# Can we include $\infty$ and $1/\infty$ into $\mathbb R$? [duplicate]

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, $$(3\times \infty)^{-1} =\frac13 1/\infty$$

• Cf. extended real number system – J. W. Tanner Sep 9 '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 9 '19 at 13:28
• – Mauro ALLEGRANZA Sep 9 '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 9 '19 at 13:29
• Wait. Let me clarify – Ma Joad Sep 9 '19 at 13:30

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.