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philosophking
Apr12-04, 10:58 PM
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.

philosophking
Apr12-04, 11:12 PM
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.

philosophking
Apr12-04, 11:29 PM
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

Stevo
Apr13-04, 03:40 AM
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.

philosophking
Apr13-04, 12:41 PM
thanks, much appreciated