Determining whether floor((n+1)/2) is 1-to-1 or onto

[leo255]

Homework Statement

[/B]
floor((n+1)/2)

Find whether this function is 1-to-1 and/or onto from Z to Z.

Homework Equations

The Attempt at a Solution

This is not one-to-one because f(1) = f(2) = 1.

Regarding onto, we need to show that f(a) = b

floor((n+1)/2) = b
2b = n + 1
n = 2b - 1

f(2b-1) = floor((2b-1) + 1)/2) =

floor(2b/2) = floor(b) = b.

I'm not sure how to take the floor function into account for my onto calculations.

 

[Samy_A]

Let f be the function defined by f(n)=floor((n+1)/2) (n∈ℤ).

I may be confused, but haven't you actually proved that for b∈ℤ, f(2b-1)=b?
That proves that f in onto.

 

[Mark44]

Homework Statement

[/B]
floor((n+1)/2)

Find whether this function is 1-to-1 and/or onto from Z to Z.

Homework Equations

The Attempt at a Solution

This is not one-to-one because f(1) = f(2) = 1.

Regarding onto, we need to show that f(a) = b

In more detail, you need to show that for each $b \in Z$, there exists an $a \in Z$ such that f(a) = b.

floor((n+1)/2) = b
2b = n + 1
n = 2b - 1

f(2b-1) = floor((2b-1) + 1)/2) =

floor(2b/2) = floor(b) = b.

I'm not sure how to take the floor function into account for my onto calculations.