1. Apr 12, 2004

### philosophking

Please help me. I'm trying to find functions where f:N-->N (the set of natural numbers to the set of natural numbers), such that:

f is surjective but not injective,
f is neither surjective nor injective

I'm really not sure how to determine these. Thanks for your consideration.

2. Apr 12, 2004

### philosophking

errr... now that I look back at my other answers ( i had to find one that is bijective and one that is not surjective but injective), i don't even know if those are right.

For bijective, could you have f(n) = n ?
For injective but not surjective, could you have f(n) = 2n + 1 ?

I'm so confused.

3. Apr 12, 2004

### philosophking

OK hehe i think i figured some stuff out, for my first original question, f(n) = (n-5) + 2 works. But I still need help on my last question! please!

Also, the most recent two questions i asked can be disregarded... haha wow sorry if i confused anyone

4. Apr 13, 2004

### Stevo

f: N -> N

If f(n) = n, f is bijective.
If f(n) = 2n, f is injective but not surjective (2n+1 also works).
If f(n) = floor(n/2), f is surjective but not injective.
If f(n) = constant, f is neither injective nor surjective.

I think these are all pretty well-known examples.

5. Apr 13, 2004

### philosophking

thanks, much appreciated