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Consider the equation $x-1=0$. The obvious solution is $x=1$. However, what is stopping us from creating more solutions by multiplying both sides by arbitrary values? For example:

$$ \begin{align} x-1 & = 0 \\ x\cdot(x-1) &= 0\cdot x \\ x\cdot(x-1) &= 0 \end{align} $$

If we substitute $x=0$ into this equation now, it gives us a solution. But 0 is not a solution to $x-1=0$. Does multiplying both sides by $x$ imply that there is a restriction $x\neq0$ or something? Why?

We could(?) go further with this, for example, multiply by $(x-2)$ giving $(x-2)(x-1)=0$, meaning $x=2$ is now an (incorrect) solution.

Gary
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Monolith
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  • Doesn't multiplying both sides of an equation by the same value ($x$) keep the equations the same? – Monolith Nov 03 '22 at 02:39
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    You're using the fact that $x - 1 = 0 \implies x(x-1)=0.$ But $x(x - 1)=0$ does not imply $x - 1 = 0.$ Not every operation you can do on an equation is reversible. That is, the equations are not necessarily "the same". – David K Nov 03 '22 at 02:40
  • We can tell the equations are not the same because they have different solution sets. – David K Nov 03 '22 at 02:44
  • Multiplying both sides of an equation with the same $x$ is reversible if you can "go back" by dividing by that same $x$. Now what happens if $x=0$? – Torsten Schoeneberg Nov 03 '22 at 02:44
  • If you continued along your logic, you could multiply both sides of $x-1=0$ by $x-a$ for any $a$ to get that $a$ is a solution to $(x-a)(x-1)=0$. But this shouldn't lead to that every $a$ is a solution to $x-1=0$. – Michael Burr Nov 03 '22 at 02:48
  • @MichaelBurr Correct, that is exactly why I asked this question. It seemed to me that the steps made make sense, so I was wondering what was incorrect with my reasoning. – Monolith Nov 03 '22 at 02:52
  • Only if you multiply by a function that is nowhere-zero do the solutions of the equation stay the same. – Ted Shifrin Nov 03 '22 at 02:59

4 Answers4

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Taking your question, every one of your steps is valid.

$$\begin{equation}x - 1 = 0\end{equation}$$

has one solution, $x = 1$, and

$$x(x-1) = 0$$

has two solutions, $x = 0$ and $x = 1$.

Where this goes wrong is in assuming that this holds for the original equation: you are implicitly saying

$$\begin{aligned}0(0 - 1) &= 0\\ \therefore \, 0-1&=\frac{0}{0} =0\end{aligned}$$

However, this fails to hold, as division by zero is undefined. It is this undefined division by zero which lets you accomplish this for any polynomial $p(x)$, to which you wish to add the solution $x = c$:

$$\begin{aligned} p(x) &= 0\\ (x -c)p(x) &= 0\\ (c - c)p(c)&=0\\ p(c) &= \frac{0}{c-c} = \frac{0}{0} \neq 0\end{aligned}$$

Lemmon
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  • Wait, but what is the issue with the equation $0(0-1)=0$? Is it not valid? – Monolith Nov 03 '22 at 02:44
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    $0(0-1) = 0$ is valid, but does not entail $(0 - 1) = 0$, it instead entails $0 - 1 = \frac{0}{0}$, which is undefined. – Lemmon Nov 03 '22 at 02:46
  • Why must $(0-1)=0$ be entailed in the equation $0(0-1)=0$? – Monolith Nov 03 '22 at 03:21
  • When we say that $x = 0$ is a solution to the equation $x - 1 = 0$, we are saying that $0 - 1 = 0$. Thus, for $0(0-1) = 0$ to entail that $x = 0$ is a solution to $x - 1 = 0$, it must entail that $0 - 1 = 0$. – Lemmon Nov 03 '22 at 03:23
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The point is that the set of solutions to an equation does not change if you apply a one-to-one function to both sides of the equation. In going from $A = B$ to $f(A) = f(B)$ in the solution process the logic always goes in reverse: if $x$ is a solution of $f(A) = f(B)$, then $x$ must be a solution of $A=B$. Multiplication by a nonzero number is one-to-one: if $ab = ad$ and $a \ne 0$ then $b = d$. But multiplication by $0$ is not one-to-one, as $0 \times (anything) = 0 \times (any\ other\ thing)$. So when you multiply both sides of your equation by something that could be $0$, you can introduce spurious solutions, as in your example.

Robert Israel
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Given functions $~f(x), g(x),~$ for a particular value $~x_0,~$ assume that $f$ and $g$ are well defined at $~x_0.$

Then,

$$f(x_0) = 0 \implies \left[f(x_0) \times g(x_0)\right] = 0.$$

However, the converse is not generally true.

That is,

$$\left[f(x_0) \times g(x_0)\right] = 0 ~~\text{does not imply}~~ f(x_0) = 0.$$

user2661923
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0

We have two equations that $x$ must satisfy:

The equation you obtained, $x(x-1)=0$
AND
The given equation, $x-1=0$

So $x=1$ is the only solution.

Dan
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