If $c$ is the only real solution of $f(x) = 0$ then prove that $c$ is the only hyperreal solution. 
If $c$ is the only real solution of $f(x) = 0$ then prove that $c$ is the only hyperreal solution.


I can't see how this is a valid question since we can have $c + \delta$ as a solution to $f^*(x) = 0$ and since $st(c+ \delta) = c$ we still have $f(c) = 0$ as only real solution.
Am I missing something ?
 A: It follows from transfer: it's true that $$(\exists c)[\ ^*f(c) = 0 \wedge (\forall x)(^*f(x) = \ ^*f(c) \to x = c) ]$$
if and only if it's true that $$(\exists c)[f(c) = 0 \wedge (\forall x)(f(x) = f(c) \to x=c)]$$
We could identify an infinitesimal if your theorem were true: if there were $\delta$ with $\delta \not = 0$ such that $^*f(c) = \ ^*f(c+\delta) = 0$, then we could identify $\delta$ as $(c+\delta) - c$. Since it's not possible to distinguish "strictly nonstandard" reals from standard reals without referring to the standard universe, there must either be both standard and nonstandard $\delta \not = 0$ making $\ ^*f(c+\delta) = 0$, or there must be no $\delta \not = 0$ with $^*f(c+\delta) = 0$.
A: Suppose $d$ is another hyperreal solution to $f(x)=0$. Then we have two distinct hyperreal solutions $c$ and $d$, which we can summarize by the formula
$$\exists x \exists y [(x\not=y) \wedge (f(x)=f(y)=0)].$$ Then by downward transfer we get that the same formula holds over $\mathbb R$, so that $c$ is not a unique real solution. The contradiction proves the uniqueness of solution in the hyperreals, as well.
