# Why use the real numbers instead of some larger totally ordered field?

If there are ordered fields of cardinality larger then the reals why not use one of them? Is it just the completeness properties that are good? If it is why not embed the reals in something bigger and then you could use completeness in that field but only when confining yourself to cases involving just real numbers, while still having the other elements in that field to use for another purpose.

• Completeness is good. Some do embed $R$ in larger fields, e.g., the "hyperreals" when doing non-standard analysis – Lord Shark the Unknown Oct 29 '17 at 5:51
• @bof Use them for what? I'm trying to ask what makes studying the reals more interesting then constructing larger totally ordered fields with the reals embeded in them? Obviously "interesting" or "useful" are subjective terms, but can you then tell me at least from your prospective on the matter then? – Johny6 Oct 29 '17 at 6:16

Suppose $\varepsilon$ is a member of such a field and is greater than $0$ and less than every real. Then $\underbrace{\varepsilon+\cdots+\varepsilon}_\text{finitely many terms}$ must still be smaller than every real, since if $\underbrace{\varepsilon + \cdots+\varepsilon}_{n\text{ terms}}$ were at least as large as some real, then $\varepsilon$ would be at least as large as that real over $n,$ and thus not smaller than all reals. Elements such as $\varepsilon,$ that are smaller in absolute value than all reals, are called infinitesimals.
The set of all infinitesimals in an ordered field that contains nonzero infinitesimals cannot have a smallest upper bound within the field. To see this, consider two cases: $(1)$ that smallest upper bound is an infinitesimal, and $(2)$ it is not. If it is not, then half of the smallest upper bound is also bigger than all infinitesimals and you have a contradiction. If it is, then twice that smallest upper bound is also infinitesimal, and you have a contradiction.
• So you are saying $\lim_{n \to \infty} n\varepsilon$ is bounded above by $1$ but it can't converge to any element. Or in other words $\mathbb{R}$ has an absolute value which is Archimedian. Larger fields (such as $\mathbb{R}(x)$) have some absolute values but they are not Archimedian. – reuns Oct 29 '17 at 6:29