## Show if a,b,a1,b1 are hyperreal with a≈a1, b≈b1, and a,b are finite, then ab≈a1b1

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

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

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

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

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

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

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