I am to prove this property of *$\Bbb R$: If $x \approx y$ and $u \approx v$ and $u,x$ are finite then $xu \approx yv$. My question is can I just use the transfer principle for the multiplication property of equality? Also since, $x - y \approx 0$ and $x$ is finite, wouldn't it be true that $y$ is finite as well?
Asked
Active
Viewed 52 times
1
-
I'm not too familiar with the conventions of non-standard analysis, so I'm wondering: does $a\approx b$ mean "$a-b$ is infinitesimal"? – Arthur Jan 26 '18 at 08:57
-
@Arthur, yes it does. – Halinka Jan 26 '18 at 09:01
2 Answers
1
One cannot use transfer directly in this case because the relation $\approx$ of infinite proximity is not an internal relation. However the proof is elementary and uses merely the fact that an infinitesimal multiplied by a finite hyperreal is still infinitesimal. The latter fact is immediate from the definition of an infinitesimal.
Mikhail Katz
- 42,112
- 3
- 66
- 131
0
Simply write $y=x+\epsilon$ and $v=u+\delta$ with infinitesimals $\epsilon,\delta$, and see what you get for $yv$.
Hagen von Eitzen
- 374,180