2

I have a continuous variable whose range is within $[-1, 1]$. I want to map the values of this variable to the range $[0, 1]$ instead. What I do is I add the value of $1$ to the the variable and divide the result by $2$`.

To me, this looks like a linear, valid and invertible operation meaning that the mapping is one-to-one between the domain and the co-domain of this operation/function.

I wonder, is my intuition correct? I would appreciate if someone can clarify this.

Henrik supports the community
  • 6,831
Amir
  • 435

2 Answers2

4

Yes, your intuition is correct. To be precise, you are describing the function $$f: [-1,1] \rightarrow [0,1],$$ $$f\left(x\right) = \frac{x+1}{2},$$ which is bijective and has inverse $$g: [0,1] \rightarrow [-1,1],$$ $$g\left(y\right) = 2y-1.$$

Sam Streeter
  • 1,556
  • Both your answer and C. Grant's answer are satisfactory for me. However, I want the other person to gain the privilege of commenting so I marked his answer as the final answer :) I Hope you don't mind – Amir Oct 31 '18 at 14:51
  • Ah, that's nice of you! Yes, I'm happy for you to spread the reps. I was initially a bit suspicious of their answer being so similar to mine, even in the wording, but I think that's just because there's a pretty canonical response here that we've both used, and their reasoning for the inverse is different, so fair enough. – Sam Streeter Oct 31 '18 at 14:57
2

Yes your intuition is correct, in more detail what you've done is defined a one-to-one and onto function $f: [-1,1] \to [0,1] $ where $f(x) =\frac{x+1}{2}$.

If you want to check that is in Infact invertible you can try solving the equation $ x=\frac{y+1}{2} $ where $x$ is the original input, and y is the output of the function $f(x)$

C. Grant
  • 220