Nonstandard part of a limited hyperreal

Question

Let $b$ be a limited hyperreal and $x$ be its standard part, i.e. the unique real number infinitely close to $b$. Is it true that one can find an infinitesimal $\varepsilon$ such that $$b = \frac{x}{1+\varepsilon}$$

Answer 1

If $b$ is a non-zero infinitesimal, then $x = \mathrm{st}(b) = 0$, and $\frac{0}{1+\varepsilon} = 0 \neq b$ for all $\varepsilon \in \:^\star\mathbb{R}$.

Otherwise, $b$ is either zero or not infinitesimal. In the first case, any infinitesimal $\varepsilon$ works. In the second case, $b$ is non-zero and so is $x$. Therefore, we can set $\varepsilon = xb^{-1} - 1$ and calculate as follows:

$$ \frac{x}{1+\varepsilon} = \frac{x}{1 + xb^{-1} - 1} = \frac{x}{xb^{-1}} = \frac{1}{b^{-1}} = b. $$

All that's left to prove is that $\varepsilon$ is actually infinitesimal. Consider the function $f: \mathbb{R} \rightarrow \mathbb{R}$ defined via $f(y) = xy^{-1} - 1$. We have $\mathrm{st}(b) = x$, so by the continuity of $f$ we get $$\mathrm{st}(\varepsilon) = \mathrm{st}(xb^{-1} - 1) = \mathrm{st}(\:^\star f(b)) = \:^\star f(\mathrm{st}(b)) = \:^\star f(x) = xx^{-1} - 1 = 1 - 1 = 0.$$

Therefore, $\varepsilon$ is actually infinitesimal.