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If we try to divide any two random arbitrarily long rational numbers like

103850.2387209375029375092730958297836958623986868349693868398659825528365...

and

127.123123123...

Is it guaranteed that the result is also a rational number?

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  • $\begingroup$ But if we divide arbitrary amount of rational numbers then it's possible to get an irrational one :) But of course it's not what you are asking. $\endgroup$
    – Mihail
    Aug 5, 2015 at 16:09

2 Answers 2

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The quotient of two rationals is always a rational.

For if $\alpha = \frac{a}{b}$ and $\beta = \frac{c}{d}$ with $a, b, c, d$ integers with none of $b, c, d$ being zero, then

$$\frac{\alpha}{\beta} = \frac{ad}{bc}$$

is a quotient of integers, and so is rational.

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  • $\begingroup$ There is a mistake: the last equation should read $\frac{\alpha}{\beta} = \frac{ad}{bc}$. $\endgroup$
    – jub0bs
    Apr 8, 2014 at 16:17
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If we get an irrational number by dividing a rational number by another rational number, then product of rationals won't be a binary operation in $\mathbb{Q}$. But we know that $\mathbb{Q}\setminus \{0\}$ is a group with respect to multiplication.

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