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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.

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

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Dennis Jaheruddin
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    What is the value of $\frac{2x}{x}$ everywhere except 0? What "should" the value then be? The limit gives you exactly this. The value you "should" get. But the expressions which approach $\frac{0}{0}$ can have multiple values so it is undefined. – 05storm26 May 21 '14 at 14:29
  • How do you interpret the string of symbols $x/x$? If you interpret it as denoting a rational function, that rational function is the constant $1$ and can be evaluated without problem at $0$. If you interpret it as a fraction of two real numbers, you have a problem at $0$. – Daniel Fischer May 21 '14 at 14:30
  • When $x$ is non-zero member of field then $x/x=1$ follows from axioms of Field. I believe this is the case in other algebraic structures when it is zero. Basically the division $/$ or the number $1/x$ is only defined on non-zero numbers. So what happens when $x$ is zero is out of question. – hrkrshnn May 21 '14 at 14:31
  • @05storm26 Do you mean that division goes before simplification? – Dennis Jaheruddin May 21 '14 at 14:31
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    @DennisJaheruddin, simplification and division are not distinct processes. Simplification is simply a question of applying division to the numerator and denominator. – Pockets May 21 '14 at 14:33
  • Note that this is equivalent to ask if functions like $3+\sqrt{x}-\sqrt{x}$ or $4+\log{x}-\log{x}$ are defined for negative $x$. – leonbloy Jun 08 '18 at 20:06

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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?

InsideOut
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snar
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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 $\color{red}{2}=1.$

JakeD
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Jika
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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.

celtschk
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"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.

Maurice Dassen
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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.

Cave Johnson
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RusselBMcDonald
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    You're misunderstanding the solution to $x^2\le 4x^3+5x^2$. From that inequality, we can deduce that EITHER $x\not=0$ and $1\le 4x+5$, OR $x=0$ - that is, it splits into cases, each of which is handled separately. The former case leads to the solution "$x\not=0$ and $x\ge -1$," and the latter case leads to the solution "$x=0$" (since $x=0$ satisfies the original equation); combining these gives the full solution, $x\ge -1$. – Noah Schweber Oct 24 '16 at 05:26
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    You do not need to divide by $x^2$ to solve that. Subtracting $x^2$ is a fine first step, and is valid regardless of what value $x$ is. \begin{align} x^2&\leq4x^3+5x^2\0&\leq4x^3+4x^2\0&\leq4x^2(x+1)\end{align} Equality holds at $x=0$ and $x=-1$. The right side is positive on $(-1,0)\cup(0,\infty)$. – 2'5 9'2 Oct 24 '16 at 05:27
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    When you say "you suddenly can't have x = 0 any more?" it sheds light on what you are misunderstanding. No one is saying that in the way you are interpreting it. If you choose to divide by $x^2$, no one is saying that $x$ cannot be $0$ as a solution to the original inequality. They are saying that either $x=0$ (which is obviously a solution to the original inequality) or $x\neq0$ and then the result of the division is valid. – 2'5 9'2 Oct 24 '16 at 05:31
  • I appreciate all of the responses! Will digest it. Had a "heated" discussion about this with the family when trying to help my son with his homework. still struggling but will review the comments. Truly appreciate it. – RusselBMcDonald Oct 24 '16 at 05:34
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    Consider instead the inequality $x^2\le -x^2$. Clearly that inequality is fulfilled by $0$ (but by no other number). Now divide both sides by $x^2$. You get $1\le -1$ which is never true, no matter what value you choose for $x$. – celtschk Oct 24 '16 at 06:18
  • Perhaps it may help to split into the two cases (x=0 and x≠0) first?  If x=0, then the inequality is trivially true; or if x≠0, then you can divide by it, etc. – gidds Oct 04 '23 at 09:47
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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.

Pockets
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  • If I read the linked answer it actually states that the value is 1. Do you mean that the answer on SO is incorrect? – Dennis Jaheruddin May 21 '14 at 14:32
  • Surely one cannot say $\frac{y}{0}$ tends to infinity. Using traditional arithmetic this is not defined no matter the $y$ so it cannot tend to anything. – DanZimm May 21 '14 at 14:32
  • I think it is bad to write "tends to infinity" for $\frac{y}{0}$. The existence of $\frac{y}{0}$ is at sake. You should define what is "tends to $\infty$ " and $\frac{y}0$ leads to that definition. – hrkrshnn May 21 '14 at 14:32
  • @boywholived, correspondingly fixed. Thank you for the correction. – Pockets May 21 '14 at 14:37
  • @DennisJaheruddin, the answer you link explains why MATLAB evaluates $f(0)$ as $1$ when defining $f(x)=\frac{x}{x}$. It is not saying that $\frac{0}{0}=1$. – Pockets May 21 '14 at 14:39