# Why is pi non-repeating?

Ok, I have just learnt the Pigeonhole Principle(PHP) and its application with decimal expansion.

To convey my question clearly, I need to convey my understanding of PHP with regards to decimal expansion so here goes...

By the long division process, we can obtain a infinite number of remainders since $$0$$ is also considered a remainder(refer to Fig 1 to get what I mean) which is $$>$$ finite number of possible values of remainders(by quotient-remainder theorem, $$0≤r)

→By PHP , we’ll definitely get a remainder whose value is same as a previous remainder’s value (In Fig 1, we score a hit when r7=r1)

→As such at some point, the sequence of remainders will infinitely repeat

→This sequence of remainders can either be zeroes or non-zeroes

→If it's a sequence of zeroes, we get terminating decimals like $$2$$(as in $$2.\color{red}{\overline{000}}\ldots...$$) or $$3.625$$(as in $$3.625\color{red}{\overline{000}}\ldots...$$)

→If it's a sequence of non-zeroes, we get a repeating decimal like $$3/14$$=$$0.2\color{red}{\overline{142857}}\ldots$$

My question: So why does PI not fall into either of these categories? Does it somehow violate PHP?

Apologies for the screenshots in advance.

Fig 1:

• The piegon hole principle says that the we will have to have a digit repeating infinitely often. A pattern in which it repeats is not necessary. Commented Oct 22, 2020 at 11:22
• I will add that should you have periodicty of the digits from some point onwards, then the number is rational. This has a proof with geometric series which I think is nice. Commented Oct 22, 2020 at 11:24
• The second → is not justified. Consider the sequence : $1010010001000010000010000001....$ The pattern is $1$ followed by $n$ zeroes, for $n=1$ to $\infty$. There is no repetition. Commented Oct 22, 2020 at 11:28
• Your logic (and use of the PHP) works for long division with integers. It doesn't apply to $\pi$ because $\pi$ is irrational: it cannot be represented as the quotient of any two integers. To assume otherwise would be begging the question. Commented Oct 22, 2020 at 11:30
• @DavidDiaz, ahh so one flaw in my thinking that π=22/7 (ie it can be represented by quotient of integers) so premise is already false then?
– Leon
Commented Oct 22, 2020 at 21:44

The second step of your inference needs to be better justified. Note that in your setup the divisor $$d$$ must be in integer in order for the application of PHP to make sense. Now there are two cases.

1. The dividend is an integer, which is represented as $$n.00000\ldots$$. Then you can use PHP to argue that at some remainder obtained after the decimal point repeats. So you have $$r_k=r_{k+n}$$ for $$k,n$$ obtained after the decimal points. Then you need to further argue that bringing down $$0$$ at $$r_{k+n}$$ yields a periodic division pattern of length $$n$$ from there on out, which yields an eventually repeating decimal in your answer.

2. The dividend is not an integer, but some arbitrary real number. In this case, PHP tells you get duplicate remainders. But you will not be able to further argue that the list of remainders eventually repeats. For example consider dividing $$2$$ into $$0.10100100010000\ldots$$ (here I have $$1$$ followed by $$n$$ zeroes for $$n=1$$ to $$\infty$$). Your list of remainders is $$0,1,0,1,0,0,1,0,0,0,1,0,0,0,0\ldots$$ with no repetition. However, even if you did get a repeating sequence of remainders, this would not necessarily mean your answer is a repeating decimal. For example, consider any real number $$x$$, represented as a decimal $$a_0.a_1a_2a_3\ldots$$. Do the trivial long division of $$1$$ into $$x$$. You will get remainder $$0$$ each time, but the decimal sequence in the quotient need not be (eventually) repeating.

Now, as for $$\pi$$, it has to fit into the second case, because if you start your long division with the dividend and divisor being integers, then you are already computing a rational number.

• So another flaw in my thinking process is that I assumed, the pattern must be a fixed sequence of digits but doesn't have to be?
– Leon
Commented Oct 22, 2020 at 21:56
• But isn't $0.10100100010000$ irrational(I googled it)? According to all $3$ Davids that said pi is irrational, shouldn't the same logic apply for $0.10100100010000$?
– Leon
Commented Oct 22, 2020 at 21:57
• Sorry I meant to say $0.10100100010000...$
– Leon
Commented Oct 22, 2020 at 22:06
• Yes that number (lets call it $\alpha$) is irrational, and the same logic does apply to it. Just like $\pi$, you would not be able to produce $\alpha$ using long division as in case 1. Commented Oct 23, 2020 at 10:08
• The other answers point out that if you do long division with two integers then your answer is rational by definition. But you could apply the same reasoning with PHP when only the divisor is an integer $d$, and the dividend is an arbitrary real number. My answer shows that there are still logical flaws in this second case, namely: (1) In an infinite sequence of remainders between $0$ and $d$, PHP implies that there are repetitions, but the sequence does not need to be periodic. (2) Even if the sequence of remainders is periodic, it need not yield a quotient (answer) that is periodic. Commented Oct 23, 2020 at 10:15

$$\pi$$ is a transcendental number and not a rational one. The PHP argument seems to apply to rational numbers.

You have assumed you can compute $$\pi$$ by dividing one number with a known, finite number of digits by another number with a known, finite number of digits.

This is the very definition of what it means for a number to be rational. And indeed your argument is a correct proof that the decimal representation of any rational number repeats.

But $$\pi$$ is not rational. It cannot be computed by dividing one number with a known, finite number of digits by another number with a known, finite number of digits.