# Thread: one-to-one and/or onto?

1. ## one-to-one and/or onto?

Define h: Z -> Z by h(x) = 7|x| – 6

a. Is h one-to-one? Explain.

b. Is h onto? Explain.

I'm having a hard time finding an explanation for this one. I think it is one-to-one, but not onto. Is this correct? An explanation would be appreciated

2. Is it true that $h(1) = h( - 1)$? One-to-one??

Is is possible that $h(x)=-1$? Onto??

3. Hi, Plato. Thanks for the response. I guess I'm just confused by this.

Originally Posted by Plato
Is it true that $h(1) = h( - 1)$? One-to-one??
Yes, I think so...but I really can't explain why

Originally Posted by Plato
Is is possible that $h(x)=-1$? Onto??
I don't believe so...???

I guess if someone can help me think through this to determine why or why not.

4. Ok, I've been trying to think my way through this. This is what I think so far.

it is one-to-one, but not onto since x and –x map to same element |x|

Am I on the right track?

5. Originally Posted by relyt
Ok, I've been trying to think my way through this. This is what I think so far.

it is one-to-one, but not onto since x and –x map to same element |x|

Am I on the right track?
Hi relyt,
what is the definition of one-to-one and onto?

6. A function f is one-to-one if and only if whenever f(x) = f(y), x = y

A function f (from set A to B) is onto if and only for every y in B, there is at least one x in A such that f(x) = y

Ok, so

It is not one to one since $h(-4) = h(4)$

and I don't believe it would be onto then either

7. Originally Posted by relyt
Define h: Z -> Z by h(x) = 7|x| – 6

a. Is h one-to-one? Explain.

b. Is h onto? Explain.

I'm having a hard time finding an explanation for this one. I think it is one-to-one, but not onto. Is this correct? An explanation would be appreciated
not 1-1
$h(2)=h(-2)$

not onto
$\not \exists c \in \mathbb{Z}$ s.t. $h(c)=-50$