How can I prove that $\frac{m^2+1}{\left|m+\frac12\right|}$ is a rational number for all $m \in \mathbb{Z}$? How can I prove that
$$\frac{m^2+1}{\left|m+\frac12\right|}$$
is a rational number for all $m \in \mathbb{Z}$?
I know that the numerator is rational number  but the denominator is always a rational number for any $m \in \mathbb{Z}$.
 A: You can either prove the rationals are closed under the four arithmetic operations, or in your specific case you can just demonstrate it.  If $m \gt 0$ your fraction is $\frac {m^2+1}{m+\frac 12}=\frac {2m^2+2}{2m+1}$ and we have displayed two integers you can divide to get your number.  The denominator is never zero because of the $\frac 12$.  The case $m \lt 0$ is similar.
A: Use the fact that $\mathbb{Q}$ is a field thus if $k  \in \mathbb{Q},k  \neq 0$ then $k^{-1}=\frac{1}{k} \in \mathbb{Q}$
and that if $a,b \in \mathbb{Q}$ then $ab \in \mathbb{Q}$
A: If you mutiply the top and bottom of your fraction by $2$, you get:
$$\frac{m^2+1}{|m+\frac12|} = \frac{2(m^2+1)}{|2m+1|},$$
which is a ratio of two integers, and we don't have to worry about the denominator being $0$, because that's not possible for an integer $m$. In general, to show that the quotient of two rationals is rational, you just need to clear denominators:
$$\frac{a/b}{c/d} = \frac{a/b}{c/d}\cdot\frac{bd}{bd} = \frac{ad}{bc}.$$
A: The quotient of two rationals is always a rational number.
