Earlier quoted context omitted.
> before the last digit, which is '1' No. There is no last digit. That's the whole point. If there were a last digit your argument would be correct, but there isn't, so it's not.
I don't see the problem with having a first digit and a last digit and an infinite number of digits in between. Edit: Infinitesimal divided by two is infinitesimal, in the same way that infinity multiplied by two is infinity. So 0.000...0001 / 2 = 0.000...0001 . Infinitesimal multiplied by any finite number is infinitesimal. Infinitesimal multiplied by infinity is every number in the interval from infinitesimal to in…
Good luck proving or calculating anything.
You can define "infinitesimal/2 == infinitesimal", but nothing good will come out of it. A definition is no good unless it lets you do something.
Letting e=infinitesimal, you have e/2==e, so e==2e so 0==2e-e so 0==e. This definition is inconsistent with being able divide by non-zero integers and subtraction.