# surjective function

• May 2nd 2012, 10:26 AM
qwerty31
surjective function
How would I prove that f(x) = [x] is surjective, where [x] denotes the bracket function
I know how to prove that something is surjective, but I'm stuck on this particular example
• May 2nd 2012, 10:41 AM
emakarov
Re: surjective function
What is the bracket function?
• May 2nd 2012, 10:42 AM
qwerty31
Re: surjective function
Quote:

Originally Posted by emakarov
What is the bracket function?

Also known as the floor function, gauss brackets?
• May 2nd 2012, 10:52 AM
emakarov
Re: surjective function
In proofs of surjection, knowing the domain and codomain is essential. For example, f(x) = [x] is not surjective if f is considered from $\displaystyle \mathbb{R}$ to $\displaystyle \mathbb{R}$, nor when it is considered from $\displaystyle \{x\in \mathbb{R}\mid x>0\}$ to $\displaystyle \mathbb{Z}$. So I'll assume that $\displaystyle f:\mathbb{R}\to\mathbb{Z}$.

We need to prove that for every $\displaystyle n\in\mathbb{Z}$ there is an $\displaystyle x\in\mathbb{R}$ such that [x] = n. Can you find several such x for n = 3?