Question

Show that the function is onto...

Answer 1

I need help with iii

Answer 2

@rational @FibonacciChick666 @jtvatsim

Answer 3

Okay, do you know the definition of an onto function ?

Answer 4

For any real number \(x\) you want to show that \(\left\lfloor\dfrac{x+1}{2}\right\rfloor=n\) for any integer \(n\). Using the fact that \(\lfloor n\rfloor=n\) for any integer, this amounts to showing that \(\dfrac{x+1}{2}=n\). What value of \(x\) will work?

Answer 5

So angus has the right idea, you need to know the definition of onto before you can do anything. Once you have the definition, proving it should be relatively simple as the floor function maps the reals to the integers.

Answer 6

well, yeah..I know what an onto function is,

an onto function has an ancestor for every element in the co-domain

and I know that this is an onto function

I guess I just don't know how to justify it in the answer...to prove it wrong I would only need a counter example...but to show that it holds true...

Answer 7

For proving the function is onto you need to show \[\forall n\in \mathbb{Z}, ~\exists x\in\mathbb{R} : \left\lfloor\dfrac{x+1}{2}\right\rfloor=n\]

Answer 8

Let \(x=2n\) and conclude the proof

Answer 9

Also recall this floor function property : \(\lfloor a+x\rfloor = a+\lfloor x\rfloor \) for any integer \(a\).

Answer 10

but how does that show it is an onto function?

Answer 11

you're showing for every \(n\) in co-domain, there exists an ancestor \(2n\)

Answer 12

because \( \left\lfloor\dfrac{2n+1}{2}\right\rfloor= \left\lfloor n+\dfrac{1}{2}\right\rfloor=n+ \left\lfloor\dfrac{1}{2}\right\rfloor=n+0=n\)

Answer 13

ooooh thanks, I get it now, it makes so much sense...

Answer 14

np :)