# 1-1, onto, both, or neither

1. Nov 19, 2012

### nicnicman

1. The problem statement, all variables and given/known data

Is the function one-to-one, onto, both, or neither?
f: Z→Z has the rule of f(n) = 4n^3 - 1

2. Relevant equations

3. The attempt at a solution

My answer: one-to-one

Is this correct?

2. Nov 19, 2012

### Zondrina

Well, how would you show it was one to one? How would you show it was onto?

3. Nov 19, 2012

### halo31

I believe you are right that it is injective. It be wise that you show a proof to confirm its 1:1 and a counterexample to show its onto.

4. Nov 19, 2012

### nicnicman

Well, the function is not onto because there is no integer n where 4n^3 - 1 = 1.

5. Nov 19, 2012

### nicnicman

One-to-one proof:

4u^3 - 1 = 4v^3 - 1
4u^3 = 4v^3
u^3 = v ^3
u = v

6. Nov 19, 2012

### Zondrina

Good. Now what about onto?

7. Nov 19, 2012

### nicnicman

Thanks.

The function is not onto because there is no integer n where 4n^3 - 1 = 1.

8. Nov 19, 2012

### Dick

That seems fine to me. Neither of those is really a full scale proof. But I doubt you are expected to provide one. They are both good arguments that what you say is correct.

9. Nov 19, 2012

### nicnicman

Yeah, the book doesn't even ask for a proof, but it's nice to know that the answer is right.

10. Nov 20, 2012

### HallsofIvy

I disagree. What he gave are "full scale proofs". He showed that if f(u)= f(v) then u= v which is a perfectly good proof that f is "one to one". And he gave a counter example showing that it is NOT "onto".

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