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I don't understand how $f : \mathbb N \to\mathbb N$ (where $0$ isn't included in the natural numbers set), $f(n) = n^2$ is not bijective. It seems both injective and surjective to me?

Thanks got it!

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    $\begingroup$ For which $n$ is $f(n) = 3$? $\endgroup$ Apr 15 '14 at 23:20
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It is surely not a surjection onto $\mathbb N$. For example, there is no n such that f(n) = 2.

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It seems as though your notion of surjective is off. A function $f : X \rightarrow Y $ is surjective if and only if $\forall y \in Y$, $\exists x \in X$ such that $f(x) = y$. That is to say the set being mapped to gets completely "covered" by the map. In your case $\mathbb{N}$ does not get "covered" by the map.

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We can only find $n$ such that $f(n)$ is a square number. Can we find an $n\in\mathbb{N}$ such that $f(n)$ is not a square number? Of course not! What does this imply about surjectivity? What does the definition of a bijection (injection+surjection) now tell us?

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