I need a little confirmation on the following proof:
The problem is this:
For an interval I, with f:I -> R where f is strictly increasing with f(I) an interval, prove that f^(-1):R -> I is strictly increasing.
My proof is this:
Because f is strictly increasing, for y,z exists in I such that y<z,
Let y,z exist in I. Then f(y),f(z) exist in R. and f(y) < f(z).
but f^(-1)[f(y)]<f^(-1)[f(z)] which implies y<z and thus f^(-1) is strictly increasing.
This proof seems trivial to me, which makes me think I did something wrong. I possibly assumed too much in my proof, not sure. Any ideas?
Oct 31st 2009, 10:55 PM
I suffer from similar habbits. Always assuming things are complicated. Your proof is correct in its simplicity. You are absolutely right to assume the existence of , its given in the problem statement. Hence you may use the identity function , by the definition of the inverse.
In fact just to reassure you that your proof makes sense, remember that f is the inverse of the function . And a function only has an inverse if it is strictly increasing or decreasing (a standard theorem from set theory). You have of course proved is strictly increasing. And thus the inverse of exists. namely, f.