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What is a formal definition of a irrational number? Usually, we say that it is a number that it is not rational. Is it enough?

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Uncle Google and auntie Wikipedia are your friends. Wikipedia correctly states:

In mathematics, an irrational number is any real number that cannot be expressed as a ratio of integers

In a way, it's not enough to say that any number that is not rational is irrational, because most complex numbers (like $i$) are neither rational nor irrational.

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    $\begingroup$ Wikipedia is wrong (not surprising). Irrational is also used in rings other than $\,\Bbb R,\,$ e.g. $\,i\in\Bbb C\backslash\Bbb Q\,$ is irrational. See here for further discussion. $\endgroup$ – Bill Dubuque Apr 22 '15 at 14:14
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A real number is irrational if is not rational.

But the definition of real number is much less simple.

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It is a number that cannot be written as $\frac{p}{q}$, for $(p,q)\in\mathbb{Z}\times\mathbb{N}^\ast$. Formally: $r$ is irrational iff (i) $r\in\mathbb{R}$; and (ii) $\forall(p,q)\in\mathbb{Z}\times\mathbb{N}^\ast, r\neq \frac{p}{q}$.

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  • $\begingroup$ But $\mathbb{Z}$ and ${N}^\ast$.are infinite. So p and q can be infinite as usual? $\endgroup$ – Erivelton Geraldo Nepomuceno Apr 22 '15 at 13:41
  • $\begingroup$ What do you mean, $p$ and $q$ infinite? Each element itself is finite (as is every real number), even though the set is infinite. (the set $[0,1]$ is infinite, yet any single real number $x\in[0,1]$ is finite). $\endgroup$ – Clement C. Apr 22 '15 at 13:49
  • $\begingroup$ When you say $(p,q)\in\mathbb{Z}\times\mathbb{N}^\ast$, we may admit $p=\infty$ and/or $q=\infty$? $\endgroup$ – Erivelton Geraldo Nepomuceno Apr 22 '15 at 13:57
  • $\begingroup$ No -- $\infty$ is not an element of $\mathbb{N}$ nor $\mathbb{Z}$ (nor $\mathbb{R}$, for that matter). $\endgroup$ – Clement C. Apr 22 '15 at 14:00
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    $\begingroup$ It has infinitely many elements, each of them being an actual value. Again, look at the set $[0,1]$ of all real numbers between $0$ and $1$: none of them is inifnite (they are all between $0$ and $1$, by definition!). Yet, you have infinitely many of them. $\endgroup$ – Clement C. Apr 22 '15 at 14:14
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Yes, irrational = "not rational". It is well-defined in any context where "rational" is well-defined, e.g. in any ring containing the rational numbers $\,\Bbb Q.\,$ For example, as I mentoned in an answer, to the question Is $i$ irrational?, many algebraic number theorists use the terminology for complex numbers $\not\in \Bbb Q.\,$ In particular, be aware that irrational need not imply real (as in the classical case).

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