# 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 $m \in \mathbb{Z}$.

• The rationals are closed under multiplication, division, addition, and subtraction by definition. – The Great Duck Aug 18 '17 at 3:56
• @JohnColeman A common definition of $\mathbb{Q}$ is as the field of fractions of $\mathbb{Z}$. Depending on how you define $\mathbb{R}$, it might not make sense to define $\mathbb{Q}$ using the concept 'real number'. – jwg Aug 18 '17 at 13:08
• @JohnColeman jwg's point is exactly what I'm saying. You don't have to necessarily say it is all numbers expressed as that ratio. It can also be the "rational set" or "field of fractions" which for all algebraic number rings is already defined to be a field. Regardless, my point was to say the question was trivial to the point that any answer will just say "by definition". I suspect that the asker might want something deeper than just the properties of rational numbers. – The Great Duck Aug 18 '17 at 14:58
• @JohnColeman Think of it this way: Define the minimal extension upon the integers such that the resulting set is a field. That's the rational numbers. Now you might say that we have to prove the two definitions are equivalent but we don't as some people will define the rational numbers by my above statement. Is it wrong? Probably...? But does it actually matter that much? After all, being closed under those things does follow directly from its definition either way. – The Great Duck Aug 18 '17 at 15:00
• @JohnColeman in all proofs there is a concept of "previous knowledge" or "context". If the author wants them to demonstrate the rationals are closed, that's one thing. However, most will just say they are as it's pretty well accepted at this point. I just had such a course this last spring and probably had a similar question (I don't know for but I'm just guessing) and never were we asked to actually show the rationals were closed. It might be brought up in-class if people didn't know that, but it was considered self-evident. – The Great Duck Aug 18 '17 at 15:04

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:

$$\frac{a/b}{c/d} = \frac{a/b}{c/d}\cdot\frac{bd}{bd} = \frac{ad}{bc}.$$

• It may also be important to note that $|2m+1|$ is never equal to $0$ for $m\in\mathbb{Z}$. – Barry Cipra Aug 17 '17 at 20:03
• Fair enough. Edited, although if we know that the original is defined at all, that's already taken care of, really. – G Tony Jacobs Aug 17 '17 at 20:11
• @GTonyJacobs I do appreciate showing that the denominator is non-zero. If the question were "For which $m\in\Bbb Z$ is $\frac{m^2+1}{|m+1|}$ a rational number?" you'd definitely have to do the check. – Hagen von Eitzen Aug 18 '17 at 9:01

The quotient of two rationals is always a rational number.

• yeah I know it bur how can I prove that? – Eii Aug 17 '17 at 20:00
• Let $p=a/b$ and $q=c/d$. Simply compute $p/q$ and use the definition of a rational number. – szw1710 Aug 17 '17 at 20:14
• @MichaelHardy yeah but also c must not be 0 right? – Eii Aug 17 '17 at 20:49
• @MichaelHardy That's not exactly what arithmetic would tell us. :p – Adayah Aug 17 '17 at 20:55
• @eli : Yes. Actually $a$ is the only one that could be $0. \qquad$ – Michael Hardy Aug 17 '17 at 23:52

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.

• Except when you are dividing by $0$. – Kenny Lau Aug 17 '17 at 20:23

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 $ab \in \mathbb{Q}$

• Except when $k=0$. – Kenny Lau Aug 17 '17 at 20:23
• Yes indeed thank you for point it out... – Marios Gretsas Aug 17 '17 at 20:26