an example of affirming the consequent: "if P then Q; Q is true; therefore P."


As to why in the first step we only have an "if" and not an "if and only if," I think that thisIs4d's answer explains this very elegantly. Here is my own explanation of why:

The first step is an "if and only if," because the argument takes care to exclude the potentially problematic case of $x=0$, thus we have:

$$ x^2+x+1=0 \iff x+1=-\frac{1}{x} $$

We can rearrange the original equation so that the part that is substituted is isolated on one side:

$$ -x^2=x+1 \iff x+1=-\frac{1}{x} $$

Now at this point, we are strongly reminded of the transitive property of equality, "things that are equal to the same thing are also equal to one another." That is, if $a=b$ and $b=c$ then $a=c$. It is tempting to just "add on" the third equality to our equivalence:

$$ -x^2=x+1 \iff x+1=-\frac{1}{x} \iff -x^2=-1/x $$

Taking the first and third terms (and doing some rearranging) would give us:

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

Which, as we saw above, is not true. Why not? Remember that an equivalence is an "if and only if:" when one side is false, the other side must also be false. And if $a=b$ and $b=c$ are both false, we could still have $a=c$ by coincidence! Therefore we have to downgrade our equivalence to an implication, a.k.a. a plain "if."


Unlike in the classic argument that $1=2$ which involves a single 'obviously illegal' step (division by zero), there is no one wrong step in this argument. Most answers identify the final substitution (or the logical argument it implies) as "the" mistake, but it is equally as valid to recognize the first substitution as the mistake (as I think Jakub Komárek's answer does). That, I think, makes this a better example of a "bad" proof.

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Substituting one equivalent equation into another generally produces extra roots. As a degenerate case, substituting an equation into itself gives a tautology, but that doesn't mean the original equation was a tautology.

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All the steps made except of the substitutions are in fact valid equivalences, meaning that the equations before and after doing the transformation step have the same solutions. For example, the equation $x^2+x+1=0$ has the same solutions as $x+1+\frac{1}{x}=0$. It is very common to use a chain of equivalent transformations to solve an equation: if we can show that the original equation has the same solutions as $x = \text{some value}$, we have found the (only) solution.

As pointed out in other answers, this is not what is happening in the substitution steps. The logic behind substitution is that "if we know that this thing is equal to this other thing, then we can use them here interchangeably and nothing can change". Notice that this says nothing about the other direction: if two things can be used interchangeably at one place, it does not need to mean that they are in fact equal. For this exact reason, the substitution steps do not give us an equivalence, but this if-then relationship is called an implication.

Because of the implications both of the substitution give us, if $x^2 + x + 1 = 0$, then it must definitely also be true that $x^3 - 1 = 0$. No real number satisfies the first equation, so the if part is therefore never true and we can never make use of the then part. However, if we allow ourselves to look for solutions in complex numbers, the former equation has a solution $x = -\frac{1}{2} - \frac{\sqrt{3}}{2}i$. If we plug this number in the second equation, it in fact is true that ${\left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i\right)}^3 - 1 = 0$.

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    $\begingroup$ Not all those steps are equivalences. After step 3, OP has $x^2-\frac{1}{x}=0$, and the solution $x=1$ has already been introduced. $\endgroup$
    – wimi
    Mar 31, 2023 at 22:32
  • $\begingroup$ I agree, from this point of view the problematic step is the first substitution. $\endgroup$ Apr 1, 2023 at 0:53
  • $\begingroup$ @wimi You are right, both substitutions only give implications. I have edited the answer to hopefully fix this mistake. $\endgroup$ Apr 1, 2023 at 9:47
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IMHO1: this is not really a problem about algebra, it is a problem about logic.

IMHO2: while other answers have explained something about the logic, I can't see that it has been clearly stated what OP has done.

So: you have correctly proved (or observed) two things: $$\displaylines{ \strut\hbox{if $x^2+x+1=0$, then $x^3-1=0$}\ ;\cr \hbox{if $x=1$, then $x^3-1=0$}\ .\cr}$$ These do not enable you to deduce any relation between the statements $$x^2+x+1=0\quad\hbox{and}\quad x=1\ .$$ In particular, your "substituting into the original equation" is, in effect, a claim that if $x=1$, then $x^2+x+1=0$, and this is an implication which you have not proved.

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  • $\begingroup$ The technical term for this is a fallacy of the undistributed middle. $\endgroup$ Apr 1, 2023 at 0:51
  • $\begingroup$ I agree with you technically, however, this is also true of every other statement in the proof from a purely logical perspective. As in, no other statement follows from the others in a logically tautological way. As a result, its my opinion that it is appropriate to use algebraic reasoning here, since the rest of the proof is also not logical, but algebraic. $\endgroup$ Apr 2, 2023 at 18:39
  • $\begingroup$ @WilliamOliver But where has the OP made a specifically ALGEBRAIC error? $\endgroup$
    – David
    Apr 2, 2023 at 23:50
  • $\begingroup$ Although it wasn't explicitly stated, errors of this kind typically stem from the reasoning "there is only one root of x^3 - 1, therefore if x^3 - 1 = 0, then also x^2 + x + 1 = 0" Because if there was only one root of x^3 - 1, then OPs reasoning would be correct. But there isn't. This is because the notion of the "nth roots of unity" is not exactly trivial unless you explicitly study mathematics, and if you look at the graph of x^3 - 1 it really looks like there should only be one root. $\endgroup$ Apr 3, 2023 at 1:04
  • $\begingroup$ In other words, although you are correct technically, knowing that you have to justify each step logically is not as helpful as knowing the algebraic reason. Knowing only logic, it will take much longer to step through the proof step by step, justifying every step in a logical way, until you figure out the error at the very end. On the other hand, if you know algebra (especially the history of algebra, in which the polynomial x^2 + x + 1, and polynomials of the form x^d - 1 play a particularly important role), then the error jumps out at you almost immediately. $\endgroup$ Apr 3, 2023 at 1:13
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I'm sure it's the "And substitute into the original equation." that caused the problem.
Before substitute, it's still $x^2+x+1=0$ that is x(x+1)+1=0.
After substitute, it became x(-x2)+1=0 and factorize into (x-1)($x^2+x+1$)=0.
You can see that the (x-1) came out.
And also note that substituting in the reverse way doesn't work, as $x^2+x+1=0$ implies (x-1)($x^2+x+1$)=0 but not vice versa.

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Observe that $x^3 - 1 = (x - 1)(x^2 + x + 1)$, therefore, by the fundamental theorem of algebra, $x^3 - 1 = 0$ has exactly one solution that $x^2 + x + 1 = 0$ does not have, and that solution is 1.

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The original $LHS = (x+ 0.5)^2 + \frac 34$, which is always greater than $0$, so the equation has no solution. Dividing by x doesn't affect the equivalence of the new equation, so the next equation also has no solution. The substitution basically introduces something false into something false, which ends up making the final equation solvable for x in R.

E.g.

$x^2 + 5 = 0$ (no solution)

$=> 5 = -x^2$

Substitute $-x^2$ for $5$:

$x^2 - x^2 = 0$ (instantly has infinitely many solutions)

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  • $\begingroup$ Welcome to Puzzling! (Take the Tour!) How does your answer add to the more complete ones already given? You should always look at existing answers before providing one of your own, to ensure you are not just repeating something that has already been said. $\endgroup$
    – Rubio
    Mar 30, 2023 at 20:54
  • $\begingroup$ I like your line of thinking here, however I don't see where you got your "original LHS" from the original post. Can you explain? $\endgroup$ Mar 31, 2023 at 16:37
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    $\begingroup$ This is true, but I don't think it fully explains the apparent paradox for two reasons; 1) the original equations has two solutions in the complex numbers; 2) we can perform essentially the same reasoning on an equation that does have real roots, e.g.: consider $x^2=x+1$. $x=0$ is not a solution so we can divide by zero: $x=1+1/x$. Substitute for $x$ in the original equation: $x^2=(1+1/x)+1$. Multiply by $x$: $x^3=2x+1$. Note that $x=-1$ is a solution. Substitute for $x$ in the original expression: $1=0$. $\endgroup$ Apr 1, 2023 at 0:45
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First, I would point out that if $x^2+x+1 = 0$, then $x^3-1$ definitely also is 0 since $x^3-1=(x-1)(x^2+x+1)$ (this is the standard difference of squares "trick"), so there is no problem with getting to this step.

As to where the actual problem lies, this is one of those situations where everything clears up if we are precise with the formulation. We should start by saying "Let $x$ be a number such that $x^2+x+1=0$". Then, we can conclude that "for the aforementioned $x$, we have that $x^3-1 = 0$". But we cannot then deduce that $x = 1$, since there are other solutions to that cubic!

A clearer way of presenting such an "argument" is as follows:

If $x^2 + x + 1 = 0$, then $(x-1)(x^2+x+1) = 0$ so $x=1$ and $3 = 1+1+1=0$.

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Trivially, a solution of this equation is...

Yeah, but you cannot then infer that that solution is also a solution of the first equation. All the steps are valid math inference except this very last step.

If you have an equation $P(x)$ and derive another equation $Q(x)$ from it, all that you can infer is that any $x$ that satisfies $P(x)$ must also satisfy $Q(x)$, not the other way around!

Many algebraic manipulations are symmetrical in terms of the direction of derivation (if equation $A$ derives equation $B$ then $B$ derives $A$), but substitution is not one of them, because it already assumes that $P(x)$ holds! (Indeed, almost all of these absurd-looking incorrect proofs seem impressive only because most people have the incorrect assumption that derivation by algebraic manipulation is always symmetrical in that sense.)

In general, there might not be any $x$ that satisfies both. In this case, there are two complex numbers that satisfy both. -1 is one of the three complex numbers that satisfies $x^3-1=0$, but it is not one of the two that also satisfy $x^2+x+1=0$.

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The mistake in the solution is when the equation was divided by x. Dividing by x is not valid when x = 0 because division by zero is undefined. This means that the step where x + 1 + 1/x = 0 is not valid for x = 0.

Additionally, when the solution substituted x = 1 back into the original equation, it did not check if x = 1 was a valid solution for the equation after the division by x, which it is not.

Therefore, the mistake was made by dividing by x without considering the possibility that x = 0 and not verifying the solutions obtained after the division.

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    $\begingroup$ The argument actually does consider the possibility that $x=0$ in the second sentence, and explains that it is not a problem because $0$ is not a solution to the equation and therefore $x\ne 0$. As for "not verifying the solutions obtained after the division," the first and second equations do have the same solution set. It is only the third and fourth equations, obtained by substitution and multiplication, respectively, that have the 'extra' $x=1$ solution. $\endgroup$ Apr 1, 2023 at 1:41
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Well, I think the first problem is that you have zero on the right hand side. That means that when you do the first step of dividing both sides by x, you're doing it incorrectly.

$x^2 + x + 1 = 0 \equiv x + 1 = \frac{0}{x}$ which immediately breaks the equation, as $0$ divided by anything is still $0$. Especially as it means the equation can be further changed to put the single x on the left hand side.

$x = \frac{0}{x} - 1 \therefore x = -1$. We can then plug this into the original equation to check.

$-1^2 + -1 + 1 \equiv 1 + -1 + 1 = 0 \equiv 1 = 0$

Another way that we could simplify the original equation is to move the $1$ $x^2 + x + 1 = 0 \equiv x^2 + x = 0 - 1 \equiv x^2 + x = -1 \therefore x < 0$

This immediately tells us that $x$ has to be a negative number, as if $x$ was postive, $x^2 + x > x^2$ and it can't be if the answer is $< 0$. This is because negative numbers have the interesting property of reversing the sign of their multiplier, so $2 * -1 = -2$ and $x^2 > 0$. That means that we could stop here and declare that $x = \pm i\sqrt{x}$, but for completeness we should continue until we get to the $\sqrt{}$

$x^2 + x = -1 \equiv x^2 = -1 - x \therefore x < 0$

Our final step would be to move the power to the otherside of the equation. $X^2 = -1 - x \equiv x = \sqrt{-1 - x}$ This is interesting, as the square root of any negative number is $\pm i\sqrt{n}$

Thank you to @TannerSwett for the corrections

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  • $\begingroup$ It's not true that the square root of any negative number is $i$. If we have any negative number $-n$, then the square roots of $-n$ are $\pm i \sqrt{n}$. $\endgroup$ Mar 30, 2023 at 11:16
  • $\begingroup$ In any case, the most straightforward way to solve the original quadratic equation is to use the quadratic formula. The roots are $(-1 \pm \sqrt{1^2 - 4(1)(1)})/(2(1))$, or $(-1 \pm i \sqrt{3})/2$. $\endgroup$ Mar 30, 2023 at 11:19
  • $\begingroup$ @TannerSwett Thank you, It's been 20 years since I've done any calculus, and that was for my software engineering degree. Updating my answer to use $i\sqrt{x}$ $\endgroup$ Mar 30, 2023 at 12:16
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    $\begingroup$ What happened to the 1/x term in the first step? $\endgroup$ Mar 30, 2023 at 12:44

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