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Solution Verification, below is how I worked it out. I'm just wondering if I'm doing this right.

A function $f:R→R$ is monotonically increasing if $ y > x $ implies $f(y)> f(x)$ for all $x$ and $y$ in $R$.

Since $f(y)$ can never be less than $f(x)$ and $f(x)$ can never be larger than $f(y)$ we can assume that each input map to one and only one unique output, that is to say $f$ is injective.

In order to be surjective, there must exist an output that has no input. Or there must be a value that cannot be achieved given the function. Consider the function $f(x) = 2^x$. The range of this function is ($0$, infinity). Since the value of the function can never take on anything equal to or less than zero, this function is not considered surjective over $R$. Therefore, $f$, a monotonically increasing function, need not be surjective.

tjblue
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Your counter example is fine, except that you're missing a logical negation in your opening clause.

Your proof feels a little garbled. You seem to assume $x<y$ but you never explicitly state that. I would start with something like "Assume $f: R \to R$ is monotonically increasing and $x, y \in R$. Either $x < y$ or $y < x$ but these are essentially the same, so, without loss of generality, assume $x<y$". Then, use monotonicity to imply injectivity, and state clearly what you've proved, before saying that you're done.

user24142
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