This seems like a pretty basic question, but I've been searching around and haven't come across the answer.

Consider two infinitesimal numbers, $\epsilon$ and $\epsilon^2$. On the one hand, it would seem that $\epsilon^2 \ll \epsilon$, because $\epsilon^2 / \epsilon = \epsilon$. And on the other hand, it would seem that $\epsilon^2 \approx \epsilon$, because the difference $\epsilon - \epsilon^2$ is infinitesimal (Keisler's definition of $\approx$).

Are both of the above "it would seem" statements generally accepted as true? My instinct is to affirm $\epsilon^2 \ll \epsilon$ and deny $\epsilon^2 \approx \epsilon$, while affirming $1 - \epsilon^2 \approx 1 - \epsilon$ because $(1 - \epsilon^2) / (1 - \epsilon)$ is approximately 1. I guess what it comes down to is that I'd rather define $\approx$ using division, not subtraction. But is this tenable?

Okay, I guess that's more than one question, and maybe they aren't basic. But anyway, that's what I'm wondering.

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    $\begingroup$ Your two statements are true, and quite compatible. If you want to deny $\epsilon^2\approx \epsilon$, that is fine, but then your notation will be different from Keisler's. Very much not a good idea if you are taking a course that uses Keisler's book. $\endgroup$ – André Nicolas May 29 '14 at 0:01
  • $\begingroup$ Fortunately, I'm not taking a Keisler-based course. This is purely out of personal interest, related to the topic of infinitesimal probabilities. Thanks! $\endgroup$ – StumpyLeg May 29 '14 at 0:08
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    $\begingroup$ The relation $\approx$ is useful, since for "finite" non-standard reals, the equivalence classes with respect to $\approx$ can be identified with the reals. $\endgroup$ – André Nicolas May 29 '14 at 0:21
  • $\begingroup$ Ah: the set of 0 and all its close neighbors corresponds to 0, the set of 0.853 and all its close neighbors corresponds to 0.853, etc. Got it. $\endgroup$ – StumpyLeg May 29 '14 at 0:34
  • $\begingroup$ That's right. And that is why for this relation we do really want $\epsilon^2\approx \epsilon$. Of course one could use another symbol for that, freeing up $\approx$ to use as you wish. $\endgroup$ – André Nicolas May 29 '14 at 0:36

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