Injective function proof involving floor function Let $f : \Bbb{Z} \to \Bbb{Z}$ and $g : \Bbb{Z} \to \Bbb{Z}$ be functions defined by $f(x)=3x+1$ and $g(x)=\lfloor\frac{x}{2}\rfloor$. Is $g(f(x))$ one-to-one?


So, $g(f(x)) = \lfloor\frac{3x+1}{2}\rfloor$. I know that $g(f(x))$ is one to one because we are only considering the integers. So for every $x$ integer there is a different $y$ integer. How do I prove this? I'm supposed to prove one-to-one functions by assuming $F(x_1)=F(x_2)$ and then proving $x_1=x_2$, but I don't know how to approach this when the function is a floor function. 

Thanks in advance!
One possible Case would be: 

Let $[x]=[y]$ in which $[x]=\text{floor}(x)$. If $x\in\mathbb Z$ then $[x]=x$ and so $[y]=x\in\mathbb Z$ so $$y\in [x,x+1)$$ Since, here, $y\in\mathbb Z$ so we necessarily have $y=x$.
Oh for god sake, 
$$\lfloor\frac{3(1.1)+1}{2}\rfloor = \lfloor\frac{3(1.2)+1}{2}\rfloor = 2.$$
Thus, $g\circ f$ is not one-to-one.