More non-standard analysis (infinitesimal calculus). The problem is from the free online book "Elementary Calculus: An Infinitesimal Approach" given to me earlier today by Plato. Please let me know if this proof is solid.


Given:
(*) a,b,a_1,b_1 are hyperreal numbers such that a \approx  a_1, b \approx b_1.

Prove that if a and b are finite, then ab \approx a_1b_1. (≈ means "are infinitely close" in this context).


1. Assume that a and b are finite, and let \varepsilon, \delta, \varepsilon_1 and \delta_1 be infinitesimal.

2. Since a \approx a_1 and b \approx b_1, then a_1 = a + \varepsilon and b_1 = b + \delta.

3. Since a and b are finite and a \approx a_1 and b \approx b_1, then a_1 and b_1 are finite.

4. Since a_1 and b_1 are finite, then a_1 \cdot b_1 = [st(a_1) + \varepsilon_1] \cdot [st(b_1) + \delta_1]

5. From 2, a_1 \cdot b_1 = [st(a + \varepsilon) + \varepsilon_1] \cdot [st(b + \delta) + \delta_1]
= st(a + \varepsilon) \cdot st(b + \delta) + st(a + \varepsilon) \cdot \delta_1 + \varepsilon_1\cdot st(b + \delta) + \varepsilon_1 \delta_1
=  ab + a \delta_1 + b \varepsilon_1 + \varepsilon_1 \delta_1.
6. The sum a\delta_1 + b\varepsilon_1 + \varepsilon_1\delta_1 is infinitesimal according to the rules for infinitesimal and finite numbers. Let
\varepsilon_2 =  a\delta_1 + b\varepsilon_1 + \varepsilon_1\delta_1. Then a_1b_1 = ab + \varepsilon_2.
7. Since \varepsilon_2 is infinitesimal, it follows that ab \approx  a_1b_1.