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We cannot include $\infty$ into $\mathbb R$ if we define $0\infty =1$. However, we don't have to do so. We just define another element $1/\infty\neq 0$ to be the reciprocal of $\infty$. In that way, can we get a consistent system of numbers?

Please read the following carefully, because this is NOT the usual way we treat $\infty$. I know perfectly why we cannot define $0\infty =1$ with ordinary rules for addition and multiplication. I am NOT doing that. Please do not downvote without reading this carefully.

To be more specific, we can do arithmetic operations with this extended set of real numbers and $\infty$ and $1/\infty$. For example, $(3\times \infty)^{-1} =\frac13 1/\infty$

Is there a branch of math about this?

Ma Joad
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    Cf. extended real number system – J. W. Tanner Sep 09 '19 at 13:28
  • What do you mean by "consistent system of numbers"? A priori $0\infty = 1$ is perfectly consistent. It's just that your ordinary rules for addition and multiplication stop working. – Mees de Vries Sep 09 '19 at 13:28
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    See Extended real number line. – Mauro ALLEGRANZA Sep 09 '19 at 13:29
  • The idea of single (or finitely many) infinitely large element is not compatible with arithmetic properties. To retain the 'ordered field' property while having infinitely large numbers, one need to consider infinitely many of them, such as in hyperreal numbers and surreal numbers. – Sangchul Lee Sep 09 '19 at 13:29
  • Wait. Let me clarify – Ma Joad Sep 09 '19 at 13:30
  • You cannot have an element $1/0$ and keep all your ordinary arithmetic. You will have to give something up. See for example this answer. – Mees de Vries Sep 09 '19 at 13:34
  • I guess what you want to do here is to add a formal element $\infty$ to $\mathbb{R}$ and make this compatible with other elements, while keeping its interpretation as 'infinity'. More elegantly, you are probably considering the transcendental extension $\mathbb{R}(\infty)$ of $\mathbb{R}$ and then making this an ordered field. This is what Arthur is describing. – Sangchul Lee Sep 09 '19 at 13:42
  • @SangchulLee Yes that's what I mean – Ma Joad Sep 09 '19 at 13:44
  • Although I agree that this is a perfectly mathematically sound approach, its application is rather limited. Because in such number system, we must have inequalities such as $\frac{1}{2}\infty < \infty < \infty^2$ in order to make the ordering compatible with arithmetic operations, and this is less likely what people expect from the notion of $\infty$. The hyperreal numbers avoid this criticism by adding all the possible mode of infinities that may arise from sequences and functions on $\mathbb{R}$, and the extended real numbers do so by giving up some arithmetic properties. – Sangchul Lee Sep 09 '19 at 13:49

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Sure you can get a consistent number system out of this. You have might even have used one without realizing.

Possibly the most common example is the field of rational functions (fractions where the numerators and denominators are polynomials) with real coefficients. We have an ordering on this as well: $f>g$ is defined to mean $\lim_{x\to\infty}(f-g)>0$.

The real numbers are contained in there as the constant functions, but in addition to the real numbers you have infinite elements like $x$ ("infinite" as in larger than any element of the form $1+1+\cdots +1$), and infinitesimal elements like $\frac1x$.

And here we do indeed have $(3x)^{-1} = \frac13\cdot \frac1x$, like your example says.

There are more ways of extending the real numbers to a number system with infinities and infinitesimals too, like the hyperreal numbers, or the surreal numbers, and certainly others.

If you want it to belong to a branch of mathematics, then possibly "nonstandard analysis" is the name to look for.

J. W. Tanner
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Arthur
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