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I have this problem where I understand the meanings of both one to one and onto but proving it with functions I become very confused.

Prove that f(m, n) = 5mn is onto, where the domain of f is R × R and the co-domain of f is R. Show that f is not one-to-one

R being any real numbers.

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    See [Surjective function](https://en.wikipedia.org/wiki/Surjective_function) : "a function f from a set X to a set Y is surjective (or onto), if every element y in Y has a corresponding element x in X such that f(x) = y". – 2017-02-11
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    You have to apply the def, showing that "every element y in R has a corresponding element (m,n) in R × R such that f(m,n) = y". Thus means to show that for every real r we can find reals m, n such that r=5mn. – 2017-02-11
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    For a one line proof: $\;x = f(x/5,1)=f(-x/5,-1)\,$ for $\,\forall x \in \mathbb{R}\,$. First equality proves surjectivity, second one proves non-injectivity. – 2017-02-12

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$f$ is onto if, for any real number $z$, there exists a pair of real numbers $(m,n)$ such that $5mn=z$. Let $z\in\mathbb{R}$ be any real number. Then $z=5\times\frac{1}{5}\times z=f(\frac{1}{5}, z)$, and since $1/5, z$ are real numbers we conclude that $f$ is onto.

$f$ is one-to-one if different elements of $\mathbb{R}\times\mathbb{R}$ have different images in $\mathbb{R}$. This is clearly false; for example, $f(0,1)=0$ and $f(1,0)=0$, and $(0,1)\neq (1,0)$.