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Given a function f, where for all $x\in \mathbb{N}$ we have $f(k+1) > f(k)$. Should i use induction to prove that for any $x_1 > x_2$, $f(x_1) > f(x_2)$ will be verified?

Additionally, how can we prove that f is injective, surjective? Thanks a lot for any hints.

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    What is the domain and range of your function $f$? It makes a big difference to the question. Also a side point: "monotonously increasing" means increasing in a boring way. I think you mean "monotonically increasing" – 2017-01-30

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Induction is suitable, yes. That said, unless you're really going basic, it is rather obvious.

If you prove that $f$ is increasing, injectivity follows automatically.

$f$ definitely need not be surjective. For instance, take $f(n)=2n$.

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    I feel it's safe to assume $f:\mathbb{N}\longrightarrow\mathbb{N}$ given the title. – 2017-01-30
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    Ah I didn't see that. Also is it really wise to describe it as "rather obvious". Any explanation will contain a hidden induction argument, and for a mathematician getting to grips with formal proof, it is essential to realise this. – 2017-01-30
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    Hi, yes that is the correct assumption here @fimpellizieri. One question: sorry if it is very naive, but how does f=2n means f doesnt have to be surjective? – 2017-01-30
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    Surjectivity always depends on the codomain. If your codomain is $\mathbb{N}$, then $f$ need not be surjective. In my example, the image of $f$ is the set of even naturals which is *strictly* contained in $\mathbb{N}$. In other words, $f$ is not surjective because its image does not contain all of $\mathbb{N}$ (namely, it does not contain the odd naturals). – 2017-01-30
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    Beautiful argument. – 2017-01-30