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  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -

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Topic review (newest first)

zetafunc.
2013-11-11 06:40:21

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

Nehushtan
2013-11-11 05:28:37


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.

zetafunc.
2013-11-11 04:08:53

Suppose that f: Z -> Z, with f(x) = 3x3 - x. Is f injective?

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

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

3a - a = 3b - b
b - a = 3(b - a)(b + ab + a)

So (b - a)[3(b + ab + a) - 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(b + ab + a) - 1 = 0 definitely doesn't lead to a = b...

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