$\pi$ can be represented as $C/D$, and $C/D$ is a fraction, and the definition of an irrational number is that it cannot be represented as a fraction.

Then why is $\pi$ an irrational number?

  • 21
    $\begingroup$ Because $C$ and $D$ cannot both be integers at the same time. $\endgroup$ – Per Manne Aug 20 '12 at 16:09
  • 6
    $\begingroup$ @Sam, following you idea, every real number would be a rational number, since $x=x/1$ is always true. $\endgroup$ – Sigur Aug 20 '12 at 16:20
  • 20
    $\begingroup$ I don't see why this is so heavily downvoted; this is a genuine misconception/confusion that I have seen expressed sincerely by some students. $\endgroup$ – ShreevatsaR Aug 20 '12 at 16:23
  • 6
    $\begingroup$ I agree with ShreevatsaR! Of course it seems silly and has an easy answer, but confusions of this species are common and shouldn't be snubbed this way. $\pi$ is by definition a ratio. Many teachers undoubtedly say "rational means RATIOnal!" and do not emphasize the "integer" part. Teachers often overestimate the ability of students to notice distinctions, and react in a way which is detrimental to the students' education. $\endgroup$ – rschwieb Aug 20 '12 at 17:01
  • 2
    $\begingroup$ Sam, I hope you don't think $\pi$ is exactly equal to 22/7. $\endgroup$ – Gerry Myerson Aug 20 '12 at 23:51

A rational number is a number that can be expressed as $p/q$, where $p$ and $q$ are integers. The number $\pi$ cannot be expressed in this form; hence it is irrational.

In other words, the definition of "fraction" does not include ratios like "circumference/diameter" in which the numerator and denominator are arbitrary numbers, not necessarily integers. In the case of "circumference/diameter" (which you denoted $\pi = C/D$), it will always be the case that if the diameter is an integer, the circumference ($C = \pi D$) is not an integer, and if the circumference is an integer, the diameter ($D = C/\pi$) is not an integer: precisely because $\pi$ is irrational.

Note that a definition of "fraction" that allowed arbitrary real numbers as the numerator and denominator would be not very useful, as it would allow "fractions" like $\pi = \pi / 1$, or indeed, for any number $x$, the representation of $x$ as a "fraction" $x = x/1$.

  • 14
    $\begingroup$ Thank you. I'm twelve and this question has been bugging me since school ended in June. $\endgroup$ – Sam Aug 21 '12 at 4:46
  • 1
    $\begingroup$ @ShreevatsaR Excellent answer! It would be more awesome if you added a proof of why $\pi$ is irrational. en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational $\endgroup$ – user93957 Nov 5 '13 at 17:18
  • $\begingroup$ @Sam math never ends on MSE. :-) $\endgroup$ – Simply Beautiful Art Jan 15 '17 at 14:55

A number is called rational when it can be represented as $m/n$, where $m$ and $n$ are both integers.

Irrational numbers are the ones that are not rational, ie. cannot represented as a fraction $m/n$ where $m$ and $n$ are both integers. It is possible to prove that $\pi$ is irrational.

If we defined rational numbers as numbers that can be represented as $C/D$, where $C$ and $D$ can be any real numbers, then every number would be rational: $X = X/1$ for every $X$.

  • $\begingroup$ Exactly what I wrote above. $\endgroup$ – Sigur Aug 20 '12 at 16:35

protected by Asaf Karagila Jan 19 '14 at 8:54

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.