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I want to know: what is the last digit of $\pi$?

Some people say there are no such thing, but they fail to mention why.

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    $\begingroup$ Sometimes even rational numbers don't have a "last digit", think of 1/3=0.33333... $\endgroup$
    – Vhailor
    Commented Jul 21, 2011 at 22:12
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    $\begingroup$ Nice Poem. ${}$ $\endgroup$
    – jspecter
    Commented Jul 21, 2011 at 22:13
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    $\begingroup$ If $\pi$ has a last digit, then $0.999\ldots\neq 1$. $\endgroup$
    – Asaf Karagila
    Commented Jul 21, 2011 at 22:13
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    $\begingroup$ This question should specify "base 10". The "no last digit" phenomenon depends on how $\pi$ is represented. To take a contrived setting, base-$\pi$ numbers, then $\pi$ is written as $1$. I'm not trying to be pedantic here: representation is a fundamental part of this question. $\endgroup$
    – Fixee
    Commented Jul 22, 2011 at 6:09
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    $\begingroup$ @jspecter: A slightly enhanced version: $$ \text{I desperately want to know}\\ \text{The last digit of $\pi$}\\ \text{Some people say there's no such thing}\\ \text{They fail to mention why} $$ $\endgroup$
    – joriki
    Commented Dec 22, 2012 at 3:49

5 Answers 5

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There is no "last" digit of $\pi$. If there was a last digit, then there could only be finitely many digits in front so that $\pi$ would be a rational number. However $\pi$ was shown to be irrational in the 18th century by Lambert.

(This Meta.StackExchange post is a joke based on the impossibility of finding such a last digit)

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    $\begingroup$ Thanks for the link to the Meta.StackExchange post: hilarious. I've got my laughs in for the day. We all need a good dose of humor now and then! $\endgroup$
    – amWhy
    Commented Jul 21, 2011 at 22:16
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Since you may have never seen the topics in my colleagues' answers, I'll try to explain them in some detail.

Suppose for the sake of argument that when $\pi$ is written as a decimal expansion ($3.1415 \dots$) it does have a final digit. This would clearly imply that there is a finite number of terms in the expansion. All real numbers with finite decimal expansions can be written in the form $\frac{a}{b}$ where $a$ and $b$ are integers (whole numbers).

By this reasoning we conclude that $\pi = \frac{a}{b}$ for some positive integers $a$ and $b$, i.e., that $\pi$ is rational. This is the starting point for this short proof given by I. Niven in 1946, which is especially easy to follow if you've had a little trigonometry and even less differential calculus. The proof concludes with an absurdity like the existence of an integer between $0$ and $1$, which implies that $a$ and $b$ do not exist and $\pi$ is irrational (and has an infinite decimal expansion). It should be noted that the irrationality of $\pi$ was first established by Lambert in 1761 by studying the continued fraction expansion of the tangent function and using the identity $\tan \frac{\pi}{4} = 1$. More generally, he proved that if $x$ is rational, then $\tan x$ is irrational.

In short, there is no final digit in the decimal expansion of $\pi$ because it is irrational.

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  • $\begingroup$ Sadly, your link to Niven's proof doesn't work. $\endgroup$
    – jprete
    Commented Jul 22, 2011 at 4:53
  • $\begingroup$ Ok, try it now. $\endgroup$
    – user02138
    Commented Jul 22, 2011 at 5:25
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Proving that $\pi$ is irrational is more difficult than proving that $e$ or $\sqrt{2}$ or $\log_2 3$ is irrational. See http://en.wikipedia.org/wiki/Proof_that_pi_is_irrational .

Proving that an irrational number has no last digit is easier than that: http://en.wikipedia.org/wiki/Irrational_number

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Even rational numbers usually have no "last digit": what is the last digit of $$ 0.1313131\dots = 0.\overline{13} = \frac{13}{99} ? $$

So what sort of numbers have a last digit?

One, numbers with a terminating decimal expansion: numbers like $\displaystyle 2.23627 = \frac{223627}{100000}$. As you can see, all such numbers can be written as a fraction with denominator being a power of $10$.

Two, depending on your definition of "last digit", numbers like $0.4677777\dots = \frac1{100}46.77777$ = $\displaystyle \frac1{100}\left(46 + \frac79\right) = \frac1{100}\frac{421}{9}$. These numbers can be written as $\displaystyle \frac1{10^k} \frac{n}9$ for some integers $k$ and $n$.

So a number $x$ has a "last digit" if and only if $(9\cdot 10^k)x$ is an integer for some $k$. Only very special numbers are of this form, and it should be no surprise that $\pi$ is not. (Admittedly, I don't actually see how to prove this without invoking $\pi$'s irrationality, but it's a much weaker property.)

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HINT $\rm\ \pi = 3.1415\ \Rightarrow\ 10^4\: \pi = 31415\ \Rightarrow\ \pi = 31415/10^4\ $ is rational, contra Lambert's proof.

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