how to prove that this function is not injective

Solution 1:

First of all, $\log_{-1}$ is not a well defined operation. Thus your solving starts to go wrong there. Instead, note that the $(-1)^x$ terms have no effect on the magnitude of the values, so we get

$$|2a-1|=|2b-1|$$

Now if the $a\neq b$, then it must be the case that $2a-1\neq2b-1$. By the above relation, we get that

$$2a-1=-(2b-1)$$ $$\Rightarrow a+b=1$$

From here, it seems that the injectivity of $f$ depends on the domain--specifically how $\mathbb N$ is defined. If in your class $0\in\mathbb N$, then we have the solution $a=1,b=0$. If $0\not\in\mathbb N$, as is more common, then this equation has no solution, and $f$ is injective.

Since your question seems to tell you that $f$ is not injective, we will assume the former.

Let $a=1,b=0$. Then $a\neq b$. Note:

$$f(a)=f(1)=\frac{(-1)^1 (2(1)-1) +1}{4}=0$$

$$f(b)=f(0)=\frac{(-1)^0 (2(0)-1) +1}{4}=0$$