Is $e$ transcendental when working with hyperreal numbers?

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?

• what is the definition of transcendental number in the hyperreal framework? if it satisfies the definition then taht is your answer. Feb 12, 2019 at 14:38
• 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. Feb 12, 2019 at 22:16

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