I received an F on this assignment and was told that this was not a bijection. The counterexample my professor wrote on my paper was, "$\frac{x}{x^2 + 1} = 1$ has no real roots, so it's not onto the natural numbers."
I'm either incorrect, or he misread how my function works, or I need to restate something in a better way.
When you say "suppose $x=y$ for some $x,y \in B$ you assume that forces $x=y$ but give no reasoning. You should suppose that $x \neq y$ for $x,y \in B$ and show that they are mapped to different naturals. Alternately, you could prove that different naturals are mapped to different members of $B$. Again, you assume the wrong thing: you should assume $i \neq j$ and show they are mapped to different members of $B$.
In fact, you don't need a bijection. You either need that $f^{-1}$ is injective or that $f$ is surjective. Either one shows that $B$ is infinite.
I would define my function the other way: let $h: \Bbb N \to B$ be defined by $h(i)=\frac i{i^2+1}$ Then show $h$ is decreasing, which proves it is injective.
It looks to me like your prof misunderstood (apparently thinking that $f$ was supposed to map $\mathbb N$ to $\mathbb N$ rather than $B$ to $\mathbb N$). But also you didn't really show that $f$ is well-defined. For that you need to show that if $i/(i^2+1) = j/(j^2+1)$ then $i = j$: you did the converse, which is trivial.
Why isn't what I have written a bijection?
Because what you have written is not a function, to begin with. More precisely, because there is no guarantee that the string of characters you wrote defines anything.
Recall how a function is often defined. One fixes some sets $X$ and $Y$ and one decides that, for each $x$ in $X$, $f(x)=[\text{some complicated formula involving}\ x]$. Then $f$ is well defined, provided that, for each $x$ in $X$, $[\text{some complicated formula involving}\ x]$ is indeed an element of $Y$.
But one never decides that $f:[\text{some complicated formula involving}\ x]\mapsto x$. Why? First thing is that it could happen that $z=[\text{some complicated formula involving}\ x]$ and also $z=[\text{some complicated formula involving}\ x']$ for some $z$ in $X$ and some $x$ and $x'$ in $Y$ with $x\ne x'$. Then what would be $f(z)$? One should have $f(z)=x$ and $f(z)=x'$. But $x\ne x'$, hence this is impossible.
Example: The function $h:[0,2\pi]\to[-1,1]$, $x\mapsto\cos x$, is well defined but the object $f:[-1,1]\to[0,2\pi]$, $\cos x\mapsto x$, does not exist. For example, $\cos(\pi/2)=\cos(3\pi/2)=0$, hence one should choose at the same time $f(0)=\pi/2$ and $f(0)=3\pi/2$.
In your case, to assert that there exists something like $f:B\to\mathbb N$, $n/(n^2+1)\mapsto n$, is to assume from the start that indeed $n/(n^2+1)\ne m/(m^2+1)$ for every $n\ne m$ in $\mathbb N$, that is, roughly the result you are asked to prove. (Thus, the grade is correct but the explanation for it is not.)
