When $f\colon\mathbb{Z}^{+} \to\mathbb{R}$ satisfys $f(2x)=2f(x)$, then is it injective?

Question

If $f\colon\mathbb{Z}^{+} \to\mathbb{R}$ satisfys $f(2x) = 2f(x)$. Is it injective?

I’m new in analysis and can’t understand how to approach this problem. Should I find an analytic expression for $f(x)$ first?

Answer 1

No. Consider a function $f:\mathbb{Z}^+\to\mathbb{R}$ with $f(1)=1$, and $f(3)=1$. The functional equation $f(2x)=2f(x)$ only determines the values $f$ takes on inputs of the form $x=2^n$ and $x=3\cdot2^n$, $n\in\mathbb{N}$; since these two sets of naturals are disjoint, this is still well-defined; and since $f(1)=f(3)=1$, this is not an injection.

Answer 2

Since the domain of $f$ is $\Bbb{Z}^+$, defining the function $f$ means choosing real numbers to represent the values $f(1), f(2), f(3)$, etc. Try to start picking some values. What can you choose for $f(1)$? Maybe you set $f(1)=5$. What about $f(2)$? Well, according to your rule, $f(2) = f(2\cdot 1) = 2 \cdot f(1) = 2 \cdot 5 = 10$, so this choice is forced. Can you pick anything you want for $f(3)$?

A function $f$ is NOT injective if it takes the same value at different points, that is $f(a)=f(b)$ for $a\neq b$. When trying to determine the function in the previous paragraph, can you get this to happen?