Recent content by allinurl

oh~ok. I got that now. If so, in this case, ⌊x₁⌋ ≠ ⌊x₂⌋ but h(x₁) = h(x₂). I can say it is not one-to-one function. Correct? I also have to say its range, how can I put? And also, I have a problem to define A ∪ [0,1]. I know [0,1] means the set [xɛR | 0 ≤x ≤1]. But then, I'm not sure what...

no, ⌊1.2⌋ is not equal to ⌊1.1⌋, but if so it means ⌊x₁⌋ is not equal to ⌊x₂⌋. In definition, the function if injective if a≠ b, then f(a)≠ f(b). So, is it one-to-one function?

I would say that there is only one which is x=1. If I always plug the same number to the x1 and x2, the answer seems to be always same. for example, if h(x₁) = h(x₂), then ⌊x₁⌋ = ⌊x₂⌋, so x₁ = x₂ This is why I'm confused. Am I correct?

Homework Statement f : R+ → R+ defined by f(x) = 2x h : R+ → N defined by h(x) = ⌊x⌋ (the largest integer ≤ x) g : 2ᴿ → 2ᴿ defined by g(A) = A ∪ [0,1] Homework Equations The Attempt at a Solution I know that every element of the domain has to correspond to exactly one element on...