42
$\begingroup$

$\pi$ 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?

$\endgroup$
  • 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
  • 21
    $\begingroup$ 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
40
$\begingroup$

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" $x = x/1$.

$\endgroup$
  • 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
8
$\begingroup$

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 $X$.

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

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