surjective function

**qwerty31** · May 2nd 2012, 10:26 AM

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

**emakarov** · May 2nd 2012, 10:41 AM

What is the bracket function?

**qwerty31** · May 2nd 2012, 10:42 AM

Originally Posted by emakarov

What is the bracket function?

Also known as the floor function, gauss brackets?

**emakarov** · May 2nd 2012, 10:52 AM

In proofs of surjection, knowing the domain and codomain is essential. For example, f(x) = [x] is not surjective if f is considered from $\mathbb{R}$ to $\mathbb{R}$ , nor when it is considered from $\{x\in \mathbb{R}\mid x>0\}$ to $\mathbb{Z}$ . So I'll assume that $f:\mathbb{R}\to\mathbb{Z}$ .

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

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