# Mapping values in the range [-1, 1] to [0, 1] in an invertible fashion

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.

• See the MathJax reference for tips on editing mathematical formulae/content. – Devashsih Kaushik Oct 31 '18 at 14:45
• Technically the function is affine rather than linear. – Matt Oct 31 '18 at 14:46
• @Matt To be fair, the term "linear", in the context of continuous functions of a single variable, can instead refer to those functions of the form $f\left(x\right) = ax + b$. However, since the OP called it a "linear operation", I agree that this frames it in such a way that it should indeed be called affine. – Sam Streeter Oct 31 '18 at 15:13

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 $$f(x)$$
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 $$g: [0,1] \rightarrow [-1,1],$$ $$g\left(y\right) = 2y-1.$$