I tried to understand the difference between rational numbers and irrational numbers. I understand what is a rational number (a number that can be expressed as the ratio of two numbers p/q).

what makes an irrational number, irrational? how do you prove in a simple way that an irrational number is irrational? why does the fact that the ratio of two numbers that can be divided by 2 is irrational? (am I right?)

I couldn't understand the chosen answer in What's the difference between rationals and irrationals - topologically? so please, bare in mind that my math skills and understanding are currently weak (understatment) and I'm working to improve them.


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    $\begingroup$ In general, there is no simple way to prove that an irrational number is irrational. The famous constant $\gamma$ has been carefully investigated since 1734, is strongly suspected to be irrational, but nobody has a proof yet. $\endgroup$ – MJD Oct 16 '13 at 21:25
  • $\begingroup$ Note that the question you refer to deals with a totally different notion of "difference between rationals and irrationals". That question looks at the whole set $\mathbb Q$ of rationals and the whole set $\mathbb R \setminus \mathbb Q$ of irrationals and wonders how those sets are different from a topological point of view. So, the question is vastly different from your question. $\endgroup$ – Magdiragdag Oct 16 '13 at 21:51

Joke (but true): The difference between a rational number and an irrational number is irrational.

Serious answer: Your question already expressed it. A rational number can be written $\frac mn$ for some integer $m$ and some positive integer $n$. An irrational number is a real number that cannot be written like that.

To show that a number is rational, the most common approach by far is to find $m$ and $n$, and prove that the number in fact equals their ratio.

To show that a number is irrational is often a good deal harder, and is usually done using some sort of proof by contradiction. For example, it took a long time for mathematicians to even prove that $\pi$ is irrational. According to https://mathoverflow.net/questions/40145/irrationality-of-pie-pipi-and-epi2, no one even knows whether $\pi^{\pi^{\pi^\pi}}$ is an integer, let alone whether it is rational (but just about anyone would bet that it's irrational).

It turns out that in several senses, almost all real numbers are irrational, and in fact even transcendental (a nastier sort of beast). There are also various techniques available for manufacturing great gobs of irrational (and even transcendental) numbers, but most of the numbers people are actually interested in are either trivially rational, trivially algebraic (not transcendental), or mysterious—no one knows for certain whether they are rational or irrational.

Part of the reason for this is that while it's very easy to put together rational numbers to get more rational numbers, you can't really put together irrational numbers to get more irrational numbers in very many ways. For example, the sum or product of two rational numbers is always rational, but the sum or product of two irrational numbers may be rational.

  • $\begingroup$ Why are transcendental number supposed to be "nastier" than non-transcendental irrationals? They are possible a little harder to work with, but charming and well-behaved nonetheless. $\endgroup$ – Malvolio Feb 15 '14 at 0:08
  • $\begingroup$ @Malvolio, I don't think I said that. $\endgroup$ – dfeuer Feb 17 '14 at 1:33
  • $\begingroup$ You wrote "even transcendental (a nastier sort of beast)..." $\endgroup$ – Malvolio Feb 17 '14 at 3:17
  • $\begingroup$ @Malvolio, you got me. $\endgroup$ – dfeuer Feb 17 '14 at 3:20
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    $\begingroup$ I think you should apologize to each transcendental number individually. If you are pressed for time, you can just apologize to the transcendental numbers between π/4 and π/2. $\endgroup$ – Malvolio Feb 18 '14 at 6:37

An irrational number is simply a real number that is not rational. In other words, it is a real number that can't be written as the ratio of integers.

A classic example is the number $\pi$; let's try to get some intuition for irrational numbers through this example. You may be familiar with the first few digits in the decimal expansion of $\pi$; \begin{align} \pi = 3.14159\dots \end{align} Notice that if we truncate this decimal expansion at any point, then the resulting number is rational. Let's say, for instance, that we truncate at the tens place, then we obtain the number $3.1$ which can be written as \begin{align} 3.1 = \frac{31}{10} \end{align} Alternatively, if we cutoff at two, three, or four decimal digits, then we get the following rational numbers that are better and better approximations to $\pi$: \begin{align} 3.14 &= \frac{314}{100} \\ 3.141 &= \frac{3141}{1000} \\ 3.1415 &= \frac{31415}{10000} \end{align} We can continue in this way, obtaining better and better rational approximations to $\pi$, but notice that we can never quite get to $\pi$ with only a finite number of decimal digits. We need the full, infinite sequence of decimal digits to get exactly the number $\pi$. In other words, although there are better and better rational approximations to $\pi$, each of which is a ratio of integers, there is no way to write exactly the number $\pi$ as such a ratio.

  • $\begingroup$ this is also an excellent, succinct and simple answer. $\endgroup$ – gideon Apr 8 '15 at 16:49
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    $\begingroup$ @gideon Well thanks, especially since this caused me to look back at the answer and correct an error. $\endgroup$ – joshphysics Apr 8 '15 at 20:20

First, the ratio of two even numbers is still rational, so that's wrong. Secondly, the proof of irrationality depends on the number. Examples are $\pi, e, \sqrt 2, \sqrt 3, \sqrt 7, \ldots$ ($\sqrt p$ is irrational if $p$ is prime)


I posted this answer to a similar question.

The question you cite titled "What's the difference between rationals and irrationals - topologically?" is really on a somewhat different topic: it's not about the difference between one kind of number and another, but rather on the difference between the set of ALL numbers of one sort and the set of ALL numbers of the other sort. You need to know some topology to understand the answers. But the question you pose is simpler.


One way of thinking about it is using decimal expansions, which you are most likely familiar with. Let $a$ be a number, for convenience sake let us consider only numbers between $0$ and $1$, and consider the base 10 expansion $a=0.a_1a_2a_3\dots$ where each $a_i$ is an integer between $0$ and $9$. The decimal expansion a rational number will either terminate (end with infinitely many $0$'s) or end by repeating the same finite string of numbers ad infinitum.

Whereas, an irrational number will not terminate and will not repeat itself infinitely many times.

Creating an irrational number is then easy, for instance, the number $0.10100100010000100000\dots$ is not rational. Going the opposite way, is often much more difficult, but there are many examples of proofs of irrationality that are fairly accessible. You may want to read up on proofs of the irrationality of $\sqrt2$ and $\sqrt3$, as they are both fairly straightforward.


To understand if a given number is irrational depends on how the given number is constructed.
Considering your mathematical skills (as you claim it is not very high) you should check for numbers that do not have repeated decimal.
For example, the number $0.1234567...$ and so on is irrational (the proof of this is not a very easy one)
You are able now to construct an irrational number on your own: choose a number with an aperiodic decimal,for example $1.24681012...$ and there it is:You managed to find an irrational!
One thing you should also know for irrationals is that if you choose a random number from the whole set of real numbers, the probality that it is irrational is 100%.(surprising fact isn't it?)

  • $\begingroup$ Could you clarify what you mean by "$0.1234567\ldots$ and so on"? After a few more digits it will become unclear what the "and so on" means. $\endgroup$ – Trevor Wilson Oct 16 '13 at 23:32
  • $\begingroup$ @Trevor Wilson i mean $0.123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960$ ''and so on'' $\endgroup$ – Konstantinos Gaitanas Oct 17 '13 at 10:02

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