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This question already has an answer here:

We say,

A number is rational if it can be represented as $\frac{p}{q}$ with $p,q \in \mathbb Z$ and $q\neq 0$.

Any number which doesn't fulfill the above conditions is irrational.

What about zero?

It can be represented as a ratio of two integers as well as ratio of itself and an irrational number such that zero is not dividend in any case.

People say that $0$ is rational because it is an integer. Which I find to be a lame reason. May be any strong reason is there. Can any one tell me please?

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marked as duplicate by Jack M, Chappers, N. F. Taussig, Dietrich Burde, Joel Reyes Noche May 19 '15 at 12:21

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    $\begingroup$ $0$ can be represented as $0/1$, therefore by your definition it is rational. $\endgroup$ – TonyK May 19 '15 at 10:30
  • $\begingroup$ And it can also be shown as $\frac{0}{\sqrt{2}}$. Then what? $\endgroup$ – Man_Of_Wisdom May 19 '15 at 10:30
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    $\begingroup$ $1=\frac{\sqrt{2}}{\sqrt{2}}$. So? $\endgroup$ – Hayden May 19 '15 at 10:31
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    $\begingroup$ Or less trivially, $2=\frac {\sqrt 8} {\sqrt 2}$ $\endgroup$ – Jack M May 19 '15 at 10:40
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    $\begingroup$ You may simply notice that $q=\sqrt(2) \not\in \mathbb{Z}$ $\endgroup$ – Manlio May 19 '15 at 10:42
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The definition of an irrational number is that it is not rational. And $0$ is by definition a rational number.

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    $\begingroup$ Why should I consider it a rational number if it can be represented as $\frac{0}{\sqrt{2}}$ $\endgroup$ – Man_Of_Wisdom May 19 '15 at 10:27
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    $\begingroup$ @Man_Of_Wisdom: What about $\sqrt2/\sqrt2$ then? Do you consider that irrational? $\endgroup$ – John Bentin May 19 '15 at 10:33
  • $\begingroup$ Nope. No. Not.. $\endgroup$ – Man_Of_Wisdom May 19 '15 at 10:45
  • $\begingroup$ @Alwin can you please roll back your edit to the fresh one. Which includes only two lines? I want to accept that because it answers at hand! $\endgroup$ – Man_Of_Wisdom May 19 '15 at 11:10
  • $\begingroup$ Do you mean this one above? $\endgroup$ – Alwin May 19 '15 at 11:25
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I think you're confusing some of the language in the definition. The definition says that if a number may be written in a certain way, namely as a fraction in which both numerator and denominator are integers, then it's rational. If it cannot be written this way, then it is irrational. There is nothing in the definition that prevents a rational number from being written as a fraction in other ways, such as having rational or irrational numerator or denominator.

The phrase "Any number which doesn't fulfill the above conditions is irrational" does NOT say "Any number which can be written as a fraction $\frac{p}{q}$ with $p,q\notin \mathbb{Z}$ is irrational". It simply says any number that can not be written $\frac{p}{q}$ with $p,q\in \mathbb{Z}$ is irrational.

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  • $\begingroup$ One could also use the irrationality critereon to show zero can't be an irrational number. $\endgroup$ – Rammus May 19 '15 at 11:49
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$r$ is rational if you find integers $p,q$ such that $r=\dfrac pq$. This is obviously the case for $r=0$.

The contraposition of this property is "$r$ is irrational if you cannot find integers $p,q$ such that $r=\dfrac pq$".

The contraposition is not at all "$r$ is irrational if you find irrationals $p,q$ such that $r=\dfrac pq$". (By the way, this would be a somewhat circular definition.)

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$\newcommand{\Reals}{\mathbf{R}}$You've got excellent explanations of the logical reasons for saying "$0$ is rational". Here are some complementary thoughts too long for a comment:

Definitions in mathematics exist to give convenient labels to useful logical distinctions. "Convenient" is a loose term, but generally refers to simplifying statements of theorems and facilitating common types of discussion.

When two criteria (such as "rational" and "irrational") are logical opposites by definition, it's never a good idea to allow some widespread mathematical concept (such as $0$) to be "both": If you do, every theorem that would apply to that object has to contain a clause explicitly excluding that object. That's inconvenient. In rare cases (see below), you might say "neither". But in the case of $0$, "rational" is the better label:

  • Literal application of the definition ("there exist integers $q \neq 0$ and $p$ such that $0 = p/q$") says $0$ is rational.

  • The set of rational numbers has pleasant algebraic properties (closed under addition, closed under multiplication) because $0$ is rational. (By contrast, the set of irrational real numbers is not closed under addition, e.g., $(1 - \sqrt{2}) + \sqrt{2} = 1$, or under multiplication, e.g., $\sqrt{2} \cdot \sqrt{2} = 2$, whether or not $0$ is rational.)

To make a case that "$0$ is not rational", i.e., that the definition of "rational number" should exclude $0$, one would want a compelling reason, such as "the statement of a useful theorem becomes awkward if $0$ is (regarded as) rational".

With due respect, the possibility of writing $0$ as $0/\sqrt{2}$ isn't compelling in the above sense; as others have explained, this representation does not contravene the definition. Further, it's useful, and causes no hardship, to agree that $0$ is rational.


For contrast, here are some other "edge cases" that crop up now and again:

  • The integer $0$ is "even" ($2$ times some integer) rather than "odd" (leaves a remainder of $1$ on division by $2$). (By the division algorithm, every integer is even or odd, but no integer is both. In this setting, "even" and "not odd" are logically equivalent for integers. I mention this example because a colleague once informed me that some teachers regard $0$ as neither even nor odd.(!!))

  • The zero function $z: \Reals \to \Reals$ is both "even" (for all real $x$, $z(-x) = z(x)$) and "odd" (for all real $x$, $z(-x) = -z(x)$). The notions of "even" and "odd" for functions are not logical opposites. Moreover, it is useful to declare the zero function to be both even and odd: The set of even functions is a vector space under "the usual operations"; the set of odd functions is, too. If $z$ were not "both even and odd (as a function)", at least one of these useful theorems would be false.

  • The integer $1$ is neither "prime" nor "composite". (Even though "$1$ has no positive integer factors other than $1$ and itself", we explicitly exclude $1$ from membership in the primes because declaring $1$ "prime" would spoil the uniqueness of prime factorization. On the other hand, $1$ is not "composite" because $1$ is not a product of smaller positive primes.)


There's a deeper point that, ironically, may seem at odds with my earlier stance: Mathematical definitions are human conventions, not absolute, immutable, incontestable features of logic, mathematics, or the physical universe. I suspect this raises unnecessary obstacles for the philosophically-minded who study mathematics. (Everything and More by David Foster Wallace is the most extreme example I've encountered; Wallace seemed tormented by the ontology of infinity.)

On the other hand, when one sees how tightly mathematics hangs together across times and cultures, how definitions lead to the same theorems, one is forced (even fully accepting the preceding paragraph) to admire the phenomenal coherence of the logical structure of mathematics. One starts to feel as if definitions are inevitable. One becomes willing to fight emphatically for the correct definitions. This last, I expect, explains the downvotes to your good question.

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As I understand it, the definition of an irrational number is that it is not rational. By definition then, $0$ is rational.

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