1

Let's assume that $N$ is even infinite hyperinteger and $N=1/dx$, $N=dx/dx^2$.

Let's assume that we have expression $dx+dx^2$. We can add up $dx^2$ to $dx$ and $dx$ is obtained: $dx+dx^2=dx$ (algebraic property of infinitisemals). So, $dx$ contains $dx^2$ $N$ times, but $(dx+dx^2)$ contains $dx^2$ $(N+1)$ times.

How is it possible that $dx$ contains $dx^2$ $N$ (even infinite hyperinteger) times and $N+1$(odd infinite hyperinteger) times simultaneously?

Thanks!

Mike_bb
  • 691
  • 6
    $dx+dx^2=dx$? That may be an "algebraic property of infinitesimals". But that does not hold in nonstandard analysis (where we are in an ordered field). – GEdgar Jun 08 '22 at 18:22
  • 2
    As @GEdgar comments, it is not correct to say that $dx+dx^2=dx$ for infinitesimals. Yes, standard_part of $1+dx$ is just $1$, and so on, but that's not an assertion of equality. – paul garrett Jun 08 '22 at 19:11

0 Answers0