Why is pi non-repeating?

Question

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...

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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<d $)

→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?

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Apologies for the screenshots in advance.

Fig 1:

Answer 1

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.

Answer 2

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

Answer 3

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.