Does $\frac{x^2}{x}=0$ have a solution? Help on undefined "solutions".

Question

Does $\frac{x^2}{x} = 0$ have a solution?

I belive that it does not have a solution, my reasoning: \begin{equation} \frac{x^2}{x} = 0 \implies x = 0 \end{equation}

We can then check to see if the solution works: \begin{equation} \frac{0}{0} = \textrm{undefined} \end{equation} Thus I believe that there are no solutions to the above equation.

I suppose in general my real question is:

Regardless of how the equations is laid out, if I substitute in a "solution" back into the original equation and its undefined or an indeterminate, then its not a solution?

And the same applies for any substituted value of $x$?

Example:

$f(x) = \frac{x(x-2)}{(x-2)}$ is not defined for $x = 2$, even though it seems that we can cancel out the $(x-2)$, giving $f(x) = x$, which appears that $f(x)$ is defined for $x = 2$, but its not because looking at the original equation: $f(2) = \frac{0}{0}$ is undefined.

Is my reasoning correct? I would greatly appreciate any insight you can offer as this has been on my mind for a while and like to "put it to bed".

Answer 1

Regardless of how the equations is laid out, if I substitute in a "solution" back into the original equation and its undefined or an indeterminate, then its not a solution?

Yes, this reasoning is correct. By definition, a solution is something that, when substituted into the original equation, yields a true equation.

The process of algebraic manipulation is a tool to help us find solutions, but its outcome is not the definition of a solution. In addition to your good example (where the algebraic manipulation yields a result that's not in the domain of the functions on one side of the equation or the other), algebraic manipulation can even yield results that are grammatically correct but simply not solutions -- for instance, squaring both sides of an equation frequently produces results that are not solutions.

Answer 2

You are right, no solution exists and we can proceed as follows.

For the original expression, we need that $x\neq 0$ since $\frac 0 0$ in undefined and then

$$\require{cancel} \frac{x^{\cancel 2}}{\cancel x} = 0 \iff x= 0$$

but we have that $x\neq 0$ as a condition on the original expression therefore no solution exists.

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For the given example, as noticed in the comments, we should proceed as follows

• $f(x) = \frac{x(x-2)}{(x-2)}$ is not defined for $x = 2$
• then we can assume $x\neq 2$ and cancel out the $x-2$ terms

$$f(x)=\frac{x\cancel{(x-2)}}{\cancel{(x-2)}}=x$$

• note that the above expression is completely equivalent to the original one since we are implicitely assuming that $x\neq 2$

Answer 3

There is an interesting analytic followup here. For of all, on a purely algebraic point of view, $\frac {x^2} x$ has of course no solution, because the function $x\mapsto\frac {x^2} x$ is only defined for $x\ne 0$, and is $\forall x\ne 0, \frac {x^2} x \ne 0$.

But on a analytical point of view, we have:

$$\lim_{x\to 0}\frac {x^2} x = 0$$

So it makes sense to define a function $f:{\Bbb R}\to{\Bbb R}$ by $f(x) = \frac {x^2} x$ if $x \ne 0$, and $f(0) = \lim_{x\to 0} \frac {x^2} x$. Because we now have a nice, continuous (and even $C^\infty)$ function for which the equation $f(x)=0$ has one solution: $x=0$.

Answer 4

There are somethings you need to know.

1. The function $\dfrac{x^2}{x}$ is only defined on $\mathbb{R}\setminus\{0\}$, the nonzero reals. Therefore, your answer is not included in the domain of the function, thats why it should be rejected. More likely, the cancellation process is same as telling $$\dfrac{x^2}{x}=x\quad\text{for $x\ne0$}$$

2. When we are solving the equations, we are using the previous part to proceed to the next step, so we use $\implies$ to connect each argument. Therefore, what we have done can only show that the result is the only possible choice of the answer, but they are not necessarily be the answer.

In your example, you have shown that $\dfrac{x^2}{x}=0\color{red}{\implies}x=0$, so the only possible value of $x$ that can be the candidate of the solution is $x=0$, but you have to show $x=0\implies \dfrac{x^2}{x}=0$ as well (And this is the substitution you mentioned).

Another example, we have $$x=2\implies x^2=4\implies x=\pm2$$ So the solution to $x=2$ can only be $-2$ or $+2$. But you know $x=-2$ cannot be the solution of the equation $x=2$.

Answer 5

\begin{equation} \frac{x^2}{x} = 0 \implies x = 0 \end{equation}

We can then check to see if the solution $x=0$ works

Regardless of how the equations is laid out, if I substitute in a "solution" back into the original equation and its undefined or an indeterminate, then its not a solution?

$$g(x)=0\iff x=c\\ h(x)=0\implies x=k$$

Given the above statements, $g(x)=0$ definitely has solution $c,$ but $k$ is merely a candidate solution of $h(x)=0.$

Answer 6

In general when you find solutions to some expression you can think of it as finding the values where some function equal to that expression is 0.

In this case $f(x) = \dfrac{x^2}{x} = 0$. However, at $x = 0$ a division by 0 happens, so, f is cannot be defined at 0 since there is no way to define division by 0. This means that 0 cannot be in the domain of $f$.

When you "cancel" the $x$, what you are doing is dividing by $x$, but in order to do so, you have to assume that $x \neq 0$ because you're not allowed to divide by 0. Simplifying that gets you $x = 0$ and $x \neq 0$. Unfortunately this is impossible, so we must've been wrong when we assumed that $f(x) = \dfrac{x^2}{x} = 0$. When we wrote that equals sign, we assumed that this equation was true for some $x$, but that assumption led us to nonsense, therefore, there are no solutions and the equation is invalid. In fact we may now write $\dfrac{x^2}{x} \neq 0$.

In your other example where you had $f(x) = \dfrac{x(x - 2)}{x - 2}$, when you cancel out the $x - 2$ what you are doing is dividing by it. Since you can't divide by zero, you are making the assumption that $x - 2 \neq 0$.

The new expression isn't actually $f(x) = x$ but rather:

$f(x) = x$ if $x \neq 2$. After all, to get here, we had to assume that $x \neq 2$, so we can't drop that assumption. So we can see that the new function still isn't defined at $x = 2$.

In short, when you are dividing by a number, you are assuming that what you're dividing by is nonzero. When you set an expression equal to 0, you are assuming that there is some value of $x$ for which the equation is true. If you get a contradiction like earlier, it means that your assumption that the equation was true for some value of $x$ was wrong and your equation should've been an inequality.