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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 $m \in \mathbb{Z}$.

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    $\begingroup$ @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'. $\endgroup$ – jwg Aug 18 '17 at 13:08
  • $\begingroup$ @JohnColeman You should distinguish between 'could be defined' and 'commonly is defined'. Defining the rationals as the field of fractions of the integers is extremely common. Defining the complex numbers as the algebraic closure of the reals is extremely uncommon (in analysis), and would require one to prove that such as set can be expressed as a vector space over R. Defining the rationals as a subset of the reals is also extremely uncommon. How do you define the reals? $\endgroup$ – jwg Aug 18 '17 at 15:17
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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:

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

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    $\begingroup$ It may also be important to note that $|2m+1|$ is never equal to $0$ for $m\in\mathbb{Z}$. $\endgroup$ – Barry Cipra Aug 17 '17 at 20:03
  • $\begingroup$ Fair enough. Edited, although if we know that the original is defined at all, that's already taken care of, really. $\endgroup$ – G Tony Jacobs Aug 17 '17 at 20:11
  • $\begingroup$ @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. $\endgroup$ – Hagen von Eitzen Aug 18 '17 at 9:01
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The quotient of two rationals is always a rational number.

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

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  • $\begingroup$ Except when you are dividing by $0$. $\endgroup$ – Kenny Lau Aug 17 '17 at 20:23
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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 $ab \in \mathbb{Q}$

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

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