Prove that the copying f:R->[1,+∞), f(x)=x²+1 is surjection but not injection?
Prove that the copying f:R->[1,+∞), f(x)=x²+1 is surjection but not injection.
How T_T, I got exam in 3 hours lol
- Scarlet ManukaLv 71 decade agoFavourite answer
For y ≥ 1, √(y-1) is in R and f(√(y-1)) = y. (You can derive √(y-1) by setting x^2 + 1 = y.) So for every y ≥ 1 there is an x in R such that f(x) = y, i.e. f is a surjection.
However, for x in R, f(x) = f(-x) and so f is not an injection (e.g. f(1) = f(-1) but 1 ≠ -1). Note that in the first part we could equally well have chosen x = -√(y-1), which is a different number to √(y-1) for all y > 1.
- Curt MonashLv 71 decade ago
It's not an injection because, for example, -1 and 1 are both mapped to the same thing (namely 2).
It's surjection because for any y, we can solve x^2+1 = y. Specifically, set x = the square root of (y-1).