Thread: Proving coordinate functions

1. Proving coordinate functions

Given that f: l --> R is a coordinate function for l.

Prove that g: l --> R defined by g(P) = f(P) + c for some constant c is also a coordinate function for l.

My textbook defines coordinate function as one that is both one-to-one and "onto" such that PQ = |f(P) - f(Q)|

I already figured out how to prove the one-to-one, but I'm having issues with proving the "onto" part of it. In class, my professor just said it was a matter of "adding c and subtracting c," but I couldn't figure out what he meant by that. Any ideas?

2. Originally Posted by spectralsea
Given that f: l --> R is a coordinate function for l. Prove that g: l --> R defined by g(P) = f(P) + c for some constant c is also a coordinate function for l.
I'm having issues with proving the "onto" part of it.
ONTO
$\displaystyle \begin{gathered} x \in R \Rightarrow x - c \in R \hfill \\ \left( {\exists P \in I} \right)\left[ {f(P) = x - c} \right] ~,f\text{ is onto}\hfill \\ g(P) = f(P) + c = \left( {x - c} \right) + c = x \hfill \\ \end{gathered}$

3. Awesome, thank you. So in general, if you were trying to prove a function was onto, you would start out with x being a real number and then trying to work it out so that some f(P) would bring you back to the x you started with?

4. Originally Posted by spectralsea
So in general, if you were trying to prove a function was onto, you would start out with x being a real number and then trying to work it out so that some f(P) would bring you back to the x you started with?
In general yes. But in this case we knew that $\displaystyle f$ is onto and we need to prove that $\displaystyle g$ is also.

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proving that a coordinate function is onto

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