Solving the One-to-One and Onto Problem of f: R→N

[nicnicman]

To me this problem doesn't seem right. Here it is:

Is the following function one-to-one, onto, both, or neither?
f: R→N f(x) = ceiling 2x/3

My answer: onto

Although, wouldn't this function be invalid since it produces negative numbers and the set of natural numbers doesn't include negatives? Consider f(-1.5) = -1.

Am I misunderstanding a concept?

 

[johnqwertyful]

A lot of people would consider the ceiling function to be f:R->Z.
It would be invalid to say it's f:R->N Unless you restrict R to R⁺

 

[nicnicman]

Well, that's the way is worded in the book, so it must be a typo. Maybe the writers meant to put Z rather than N.

Would my answer be correct if were R to Z?

Thanks for the help.

 

[johnqwertyful]

Agreed.