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## #1 2013-11-10 05:08:53

zetafunc.
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### Injectivity Proof

Suppose that f: Z -> Z, with f(x) = 3x[sup]3[/sup] - x. Is f injective?

A function is injective if f(a) = f(b) => a = b.

Suppose that f(a) = f(b). Then:

3a³ - a = 3b³ - b
b - a = 3(b - a)(b² + ab + a²)

So (b - a)[3(b² + ab + a²) - 1] = 0.

Here we have a = b as a solution, which would verify that f is injective over the integers. But I'm not convinced because of that quadratic in the other parentheses -- does that matter? Solving 3(b² + ab + a²) - 1 = 0 definitely doesn't lead to a = b...

## #2 2013-11-10 06:28:37

Nehushtan
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Registered: 2013-03-09
Posts: 897
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### Re: Injectivity Proof

You have to determine whether there exist integers a and b such that
and
. If so, then the function defined is not injective. If not, then it is.

Last edited by Nehushtan (2013-11-10 06:32:24)

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## #3 2013-11-10 07:40:21

zetafunc.
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### Re: Injectivity Proof

I see, that simplifies the problem quite a bit -- thanks!