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So I have this false proof and I am honestly confused why this is happening.

Consider $x^2+x+1=0$, then $x+1=-x^2$.

Now by simply dividing the equation by $x$ we get $x+1+1/x=0$. Substituting $x+1=-x^2$ we get $-x^2+1/x=0$ we get $x=1$ as a solution. Why is this substitution introducing a new solution, 1? Is it that all the solutions must fulfill this equation, but it is not a sufficient condition to actually be a solution? Why?

I think this might be a really silly question but it is stomping me :P

Sorfosh
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    $x^3-1=(x-1)(x^2+x+1)$ – Angina Seng Jul 22 '18 at 17:09
  • I guess the problem is somewhere there that you rewrite an equation and then substitute something you have made out of the equation into the original. – mrtaurho Jul 22 '18 at 17:11
  • @LordSharktheUnknown I am not sure how that answers my question, I am asking why is this happening. – Sorfosh Jul 22 '18 at 17:12
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    @Sorfosh Actually you multiplied by $1-1/x$, but the principle is the same. – Angina Seng Jul 22 '18 at 17:13
  • He refered to complex units roots which satify the equation $x^n-1=0$ with the trivial solution $1$ and $n-1$ other unit roots. Dividing such an equation by $x-1$ gives $\frac{x^n-1}{x-1}=x^{n-1}+x^{n-1}+\cdots+x+1$. You are dealing with the special case where $n=3$. By muliplying the equation $x^2+x+1$ by $x-1$ you are adding the solution $x=1$. – mrtaurho Jul 22 '18 at 17:15
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    You could make even worse examples. $x=-(x^2+1)\implies x^2=x^4+2x^2+1$ which we could use to rewrite the original as $x^4+2x^2+x+2=0$. That one has four separate solutions, including $\frac 12\times (1\pm \sqrt {-7})$. But all these rewrites do is to find other equations that our original roots must satisfy. – lulu Jul 22 '18 at 17:21
  • A simple way to detect that you have (likely) introduced extraneous "solutions" is that you went from a simple quadratic to something equivalent to a cubic. The original equation had two roots in complex numbers, your revised equation has three. So you have one new root. Unless that root happens to be equal to one of the original two roots, you've introduced a new distinct "solution". In this case the new "solution" was $x = 1.$ – David K Jul 22 '18 at 21:33

9 Answers9

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Is it that all the solutions must fulfill this equation, but it is not a sufficient condition to actually be a solution?

Yes, exactly right. Substituting one equation in another has the potential to introduce more solutions, because you've thrown away information. Specifically, you threw away the original equations you were given. A solution to the original equation will be a solution to your new one, but the new one may have others.

For a silly example, consider $$ x + x = x + 4. $$ There is only one solution to the equation: $x = 4$. But suppose we substitute $x = 4$ on the left side for one of the $x$'s; then we get $x + 4 = x + 4$, or $4 = 4$, which has infinitely many solutions (every real number $x$).

In your example, you ended up with the equation $−x^2+\frac{1}{x}=0$, which is equivalent to the equation $-x^3 + 1 = 0$ (assuming $x \ne 0$), or $x^3 - 1 = 0$. This factors as $x^3 - 1 = (x - 1)(x^2 + x + 1)$ (as was pointed out in a comment), so you have introduced exactly 1 new solution.

Caleb Stanford
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    @6004 I see, so how do we know which manipulations make us lose information and which do not? Clearly simple adding subtracting, or even multiplying by constants (But not by $x$ as that would introduce $x=0$ as a solution) leave the solutions alone. Is there a name for operations that conserve solutions? – Sorfosh Jul 22 '18 at 17:23
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    @Sorfosh I think it's a good question. A good way to tell is, "can I undo this operation and go back to the original equation(s) from before?" For adding and subtracting from both sides, or multiplying by a nonzero constant, you can see that you can undo the operation. For multiplying by x, you cannot undo the operation, because if x is 0 you can't divide by it. For substitution, in general it cannot be undone. But if you substitute and keep the original equation, that is OK. – Caleb Stanford Jul 22 '18 at 17:29
  • Also, some substitution can be undone. It is common if you have two equations and two variables to substitute one equation in the other. For example if we have $x + y = 3$ and $y = 4$ we substitute $y = 4$ into the first equation. This can be undone even if we throw away the old first equation, but we need to keep around the second equation $y = 4$. – Caleb Stanford Jul 22 '18 at 17:30
  • @Sorfosh One important partial answer is: if $f$ is an injective operation, then doing $f$ to both sides of an equation gives an equivalent equation. For example, we can always add $x$ to both sides of an equation because, no matter what number $c$ you pick, the function $f(z)=z+c$ is injective, so - no matter what $x$ - is the "add $x$ to both sides" operation is undoable. By contrast, the function $f(z)=x\cdot z$ isn't injective if $x=0$, so multiplying both sides of an equation by $x$ loses information unless we already know $x\not=0$. – Noah Schweber Jul 22 '18 at 17:38
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Is it that all the solutions must fulfill this equation, but it is not a sufficient condition to actually be a solution?

Yes. One important fact about mathematics is that when we manipulate an equation (or proposition, or ...), we can lose information along the way. E.g. from $$x^2=4$$ we can deduce $$0\cdot x^2=0\cdot 4,$$ and from that deduce $$0\cdot x^2=0.$$ $x=17$ is certainly a solution to this new equation, but not the original equation. Basically, we haven't done anything wrong - if $x^2=4$ then $0\cdot x^2=0$ - but we also haven't gotten an equivalent statement.

As a slightly less trivial example, from $$-y=4$$ we can deduce $$\vert -y\vert=\vert 4\vert,$$ or equivalently $$\vert y\vert=4;$$ $y=4$ satisfies this new equation, but not the original one. Again, we haven't made any mistakes, but we've lost information: the new equation $\vert y\vert=4$ tells us less than the old equation $-y=4$.

Noah Schweber
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The equation you have ended up with is not the same as the original. In fact, multiplying each side by $-x$, you get:

$$ x^3 - 1 = 0 $$

And it is clear that $1$ is a solution of this equation. The way you ended up with this fact is that, as Lord Shark pointed out, we can factor $x^3 - 1 = (x-1)(x^2+x+1)$, and you essentially showed that:

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

Which is pretty clear when you restate it like that.

Sambo
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You cheated yourself in a very interesting way! This is really illuminating!

When we solve equations, we usually just open a new line and write down some consequence of the condition. For example:

$$2x-3=2$$ $$2x=5$$ $$x=5/2$$

And then we make a note that the solution is $5/2$, without really thinking about what was happening. In fact, opening a new line and write an equation below the previous one has a very definite meaning: it is a consequence of the previous line. So usually, fake roots can sneak in, as we do not necessarily want to apply equivalent transformations, we only want to collect logical consequences of the equation. The strategy is that if we find a logical consequence that we can solve, then it simplyfies our task, as the solutions of the original equation must be among the solutions of the consequence. Throwing away extraneous roots are not so hard, after all...

Now back to your problem. You correctly deduced that if $x$ is a root, then it must also satisfy

$$E_1: x+1=-x^2$$ $$E_2: x+1+1/x=0$$

And then in particular, it also has to satisfy

$$E_3: -x^2+1/x=0$$

This is all true. But not in the other way around. Note that your original equation is equaivalent to $E_1$ which is also equivalent to $E_2$. You can also take the conjunction of two equivalent statements, to obtain an equivalent statement again. So $E_1\wedge E_2$ is also equivalent to your original condition. However, $E_3$ is just a consequence of $E_1\wedge E_2$, but from $E_3$, you cannot deduce $E_1$ or $E_2$ separately.

So long story short: not all your transformations were equivalent ones, hence extraneous roots can (and in this case did) occur.

A. Pongrácz
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The problem lies within an implicit assumption, i.e. that your equation holds.

First, we need to specify the domain of $x$. So let $x\in\mathbb{R}$.

We're now searching for all solutions to the equation $x^2+x+1=0$, i.e. the equations solution space. By substituting in a few random values for $x$, we can easily deduce that this solution space, let's call it $L$, is a strict subset of the real numbers $\mathbb{R}$, i.e. $L\subsetneq \mathbb{R}$.

So while both $x+1+1/x =0$ and $x+1 = -x^2$ still have the same solution space $L$
(all actions taken to this point are invertible, as $x=0$ is no solution)
the key point is that $L\subsetneq \mathbb{R}$ means the equations don't hold for all values.

This means, as long as $x\in L$, the equation $x+1 = -x^2$ holds, and we can substitute $x+1$ by $-x^2$.
However, for all $x\not\in L$, the equation $x+1 = -x^2$ does not hold, and therefore, your substitution may create new, erroneous solutions.

Sudix
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  • Thank you, Sudix, this is the only answer which really explains why substitution can introduce extraneous solutions. – Mike Earnest Jul 23 '18 at 15:49
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LOGIC. Or maybe just GRAMMAR. $ \;x^2+x+1=0\iff$ $ (x\ne 1\land x^2+x+1=0)\iff$ $ (x\ne 1\land -x^2+1/x=0)\iff$ $ (x\ne 1 \land x^3=1).$

What you have shown is that $x^2+x+1=0\implies x^3=1$ but you have NOT shown that $ x^2+x+1=0\iff x^3=1.$

SO $x^3=1$ does NOT "give a solution $x=1$" to the original equation. $x^3=1$ is only a consequence of $x^2+x+1=0$..

If you omit the "$\implies$" or "$\iff$", either in symbols or in words, in your writing, then you will lose track of whether you have $A\implies B$ or $B\implies A$ or $A\iff B.$

DanielWainfleet
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A tricky problem:

Let $x \not =0$, real, henceforth :

1) Originally : $f(x) = x^2+x+1;$

2) $g(x) = (1/x)f(x) = x+1+1/x.$

Note:

a) $f(x)=0$, $g(x)=0$ have no real zeroes.

b) $f(x) \not = g(x)$, if $x \not =1$.

Consider the difference :

$d(x):=f(x)-g(x) =$

$ xg(x)-g(x)=(x-1)g(x)=$

$x^2-1/x=0.$

1) $x \not =1$:

$d(x): =f(x)-g(x)=$

$ (x-1)g(x)=0$ , has no real zeroes.

2) $x=1$ :

$d(1)= f(1)-g(1)=0=$

$ 0g(0)=0$; where $g(0) \not =0$.

Combining :

$d(x) = 0$ is equivalent to $g(x) = 0$, if $x\not =1$.

By considering $d(x)$ you have introduced an additional zero at $x=1$.(Lord Shark's comment)

Peter Szilas
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Substituting a function of $x$ in an equation obviously gives results other than desired.

For instance:

$$x-1=4$$

we have $$x=5$$ $\implies$

$$x^2=25$$ $\implies$

$$4=\frac{100}{x^2} \tag{1}$$

Putting $(1)$ in original equation we get

$$x-1=\frac{100}{x^2}$$ which leads to cubic giving you other solutions including your $5$

Ekaveera Gouribhatla
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You need always to have in mind that "if there is a solution" then it should be 1 as you guessed. But this poly does not have real roots....

dmtri
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