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The function $f: ℝ → ℝ$ defined by $f(x) = x^{3}$ is onto because for any real number $r$, we have that $\sqrt[3]r$ is a real number and $f(\sqrt[3]r)=r$. Consider the same function defined on the integers $g: ℤ → ℤ$ by $g(n) = n^3.$ Explain why $g$ is not onto $ℤ$ and give one integer that $g$ cannot output.

I can't think of any integer that cannot be cubed, so this problem has me confused.

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The problem is not that we can't cube some integer, its that not every integer has an integer cube root. Consider $5\in\mathbb{Z}.$ To say that $5$ is in the image of $g$ is to say that $5^{1/3}\in \mathbb{Z}.$ As we know, $5^{1/3}\approx1.71\not\in \mathbb{Z}$.

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You don't have to think of an integer that cannot be cubed, but of an integer from which you cannot extract a cubic root (staying in $\mathbb{Z}$, of course, it will have a cubic root in $\mathbb{R}$).

For instance, $g(x)=2$ doesn't have any solution if you ask that $x\in \mathbb{Z}$.

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You've got your definitions wrong. "Not onto" means that there exists an integer that is not a cube.

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