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$\def\st{\operatorname{st}}$ I'm studying non-standard calc from Keisler's book. Taking "standard part" rule doesn't make sense... its not commutative.

e.g.

$a$ is finite non infinitesimal

$b,c$ are infinitesimal

$\st(a\cdot{b\over c}) = \st(a) \cdot \st({b\over c}) = a \cdot \st({b\over c}) $ [$\st({b\over c})$ is indeterminate so we leave it as is]

$\st(b\cdot {a\over c}) = \st(b) \cdot ...... = 0$ [$b$ is infinitesimal so $\st(b)$ is $0$]

$2$ very different answers for the same expression.. what's the deal here ?

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You are applying the property that $$st(xy)=st(x)st(y)$$ However this applies only when $x,y$ are finite; see Theorem 3 on p.37.

In your second calculation, $a/c$ is not finite. Your first calculation might be incorrect as well, depending on whether or not $b/c$ is finite.

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  • $\begingroup$ Page 40 of the pdf (full version of the book): "each infinitesimal is finite". Page 72 he carries out very similar operations to mine above. $\endgroup$ – user111824 Mar 4 '15 at 23:38
  • $\begingroup$ just to be even clearer.. page 40.. definition of: finite = between 2 real numbers (0 and smallest real are both real) $\endgroup$ – user111824 Mar 5 '15 at 0:17
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    $\begingroup$ @user111824: There is no smallest real number. Yes, infinitesimals are finite, but that’s beside the point: $\frac{a}c$ is not finite, so the theorem doesn’t apply to $b\cdot\frac{a}c$, and $\frac{b}c$ may or may not be finite, so the theorem may or may not apply to $a\cdot\frac{b}c$. $\endgroup$ – Brian M. Scott Mar 5 '15 at 0:28

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