What happens if to introduce infinite and infinitesimel quantities this way?

Question

What if to introduce $\varepsilon$ and $\omega=1/\varepsilon$ from the following equation as a definition?

$$\left(1+\varepsilon\right)^{1/\varepsilon}=e$$

or such $\varepsilon$ that $e^\varepsilon=1+\varepsilon$ and $\varepsilon\ne 0$ as a definition.

Can we deduce other properties of $\varepsilon$ from here?

Answer 1

I cannot answer if this leads to any inconsistencies, although it probably does, but I know that Euler handled limits in this way. In his work "Introductio in analysin infinitorum", he wrote proofs like:

Assume $\epsilon$ is infinitesimally small and $\omega$ is infinitely large. Then $\mathfrak{l}(1+\epsilon)=\epsilon$ ($\mathfrak{l}$ was the shorthand for logarithm) and therefore $\mathfrak{l}\left((1+\epsilon)^\omega\right)=\omega\epsilon$. But if we set $x=(1+\epsilon)^\omega$, we obtain $1+\epsilon=x^{\frac{1}{\omega}} \iff \epsilon=x^{\frac{1}{\omega}}-1$. Therefore $\mathfrak{l}(x)=\omega\epsilon=\omega\left(x^{\frac{1}{\omega}}-1\right)$.

With this argumentation, he proved $\ln(x)=\lim_{n\to\infty}n\left(x^{\frac 1 n}-1\right)$, and it is truthfully astonishing, how he could proof these things, if we consider that it isn't at all mathematically rigorous by means of todays standards.

So I think, if you have the right intuitions, you can obtain really nontrivial results using such definitions, but a lack of precision is undeniable.

Nevertheless, such assumptions even lead Euler to make some questionable statements, like: $$ \sum_{p\space prime} \frac{1}{p}=\ln\ln(+\infty) $$

Answer 2

Lets say you define the ring $K$ as $ \mathbb{R} $ with $\varepsilon$ and all the elements generated by $\varepsilon$

Assumptions:
1) $\varepsilon$ is positive
2) The rule that "A finite number of additions or products of positive numbers results in a positive number" that applies to $ \mathbb{R} $ also applies to $K$
3) Series expansions that apply to $ \mathbb{R} $ also applies to $K$

If you look at the series expansion for the exponential function

$$exp(x) = 1 + x^1/1! + x^2/2! + \dots $$

then we would have $$exp(\varepsilon) = 1 + \varepsilon + y$$ where $y$ is "positive" which would seem to contradict the definition.

I guess my conclusion is that series expansions would not be well behaved when using this new $ \varepsilon$.