YHK - yet another hacker news clone

The formalism is very simple symbolically. But the mathematical machine behind it is very complex.

Also conceptually it feels just right to use nilpotents to probe the smooth structure. In a way nilpotents are violently smaller than even non standard analysis infinitesimals, as the laters’ powers are incredibly small but never vanishing.

Another way to see this is that it makes Taylor expansion exact by killing terms above a bound so it works naturally with the ecosystem surrounding it

Finally duals are very similar to complex in a way. i can be defined as root of X^2 + 1 = 0 even if it felt impossible initially, the dual number as a non nul solution of X^2 = 0 even if it is as counterintuitive.

No. Let's take a nonstandard proof of the intermediate value theorem on [0,1] by Nelson.

By the transfer principle it is enough to prove this for a standard continuous function f on [0,1] with f(0)<0<f(1).

Take a finite subset of [0,1] containing every standard point. Colour its points blue, green, or red according to whether f is negative, zero, or positive at tha point.

The first point of the interval is blue and the last red. Hence either awe can find some green point, or we can find two neighbouring points that have different colours, the first blue and the second red.

In the first case there is a zero, so we are done. In the second, let the neighbouring points be p and q. By the completeness of the real numbers, every nonstandard real in [0,1] is infinitesimally close to exactly one standard real. So p and q are infinitesimally close to some standard real number, let's call it z.

Standard continuous functions send infinitesimally close points to infinitesimally close points. So f(p) and f(q) are both infinitesimally close to f(z). But f(p) is negative and f(q) is positive. The only standard number infinitesimally close to both positive and negative numbers is zero. Thus f(z) is zero. This proves the theorem.

You tell me, which standard proof is this? It's certainly not the nested interval proof. Not the supremum proof. Not the bisection proof in disguise. Which argument does it wrap in slightly different language? Can you point to a single textbook, course note or lecture that gives such an argument?

No. One could of course argue that this is not simpler/shorter than the usual arguments. But it is very different from them. Saying that it's the same arguments repackaged in a different language is just wrong, and detracts from an otherwise valid point.