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Why is this statement false?

If $x$ is a rational number, i.e. $\frac{p}{q}$, then shouldn't it be obvious that $\frac{q}{p}$ is also a rational number, by definition of rational numbers?

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    $\begingroup$ The rationals are a field after all. But watch out for zero. $\endgroup$
    – Mr.Fry
    Jun 25, 2014 at 3:12

3 Answers 3

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$x = 0$ is a rational number but $\dfrac{1}{x} = \dfrac{1}{0}$ isn't defined.

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  • $\begingroup$ Would you call 1 a rational number or not? $\endgroup$
    – MonK
    Jul 6, 2014 at 13:23
  • $\begingroup$ @Sid $1=\frac{1}{1}$, so it is rational. $\endgroup$
    – Casteels
    Jul 6, 2014 at 13:54
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Strictly speaking the statement has undefined meaning (because "$1/x$ is rational" has undefined meaning when $x$ can be$~0$) rather than that it is false. To make the statement meaningful, you need to ensure it is not talking about the undefined quantity $1/0$, and depending on how you do that this can make the statement false or true.

Since in mathematics we are more interested in truths than in falsehoods, when we assert a statement, one usually takes it to also implicitly affirm that everything it talks about is well defined. Applying this to your case would transform the statement into

(For any real number $x$,) if $x$ is a rational number then $1/x$ is defined and it is a rational number.

which statement is false (it fails for $x=0$).

However an other way to repair the statement is to recognise from the outset that $x=0$ is going to cause an undefined expression to turn up in the formulation, and therefore exclude it, This leads to

For any nonzero real number $x$, if $x$ is a rational number then $1/x$ is a rational number.

which statement is true (since $\Bbb Q$ is a sub-field of $\Bbb R$).

The example shows one must always be careful in handling statements that contain potentially undefined expressions.

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  • $\begingroup$ How about For any real number x, if x is a rational number and 1/x is defined then 1/x is a rational number. $\endgroup$
    – Cthulhu
    Jun 25, 2014 at 5:35
  • $\begingroup$ The first paragraph is wrong. "1/0 is rational" is not undefined. It's false. You can see this by comparing it with the definition of natural numbers. 1/0 can not be written as a fraction p/q where p is an integer and q is a non-zero integer. $\endgroup$
    – Taemyr
    Jun 25, 2014 at 9:04
  • $\begingroup$ @Taemyr: no. Division of rational numbers is only defined if the denominator is nonzero, so any statement involving division by zero is automatically undefined and thus can't be false. There simply isn't a result of this operation which you could try to write as some fraction. Note that fraction and division are not quite the same thing here. — Of course, you could embed the rationals into a number set where $x/0$ is defined, then prove the result isn't in the rational subset. But this doesn't work in the rationals themselves. $\endgroup$ Jun 25, 2014 at 9:37
  • $\begingroup$ @Taemyr: wait a moment, if you stay in the rationals then any defined result would of course be trivially true. We're actually using $\mathbb{Q}$ embedded in $\mathbb{R}$ all the time here, where $\mathbb{R}$ also doesn't have division by 0. But as in this answer, the embedding should be made explicit. $\endgroup$ Jun 25, 2014 at 9:46
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HINT: What happens in the case that $p=0$?

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