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Is this function injective or surjective: $g: \mathbb{N} \rightarrow \mathbb{N}, n \mapsto 2n^{3}-1$

I don't know how I can say this. I have to find some values for which we see it cannot be injective / surjective? That sounds too general and cheap, is it really done like that?

If so I would say that this function is surjective because we will always get at least one $n$ value for every $f(n)$ value.

It will also be injective because we will not get more than one value $n$ for every $f(n)$.

Thus the function is bijective..?

berndgr
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    See whether this helps you: http://math.stackexchange.com/questions/557309/proving-functions-are-injective-or-surjective –  Nov 27 '16 at 12:16

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It is obviously injective. It is not surjective because not all integer has the form $2n^3-1$. For example $1$ gives $1$ as image as $2$ gives $15$ and clearly then $2,3,4,\cdots,14$ are not of the form $2n^3-1$ because the function $f(x)=2x^3-1$ is increasing.

Piquito
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Hints:

$$2n^3-1=2m^3-1\iff n^3=m^3\iff (n-m)(n^2+nm+m^2)=0$$

Prove that the rightmost quadratic has no solutions in $\;\Bbb R\;$ and thus that real function is inyective.

Now, suppose

$$w=2n^3-1\implies 2n^3=w+1\implies n^3=\frac{w+1}2$$

The problem is whether the rightmost expression is natural or not...For example, what happens if $\;w=3\;$ ?

Arnaud D.
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DonAntonio
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  • I don't understand why we have to do these steps : / Maybe there is another way, for example using derivatives? Or it will not work for this task? Or if they just ask "is this function injective, say why" you think I'm allowed to draw and reason it with the draw? – berndgr Nov 27 '16 at 12:48
  • You can't use derivatives with discrete variables...but you don't need them, either: the first part is to prove injectivity and the second to disprove surjectivity, and it all is basic algebra... – DonAntonio Nov 27 '16 at 12:53