Prove that the given linear map is injective Let $n \in \mathbb{N}$ be a fixed positive integers and $B \in \mathbb{R}_+$ also be fixed. For a fixed $M>0$, let $f:[-B,B]^n \to \mathbb{R}$ be given by $f(x_1,\ldots,x_n)=\sum_{i=1}^n x_i M^i$. My aim is to prove the following:
Prove that one can always find a fixed positive real number $M$ (dependent only on $B$ and $n$) such that $f$ is an injective map.
If such an $M$ does not exist in the above question, one can assume that each $x_i$ can take only  a fixed set of countable values in $[-B,B]$. In other words, the map $f$ is defined only for a countable subset of $[-B,B]^n$. In this reduced case, can $f$ be injective?
My attempt: Let $f(x_1,\ldots,x_n)=f(y_1,\ldots,y_n)$ for some $(x_1,\ldots,x_n),(y_1,\ldots,y_n) \in [-B,B]^n$. Thus, $\sum_{i=1}^n (x_i-y_i)M^i=0$. After this step, I am stuck and unable to procced further about how to choose such an $M$ such that the above equation does't have any solution.
 A: The general statement is definitely false for $n > 1$. $f(0, 0)=0$, but we can also choose some $0 < \epsilon < \min(B, BM)$ so that $\left(-\epsilon, \frac \epsilon M\right) \in [-B, B]^2$ and thus in the domain of $f$. Thus, we have the following:
$$f\left(-\epsilon, \frac \epsilon M\right)=-\epsilon M+\frac \epsilon M M^2=-\epsilon M+\epsilon M=0$$
Therefore, $f(0, 0)=f\left(-\epsilon, \frac \epsilon M\right)$ and $f$ is not injective.
However, I think the reduced statement is actually true. Let $D$ be the domain of $f$. For any two unequal ${\bf x}, {\bf y} \in D$, $f({\bf x})=f({\bf y})$ means that $\sum_{i=1}^n ({\bf x}_i-{\bf y}_i)M^i=0$. Since the left side of this equation is a non-zero polynomial of degree at most $n$ in terms of $M$, there are at most $n$ values of $M$ which satisfies this equality.
We have the following:


*

*Since $D$ is countable, there are a countable number of pairs of unequal ${\bf x}, {\bf y} \in D$.

*Given a specific pair of ${\bf x}, {\bf y} \in D$, there are a finite number of $M$ where $f({\bf x})=f({\bf y})$.


Therefore, overall, there are a countable set of $M$ where $f({\bf x})=f({\bf y})$ is true for any ${\bf x}, {\bf y} \in D$. On the other hand, there are an uncountable number of positive real numbers, so choose a positive number that is not in the described countable set. For this $M$, $f({\bf x}) \neq f({\bf y})$ for all ${\bf x}, {\bf y} \in D$, which means $f$ is injective.
