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I was given a question saying:

"One can show that the union of two countable sets is countable. Is the set of irrational real numbers countable?"

I don't know what irrational real numbers are. Can someone please give me an example and a definition please?

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  • $\begingroup$ $\sqrt2$ for the example part. All the real numbers which cannot be expressed in the form of $\frac pq$ where $p$ and $q$ are integers. Also, it's decimal expansion is non-terminating and non repeating. $\endgroup$ – Aditya Agarwal Sep 30 '15 at 10:02
  • $\begingroup$ Or you can see it as the real numbers which are not rational. $\endgroup$ – Aditya Agarwal Sep 30 '15 at 10:03
  • $\begingroup$ Rational numbers can be expressed in the form $\frac pq$. $\endgroup$ – Aditya Agarwal Sep 30 '15 at 10:03
  • $\begingroup$ And rational numbers' decimal expansion can be non-terminating repeating (with repeating decimal digits), for example $0.505050...$, or terminating. If the fractional form of a rational number is of the form $\frac{1}{2^m5^n}$, then it terminates other wise it is non-terminating repeating. $\endgroup$ – Aditya Agarwal Sep 30 '15 at 10:05
  • $\begingroup$ I wrote all this in the comments section because I believe that there can be a better answer, with an intuition and a diagram for example, google.co.in/imgres?imgurl=https://mcourses.files.wordpress.com/… $\endgroup$ – Aditya Agarwal Sep 30 '15 at 10:06
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Here is the definition (in terms of sets) of an Irrational number: The set of Real numbers $\mathbb{R}$ minus the set of Rational numbers $\mathbb{Q}$ is the set of Irrational numbers which is written $\mathbb{R} \setminus\mathbb{Q}$.

Less formally, a definition of an Irrational number is a number that cannot be written in the form $\cfrac{p}{q}$ where $p\in \mathbb{Z}$ and $q\in \mathbb{N^{+}}$.

In simple English a Rational number is any fraction such as $\cfrac{3}{4}$, $\cfrac{6}{1}=6$ etc.

The other way to tell if a number is rational is to see if it's decimal digits recur (repeat) such as $\cfrac{1}{9}=0.111111$ and $\cfrac{2}{15}=0.13333333$, also $\cfrac{1}{7}=0.$$\color{blue}{142857}$$142857$$\color{blue}{142857}$$142857$ $\implies$ ($\color{red}{142857}$ recur in this case)

An Irrational number is $\sqrt{3}$ for example, from which it can be seen that its decimal digits do not recur (although no-one of course has "thoroughly" checked this throughout the infinite decimal): $\sqrt{3} = 1.73205080756887729352744634150587236 ....$


A word of caution:

If you are simply told that a number is not rational and nothing else, that does not mean that it is irrational, and vice versa.

However, since $\left(\mathbb{R} \setminus\mathbb{Q}\right) \subset \mathbb{R}$

It is okay to say that if a real number is not irrational then it must be rational, and vice versa.

This is because we are referring to a subset of the real numbers.

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    $\begingroup$ I'm not sure to understand what you mean by “If you are told that a number is not rational, that does not mean that it is irrational”. Also “$\pi$ for example is Irrational since none of its decimal digits recur” doesn't seem the best way to put it. $\endgroup$ – egreg Sep 30 '15 at 14:34
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    $\begingroup$ I think what egreg is suggesting is that the sentence "$\pi$ is irrational because its digits do not recur" seems backwards, since in reality we know that the digits of $\pi$ do not recur because it has been proved irrational. Nobody checked all of the digits of $\pi$ and found that they didn't eventually repeat. Same goes for $\sqrt{3}$. $\endgroup$ – preferred_anon Sep 30 '15 at 14:47
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    $\begingroup$ I'm pretty sure that Blaze is thinking about the fact that a complex number, e.g., $i = \sqrt{-1}$, is neither rational nor irrational; those terms only apply to real numbers. $\endgroup$ – Daniel R. Collins Sep 30 '15 at 14:53
  • $\begingroup$ Thread necromancy: Note that we do identify some numbers as irrational because they are non-repeating—for instance, $1.01001000100001000001\ldots$, where the number of $0$'s between successive $1$'s grows by one each time. But by and large this is not how we show that a number is irrational. $\endgroup$ – Brian Tung Jan 14 '16 at 17:57
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    $\begingroup$ @BLAZE: Thread necromancy in the context of Internet forums simply refers to the practice of reviving discussions after they've been dormant for a long time. There's nothing actually macabre about it. :-) $\endgroup$ – Brian Tung Jan 14 '16 at 23:55
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Irrational numbers are real numbers that cannot be expressed a fraction of two integers. Examples include $\pi$ and $\sqrt{2}$ etc.

See https://en.wikipedia.org/wiki/Irrational_number

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By the way, with respect to the OP question: Since the union of the rational reals $\mathbb{Q}$ and the irrational reals $\mathbb{P} \equiv \mathbb{R} \setminus \mathbb{Q}$ is simply the reals $\mathbb{R}$, and $\mathbb{Q}$ is countable, then if $\mathbb{P}$ were also countable, what would we be able to say about $\mathbb{R}$? And what is actually true about $\mathbb{R}$, and what can we therefore conclude about $\mathbb{P}$?

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