# How to prove that this function is not injective

functions

I'm very new to proving that a function is injective, surjective, bijective, invertible etc.
So I'm supposed to prove that the function below is not injective.
$$f(x) = \frac{(-1)^x (2x-1) +1}{4}$$
Where $$f:\Bbb N\to\Bbb Z$$.

Well, I tried to consider: $$f(a) = f(b)$$
Which gives:
$$\frac{(-1)^a (2a-1) +1}{4} = \frac{(-1)^b (2b-1) +1}{4}$$
Then
$$(-1)^a (2a-1) = (-1)^b (2b-1)$$
And using log base $$-1$$:
$$a(2a-1) = b(2b-1)$$
Annnd finally:
$$2a^2 -a = 2b^2 -b$$

I know that if $$a=0$$ and $$b=1/2$$ they both get $$0$$ as the answer, but $$b= 1/2$$ isn't a natural number so I can't use that right?

So I guess I'm stuck here. Is there a way to break it down further to show that $$a\neq b$$?

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:

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

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