When working with strictly real numbers, there are a number of proofs that $e$ is transcendental. However, when dealing with non-standard analysis, one can express $e$ as $(1 + \frac{1}{H})^H$ for any infinite hyperinteger $H$. Does this mean that $e$ is not transcendental within the hyperreal framework? Or am I missing something when it comes to the definition of transcendental or hyperreal numbers?

  • $\begingroup$ what is the definition of transcendental number in the hyperreal framework? if it satisfies the definition then taht is your answer. $\endgroup$
    – jimjim
    Feb 12, 2019 at 14:38
  • $\begingroup$ Actually, $e$ even corresponds to hyperrational number (or any hyperS if the S are dense in the reals). However, as the Wofsey's answer says, they are not actually equal, only infinitely close. $\endgroup$ Feb 12, 2019 at 22:16

1 Answer 1


This is totally wrong. You cannot express $e$ as $(1 + \frac{1}{H})^H$; rather, $(1 + \frac{1}{H})^H$ will be infinitely close to (but NOT equal to) $e$. After all, $(1+\frac{1}{n})^n$ is never exactly equal to $e$ for any finite $n$, so the same holds for arbitrary hyperreal $n$ by transfer.

More generally, any transcendental real number remains transcendental in the hyperreals (assuming you are using a version of the hyperreals rich enough to express the concept of being transcendental, which is second-order over the set $\mathbb{R}$). This is again just by transfer. To be clear, in this context by "$\alpha$ is transcendental in the hyperreals" I mean "there does not exist any $n\in{}^*\mathbb{N}$ and an internal hyperfinite sequence $(a_m)_{m\leq n}$ of hyperintegers which are not all $0$ such that $\sum_{m=0}^n a_m\alpha^m=0$ where that sum is an internal hyperfinite sum".

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    $\begingroup$ Ah, that's what I was missing- that $(1 + \frac{1}{H})^H \approx e$, not $(1 + \frac{1}{H})^H = e$. $\endgroup$ Feb 12, 2019 at 17:25
  • $\begingroup$ @PixelArtDragon When doing nonstandard calculus, that is still a useful property however. For example, if a $f$ is continuous, $f((1 + \frac 1H)^H) \approx f(e)$. $\endgroup$ Feb 12, 2019 at 22:19

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