Thread: one to one correspondence

Originally Posted by Madspeter

How can I determine wether this function f(x) = x^2 is a one-to-one correspondence. For Z -> Z that is.

Consider . Are they equal?
So is a bijection?

both -1 and 1 give the same value. So my guess would be that it's not a bijection.

sounds more like it's surjective?

Originally Posted by Madspeter

both -1 and 1 give the same value. So my guess would be that it's not a bijection.

sounds more like it's surjective?

OH? .

Originally Posted by Plato

OH? .

madspeter, after you answer this question, ask yourself this.

If the original question were from R -> R, would it be injective, surjective, or neither?

neither.

right?

Originally Posted by Madspeter

neither.

right?

yup.

what if we change it to [0, ) -> [0, )?

I suppose that could solve F(?) = 2 with an ugly number like 1.4xxxx? Not really sure.

Originally Posted by Madspeter

I suppose that could solve F(?) = 2 with an ugly number like 1.4xxxx? Not really sure.

The function maps integers to integers.
1.4xxxx is not an integer.

even with [0, ) -> [0, )?

excuse me for not knowing basics, I'm trying.

Originally Posted by Madspeter

I suppose that could solve F(?) = 2 with an ugly number like 1.4xxxx? Not really sure.

if it were from R -> R, but in your question its from Z -> Z.

Originally Posted by Madspeter

even with [0, ) -> [0, )?

excuse me for not knowing basics, I'm trying.

with that, x^2 would be bijective. because we restricted the domain and range.

allright. Thanks to both of you.

now if you have some ekstra spare time.
I have the answers lying around for these questions..

F(x) = -x is bijective
F(x) = 2x is not bijective

however, I fail to see why 2x is not bijective.

Originally Posted by Madspeter

allright. Thanks to both of you.

now if you have some ekstra spare time.
I have the answers lying around for these questions..

F(x) = -x is bijective
F(x) = 2x is not bijective

however, I fail to see why 2x is not bijective.

from what to what?

R -> R?
Z -> Z?
N -> N?

Z -> z