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Is there a name for the mathematical rule/axiom/property $x/x = 1?$

What are the conditions for it to apply?

For instance, the rule does not apply where $x = 0$ or $x = \inf$. I saw one site that claimed it only applied to real numbers, but it does apply to imaginary ones too, so that rule is not complete.

What is the complete rule and what is its name?

Mike Fairhurst
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  • I figure it's pretty obvious for natural numbers, then extrapolated up from there to other types of numbers. – jdods May 05 '16 at 22:23
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    I would say it's "existence of inverse"; if you have a group $(G,\bullet,e)$, then for every $g\in G$ there exists an $h$ such that $g\bullet h=1$, and we write $h=g^{-1}$, to accommodate the familiar notation of $g\bullet g^{-1}=e$. Inverses are defined with respect to some binary notation; in a ring there exists for every element an additive inverse, but not necessarily a multiplicative inverse; in a field (such as the reals) it applies to every non-zero number (Note for you inf, which I take means $\infty$, is NOT a real number, so don't treat it as one). – B. Pasternak May 05 '16 at 22:26

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Rather than viewing division as an operation in its own right (that would take a dividend and a divisor to a quotient), mathematicians think about inverses of multiplication. So one thinks of $\frac x y$ as $x · y^{-1}$ where $y^{-1}$ is, by definition, a number inverse to $y$, i.e. fulfilling $y·y^{-1} = 1 = y^{-1}·y$.

For example $\frac 3 2$ is rather thought of as “three halfs” (where a half is the inverse of two) than “three divided by two”. (Well, but one still uses the latter parlance, in fact.)

Therefore, I don’t think there is a name for this arithmetic law itself, at least I don’t know of one. But there’s of course one for the concept of inverse elements. A more fundamental concept is that of neutral elements, on which the concept of inverses depend.

And by the way: Division isn’t viewed as an operation by itself because it behaves badly: It is not totally defined as a map, say $ℚ × ℚ → ℚ$ (you can’t divide by zero), it fails to be associative, let alone commutative, and it doesn’t have a neutral element.

k.stm
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    That's a well formulated answer. – Mathematician 42 May 05 '16 at 22:36
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    so what is $y^{-1}$, for example $0^{-1}$ or $\infty^{-1}$. Isn't it true that $0^{-1}=\infty$ and $\infty^{-1}=0$, and is it true that $0\cdot\infty=1$? – Mirko May 05 '16 at 23:53
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    @Mirko: no, many times no. $\infty$ is not a number! Mathematicians use $\infty$ as a shorthand for various phenomena, and sometimes use "equations" involving $\infty$ to refer to related phenomena; but these are not actual equations of numbers, because $\infty$ is not a number. $0^{-1}$ doesn't exist. – Greg Martin May 06 '16 at 03:38
  • @GregMartin But if $0^{-1}$ doesn't exist, then inverses "behave badly" and I do not follow the point in the above answer to replace division with inverses, how does that really help answer the original question? – Mirko May 06 '16 at 03:43
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    @Mirko: The inverse $x^{-1}$ exists for any $x$ you can divide by. So it simplifies things: instead of asking "is $x/y$ defined" for arbitrary $x$ and $y$, we can just ask "is $y^{-1}$ defined" for a single $y$ and then $x/y$ is (or is not) defined for all $x$. $x/y$ can't be defined for all $x$, $y$, for reasons better explained elsewhere. – Jonathan Cast May 06 '16 at 04:08
  • @jcast Perhaps $\frac00$ is defined (and $\frac00=1$) even if $0^{-1}$ is not defined? – Mirko May 06 '16 at 04:25
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    @Mirko: let's try that: $$2=2\times1=2\times\frac00=\frac {2\times0}0=\frac00=1. $$ – Martin Argerami May 06 '16 at 05:28
  • @MartinArgerami But aren't you implicitly assuming that $0^{-1}$ is defined when you write $2\times\frac00=\frac {2\times0}0$? This looks to me like application of associativity, $2\times\frac00=2\times(0\times0^{-1})=(2\times0)\times0^{-1}=\frac {2\times0}0$. – Mirko May 06 '16 at 11:28
  • @Mirko: so you want to define $\frac00=1$ but postulate that associativity does not hold? – Martin Argerami May 06 '16 at 14:33
  • @Mirko This is also discussed here, see for example Goblin’s answer to this. – k.stm May 06 '16 at 15:18
  • @k.stm Thank you, I am happy with the link you supplied, it reaches a contradiction without mentioning $0^{-1}$. Copied from there:

    Proposition 3: $\frac{0}{0}$ is not a real number

    Proof: Suppose that $\frac{0}{0} = x$, where $x$ is a real number.

    $\frac{0}{0}=x \Leftrightarrow x \cdot 0 = 0 = (x + 1) \cdot 0 \Leftrightarrow \frac{0}{0}=x+1$

    $ \therefore x = x + 1 \Leftrightarrow 0 = 1 \Leftrightarrow \bot$

     – Mirko May 06 '16 at 21:20
  • @MartinArgerami I wanted to define $\frac00$ without reference to $0^{-1}$ (or, alternatively, explain why $\frac00$ does not exist, without reference to $0^{-1}$), as I did not feel that talking about inverses on its own answers the question. – Mirko May 06 '16 at 21:23
  • @Mirko: the whole point of fractions it to talk about inverses.And in the argument you posted you posted you are using inverses, to get $x\cdot0=0$. – Martin Argerami May 06 '16 at 22:20
  • @MartinArgerami I do not follow what you are saying. "The whole point" seems too philosophical to me. In my previous comment I copied an argument showing $\frac00$ is not a real number without using the notation $0^{-1}$, so I do not see how I am using inverses. – Mirko May 06 '16 at 23:24
  • @Mirko: First, "the whole point" is not a philosophical matter. Fractions are created so that the expression $1/n$ (that is, the inverse of $n$) makes sense. Writing $\frac1n$ or $n^{-1}$ is only a matter of notation, but there is no difference. And in your argument you use "the inverse of $0$" to go from $x=\frac00$ to $x\cdot 0 = 0$ (otherwise, there is no reason for those two expressions to be equivalent). – Martin Argerami May 06 '16 at 23:28
  • @MartinArgerami By definition $\frac a b =c$ if $a = b\cdot c$. If I wish I may proceed and define and talk about inverses, but I do not have to. – Mirko May 06 '16 at 23:49
  • @Mirko: you may, if you wish. Now, what would exactly be the point of dealing with fractions but not doing arithmetic with them? – Martin Argerami May 07 '16 at 00:38
  • @MartinArgerami I do not know the answer to your question, but that is not what the OP asked. Say you look at all even integers $E$ with multiplication. There is enough structure to say that (a) $\frac82=4$ since $8=4\cdot2$, (b) $\frac62$ is undefined, as there is no $b\in E$ with $6=b\cdot2$, (c) no unit, not even possible to talk about $2^{-1}$. Is $\frac62$ undefined because $2^{-1}$ isn't (and if so,then how is $\frac82$ defined)? In $\mathbb Z$ there is $1$ and $\frac62=3$, yet no $2^{-1}$. Proof that $\frac00$ is undefined (and couldn't be) works nevertheless, without involving inverses – Mirko May 08 '16 at 02:02
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If $0/0 = x$ then $ 0=0x$ which is true for any value of $x$, meaning there is no unique solution: $x$ is undefined for this kind of ratio.

So then y/y is undefined unless it is specified that y≠0

user643172
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