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If I have a field of all real numbers on $[0,p) (\mod p)$ under normal addition and multiplication where $p$ is a prime number, it can be shown that all integers have multiplicative inverses that are also integers. This means all rational numbers have integer equivalents E.g. $$ 3\times2 ≡ 1\mod5$$ Therefore $2$ is the multiplicative inverse of $3$ in $\mod 5$. This means that $\frac13 ≡ 2 \mod 5$. I then tried to extend this logic to irrational numbers.

As an example, I took $\pi$. We can represent $\pi$ by the infinite series $$\frac{3}{1}+\frac{1}{10}+\frac{4}{100}+\frac{1}{1000}...$$ Clearly this representation would not work for $\mod 5$ because one would be dividing by $0$. However, in $\mod 3$ and in $\mod 9$, this fraction can be reduced to $$\frac{3}{1}+\frac{1}{1}+\frac{4}{1}+\frac{1}{1}... = 3+1+4+1...$$ This is where I am stuck. Because we are summing an infinite list of integers, it would seem the sum should be an integer. However, as the series gets longer and longer it clearly just cycles through all of the integers and never approaches a definitive answer. So is $\pi$ equivalent to all the integers at once, or do irrational numbers not have integral equivalents in modular arithmetic? Does my logic have any fatal flaws that may account for my results?

Kenta S
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Zach L
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    In order to talk about infinite sums you need a topology. What is your topology? – Plutoro Sep 25 '15 at 18:14
  • Notice that if $p=2$, then $\sqrt{2}\sqrt{2}=2=0$, so the ring you describe is not a field. – Plutoro Sep 25 '15 at 18:17
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    @AlexS O ok, thanks. I'm only in high school and I was just doing this in my spare time, so I'm afraid I don't know much about these things. – Zach L Sep 25 '15 at 18:19
  • Something to think about: you could write $\pi$ in any base, not just in base $10$, and you would get a different infinite series. – Julian Rosen Sep 25 '15 at 18:20

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Basically your question boils down to "what is an infinite sum in modular arithmetic?". This issue is one of convergence. Basically, in order to talk about infinite sums, you need some notion of what it means to converge. This is called a topology. Now if you are in high school, topology is a little advanced for you so don't worry about it.

The easiest way to discuss topology/convergence is in terms of metrics. A metric is a notion of distance. For example the standard metric on $\Bbb{R}$ is $d(x,y)=|x-y|$. For example $d(0,4)=4$. So what metric do you want to use in $\Bbb{Z}/5$ for example? If you use the same one, then convergence is pretty boring, as only (eventually) constant things converge! There is no way to get "close" to a number! The closest numbers can be is $1$, or $0$ if they are the same. In order to talk about convergence in modular arithmetic, you need to cook up a non trivial topology/metric. This is not so easy.

What $\pi^{-1}$ is, or if it even exists, depends on your topology. As far as I know there is no natural/canonical topology in modular arithmetic. If there is, it is much too advanced.

This is an interesting question by the way. If you are thinking about this stuff in high school I am impressed and I hope you continue your studies in math and I'm sure you'll do well.

  • Thank you! This answer was extremely helpful! However, it still doesn't address the place of irrational numbers in modular arithmetic. – Zach L Sep 25 '15 at 18:42
  • @ZachL It does, just in a roundabout way. :) Basically it's not very well defined the way you're attempting to. –  Sep 25 '15 at 18:50
  • Sorry, I think I just got carried away with the whole "using comments to ask for more information and not just saying thanks" thing. I reread your answer and it explains very concisely the place of irrationals in modular arithmetic. :) – Zach L Sep 25 '15 at 18:58
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This is an interesting idea. Unfortunately, decimal expansions don't behave very well when we consider them modulo $p$.

As an example, let's take $p=3$ and consider the rational number $\frac{1}{11}$: $$ \frac{1}{11}=0.090909\ldots=\frac{9}{100}+\frac{9}{10000}+\ldots $$ Modulo $3$ we have $1/11\equiv 1/2 \equiv 2$, but if we consider the decimal expansion modulo $3$, we get $$ \frac{9}{100}+\frac{9}{10000}+\ldots \equiv \frac{9}{1}+\frac{9}{1}+\ldots\equiv 0+0+\ldots\equiv 0\mod 3. $$ This shows that even if the decimal expansion mod $p$ has a well-defined sum, the sum might be the wrong value. So if there is a reasonable value of $\pi$ mod $p$, we shouldn't necessarily expect to find it expressing $\pi$ as a decimal and summing the terms modulo $p$.

By the way, I think "What is/ought to be the value of $\pi$ modulo $p$" is a very interesting question.

Julian Rosen
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In response to the question "do irrational numbers not have integral equivalents in modular arithmetic?"

Just like the rational numbers have a algebraic closure, namely the algebraic numbers, finite fields have algebraic closures as well. If we consider, say, $\overline{\mathbb{F}_7}$, i.e. the algebraic closure of the integers (mod 7), this contains two square roots of five, just like the algebraic numbers contain two square roots of five (namely, $+\sqrt{5}$ and $-\sqrt{5}$). These square roots of five are irrational -- they aren't quotients of integers (mod 7) -- but they're not in any meaningful sense the same square roots of five as the real square roots of five.

You'll have to take this kind of phenomenon into account here.

Daniel McLaury
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