# hyperreals standard part inconsistency

$\def\st{\operatorname{st}}$ 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 ?

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.
• @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 $a\cdot\frac{b}c$. – Brian M. Scott Mar 5 '15 at 0:28