Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

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

zetafunc.
Guest

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
Member
From: London
Registered: 2013-03-09
Posts: 615
Website

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)


158 books currently added on Goodreads

Offline

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

zetafunc.
Guest

Re: Injectivity Proof

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

Board footer

Powered by FluxBB