Is $x/x$ equal to $1$

Question

My question is whether $x/x$ is always equal to $1$. I am mostly intersted in real numbers and particularly wonder whether $x/x$ is defined at $x=0$.

On one hand the division should simplify to $1$, on the other hand you should not be allowed to divide by zero.

I have been trying to find whether the simplification 'goes first' or whether the division causes trouble first, but it has proven impossible for me to find usefull search terms.

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Note that this question arose after reading this answer and its first comment.

Answer 1

The function $f(x) = \frac{x}{x}$ is defined for all $x \in \mathbb{R}\setminus\{0\}$. Its limits exist from the left and from the right at $0$, but it is not defined at $0$. It doesn't matter if you "simplify" first and then "check" or the other way around, because as you pointed out you're not allowed to divide by $0$. Thus $$f(x) = \begin{cases} 1 & x \neq 0 \\ \text{undefined} & x = 0\end{cases}.$$ Consider the same question, $$g(x) = \frac{x^2 - 2x + 1}{x-1}.$$ What is this function?

Answer 2

As linked to in the post you link to, $\frac{0}{0}$ is an indeterminate form - that is, there is no answer at all. It's particularly curious because it's an example of both $\frac{x}{x}$ and $\frac{y}{x}$, the former of which equals $1$ when $x\neq0$ and the latter of which has a limit that tends to either positive or negative infinity when $y\neq0$ as $x$ tends to 0.

Answer 3

For any real number $x\neq0$, $$x=x,$$$$\Leftrightarrow$$$$\dfrac{x}{x}=1.$$ $x=0$ $\Rightarrow x=x,$ but now we cannot divide by $x$. (Why?)

In fact, this is an indeterminate form.

Suppose that, $\dfrac{0}{0}=\color{red}{1}$. Hence, $0=\color{red}{1}\cdot0$.

Therefore, $\color{red}{2}\cdot0=\color{red}{1}\cdot0$ since $\color{red}{2}\cdot0=0$.

So, $\color{red}{\dfrac{2}{1}}=\dfrac{0}{0}$.

Thus $\color{red}{2}=1.$

Answer 4

I disagree with all answers I've seen here, on the internet and even at MIT. What if you have $x^2 \le 4x^3 + 5x^2$. To solve you can divide both sides by $x^2$ which gives you $1 \le 4x + 5 \Rightarrow -4 \le 4x \Rightarrow -1 \le x$. But now since you divided by a power of $x$ to solve it you suddenly can't have $x = 0$ any more? It is inconsistent. I believe in simplifying instantaneously before division occurs.

Answer 5

Note: I'm assuming $x\in R$ in this whole post.

Indeed $x/x$ is only defined when $x\ne 0$. And wherever it is defined, its value is $1$.

However when people (especially in fields other than mathematics where mathematics is used) talk about such expressions, often what they really mean is: "The continuous (or, more generally, "well-behaved") function determined by the values of the given expression where that expression is defined". And there exists indeed exactly one continuous function $f:\mathbb R\to\mathbb R$ so that $f(x)=x/x$ for $x\ne 0$, and that is the constant function $f(x)=1$ for all $x\in\mathbb R$.

As a concrete example, consider the function $\sin(x)/x$ which appears in physics in the amplitude of a wave when describing diffraction on a slit. This expression is clearly not defined at $x=0$. However is is of course silly to assume that the amplitude of a wave is not defined at some point. What physicists actually mean is the continuous function $$f(x)=\begin{cases} \frac{\sin x }{x} & x\ne 0\\ 1 & x=0 \end{cases}$$ But usually that function isn't written that way; the interpretation is implicitly assumed.

Answer 6

"On one hand the division should simplify to 1, ..."

No it does not: When you divide something by x (which is what you do in your simplification if you think about it) it is undefined for x=0. So when you simplify correctly you will also have the answer to your question.