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Is it correct to say that some number $A$ is rational because $A$ is $\frac{A}{1}$?

Because the only restriction I had see for rationality is $A=\frac{a}{b}$ with $b\neq0$, but in that way I can say $\pi$ is rational because $\pi=\frac{\pi}{1}$. What's the problem here?

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    $\begingroup$ $a$ and $b$ must be integers $\endgroup$ – angryavian Aug 16 '16 at 5:37
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    $\begingroup$ The problem is ... you forgot the definition of "rational". $\endgroup$ – zhw. Aug 16 '16 at 6:38
  • $\begingroup$ See 'the restriction for rationality' once again. $\endgroup$ – CiaPan Aug 16 '16 at 14:04
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$a$ and $b$ must be integers. $\pi$ is not

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    $\begingroup$ Next question: "are all numbers integers? The only restriction I've seen for integers is $a = a$, but in that way I can say $\pi$ is an integer ;-) $\endgroup$ – Steve Jessop Aug 16 '16 at 11:00
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    $\begingroup$ @SteveJessop You've been doing too much transcendental meditation. $\endgroup$ – TripeHound Aug 16 '16 at 11:33
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    $\begingroup$ @SteveJessop: as a matter of fact, it can be argued that $a=a$ doesn't quite make sense for reals. For similar reasons you can not always reliably compare floating-point numbers for equality. $\endgroup$ – leftaroundabout Aug 16 '16 at 14:33
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    $\begingroup$ @leftaroundabout, I think that is not quite an accurate representation (on re-reading: no pun intended). The concern in ASD, and computational approaches to arithmetic in general, are about the decideability of arithmetic, not (I think) its sensibility. That is, if I hand you two potentially different numbers $a$ and $b$, then you (or a computer) might have trouble deciding whether they are equal, and so it is a good idea to avoid asking the question; but I think few people would argue that it doesn't make sense to ask whether they are equal. $\endgroup$ – LSpice Aug 16 '16 at 15:50
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The definition of a rational is: $ a \in \mathbb Q \Leftrightarrow \exists (n,m) \in \mathbb Z \times \mathbb N ^ *, a = \frac n m$

And no not all real numbers ($\mathbb R $) are rational. It is easy to show that $ \sqrt 2 $ is not (ref. on Wikipedia)

  • assume that $ \sqrt 2 $ is a rational number, meaning that there exists a pair of integers whose ratio is $ \sqrt 2 $
  • if the two integers have a common factor, it can be eliminated using the Euclidean algorithm
  • then $ \sqrt 2 $ can be written as an irreducible fraction $ \frac a b $ such that $a$ and $b$ are coprime integers (having no common factor).
  • it follows that $ \frac {a^2} {b^2}=2 $ and $ a^2 = 2 b^2 $
  • therefore, $a$ is even because it is equal to $2b^2$ and $b^2$ is integer. Say $a=2k$
  • but then $4k^2 = 2b^2$ so $b^2=2k^2$ - for same reason b is even
  • $a$ and $b$ share 2 as common divisor which is excluded since they a coprime

    Conclusion: $\sqrt 2$ cannot be rational

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  • $\begingroup$ I'm wondering what hope someone who's confused about what a rational number is has of understanding the first sentence. Even though it's accurate. $\endgroup$ – timtfj Feb 5 at 16:46
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$A=\frac A1$ proves rationality of (left-hand side) $A$ iff (right-hand side) $A$ is integer.

Which implies integers are rational.

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  • $\begingroup$ It seems that you probably want just 'if' (not 'iff'), since of course it is not true that $A$ is rational if and only if it is an integer. Or maybe you mean that the argument presented is only a proof of rationality if $A$ is an integer? $\endgroup$ – LSpice Aug 16 '16 at 15:51
  • $\begingroup$ @LSpice Nope. Any number $q$ is rational iff it is a ratio of two integers: $q=\frac mn$ fo some $m,n\in\mathbb Z$. So $A=\frac A1$ fits the definition's condition for integer numerator value $A$ only, and in such case (say, $A=5$) the equality ($5=\frac 51$) proves rationality of $A$ ($5$). OTOH for non-integer $A$ (say $2.71$) the definition's condition is not satisfied, so the equality ($2.71 = \frac{2.71}1$) does not prove the rationality of $A$ (even though $2.71$ actually is rational, because $2.71=\frac{271}{100}$). Hence 'iff'. $\endgroup$ – CiaPan Aug 16 '16 at 21:06
  • $\begingroup$ It seems that we are saying the same thing. 2.71 is rational, although writing it as 2.71/1 doesn't prove that it is so. That is, the parentheses are "($A = A/1$ proves rationality of $A$) iff ($A$ is an integer)", rather than (as I originally read it) "$A = A/1$ proves ((rationality of $A$) iff ($A$ is an integer))". $\endgroup$ – LSpice Aug 16 '16 at 21:08
  • $\begingroup$ @LSpice That's right, apparently I wasn't clear enough. $\endgroup$ – CiaPan Aug 17 '16 at 6:17
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I'm not sure there's a consensus on what the word "fraction" means. I think maybe some people would say $\frac{1}{2}$ and $\frac{\pi}{1}$ are fractions but 0.5 and $\pi$ aren't. That might be because English is flexible and people sometimes say something with the intent of what they say not meaning anything, but rather for the purpose of enabling other people to figure something out from what they said, or something like that. As a result, I will invent my own definition of the word "fraction." According to that definition, the word "fraction" and the phrase "rational number" have different meanings. I think a fraction is any notation of the form $\frac{a}{b}$ where "$a$" and "$b$" are expressions. A rational number on the other hand is a real number such that for some integer $p$ and nonzero integer $q$, $p \div q$ is that number. $\frac{1}{2}$ and 0.5 are both rational numbers because they both equal $1 \div 2$ whereas "$\frac{1}{2}$" is a fraction but "0.5" is not.

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Other answers and comments have already answered this, but I'd like to try to express the key ideas more clearly and in one place.

The main problem is that you've misunderstood the definition of a rational number. The actual definition is

A rational number is one which can be expressed as a fraction $\frac ab$, where $a$ and $b$ are integers and $b≠0$.

Or purely in words,

A rational number is the result of dividing one integer by another, non-zero integer.

So expressing $π$ as $\fracπ1$ doesn't make it rational, since the fraction's numerator isn't an integer. You'd have to find some other fraction just using integers. (It can be proved that there isn't one, but $\frac{355}{113}$ comes surprisingly close.)

However, if A is an integer, then $\frac A1$ is the result of dividing one integer by another, making $A$ rational. So every integer is a rational number, as CiaPan's answer says. That's what your argument really shows.

The main thing this all tells us is that it's important to be sure of the definition of whatever you're dealing with. When it comes to different kinds of number, everyday usage can get in the way: for instance if someone says "Pick a number!" they invariably mean "Pick a positive integer under $100$ !" and would protest if you chose something like $2.987$ or $\sqrt[3]{19}$.

But number doesn't just mean integer, and the definition of a rational number uses integers.

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