# Integers in base Pi

• I
A proper number is expressed in $\pi$ in a similar way as a decimal integer is expressed in base 2. For example, $4375_{\pi} = 4{\pi^3}+3{\pi^2}+7{\pi}+5$. The only exception I make is that the 10 digits are included when expressing a number with $\pi$. To clarify, the first positive such numbers are: $0,1,2,3,4,5,6,7,8,9,{\pi},{\pi}+1,{\pi}+2,...,2{\pi},2{\pi}+1,...5{\pi},5{\pi}+1,5{\pi}+2,...,9{\pi}+8,9{\pi}+9,{\pi}^2,...$.

Denote this set of numbers of form $a_{n}{\pi}^n+a_{n-1}{\pi}^{n-1}+...+a_{1}{\pi}+a_{0}$ with $Z_{\pi}$. Obviously $Z_{\pi}$ is ordered, for example $-2{\pi}^2 < 3{\pi}+2$. $Z_{\pi}$ is countable, since the elements can be arranged in this way: $0,{\pi},-{\pi},2{\pi},-2{\pi},3{\pi},-3{\pi},...$

Addition and substraction are defined similarly as they are for $(Z,+)$. Following from the definition of a group, $(Z_{\pi}, +)$ is clearly a group. However, the elements of $(Z_{\pi}, +)$ are not rational numbers.

Does this imply the set of rational integers and irrational integers are isomorphic?

Office_Shredder
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How do you express ##11\pi## in this setup?

Edit to add: also, how do you express 11?

The elements of the group are not rational numbers. Hence 11 is not in the group.

$11\pi$ in $Z_{\pi}$ equals $\pi^2+\pi$ and maps to 11 in $Z$.

Office_Shredder
Staff Emeritus
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It's not a group then. If 1 is in the group, 1+1+1+1+1+1+1+1+1+1+1 must be in it also.

pbuk
It's not a group then. If 1 is in the group, 1+1+1+1+1+1+1+1+1+1+1 must be in it also.
The rational numbers are not in the group; integers are rational numbers. On the other hand, 1+1 and 2 are the same number in, say for example, group $(Z_5,+)$.

Office_Shredder
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The rational numbers are not in the group; integers are rational numbers. On the other hand, 1+1 and 2 are the same number in, say for example, group $(Z_5,+)$.
But you said 1 is in the group. If 1 is in the group, and it's closed under addition, then 11 must be in the group. Unless you are defining addition weirdly. What is 1+1+1+...+1 supposed to be equal to?

But you said 1 is in the group. If 1 is in the group, and it's closed under addition, then 11 must be in the group. Unless you are defining addition weirdly. What is 1+1+1+...+1 supposed to be equal to?
This is how I thought to set up the group, though I did not go through it thoroughly.

0 in $Z_\pi$ corresponds to 0 in $Z$.
$\pi$ in $Z_\pi$ corresponds to 10 in $Z$.
$2\pi$ in $Z_\pi$ corresponds to 20 in $Z$.
$3\pi$ in $Z_\pi$ corresponds to 30 in $Z$.

$\pi^2$ in $Z_\pi$ corresponds to 100 in $Z$.
$2\pi^2$ in $Z_\pi$ corresponds to 200 in $Z$.
$3\pi^2$ in $Z_\pi$ corresponds to 300 in $Z$.
$3\pi^2+\pi$ in $Z_\pi$ corresponds to 310 in $Z$.
$3\pi^2+5\pi$ in $Z_\pi$ corresponds to 350 in $Z$.
$3\pi^2+5\pi+2$ in $Z_\pi$ corresponds to 352 in $Z$.

$\pi^3$ in $Z_\pi$ corresponds to 1000 in $Z$.
And so on...

Office_Shredder
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So wait, what's even the point of ##\pi## here? You've just come up with a complicated way of writing down the integers.

Here's a fact. If you construct an injection f from Z to R, then define a + which I'll call +' by f(x)+'f(y) = f(x+y), then f(Z) with +' will have a group structure, but it won't tell you anything interesting. It looks like that's all you've down here.

pbuk
Gold Member
You don't seem to be answering @Office_Shredder's question; I'll put it slightly differently. 5 is a member of the set; if the set is a group then 5 + 5 must also be a member of the set. Is it?

You don't seem to be answering @Office_Shredder's question; I'll put it slightly differently. 5 is a member of the set; if the set is a group then 5 + 5 must also be a member of the set. Is it?
Yes.

pbuk
Gold Member
How is it written in the notation you are using?

If I remember correctly, given a decimal expansion of a number, substitute powers of $\pi$ for powers of ten. In other words, 127 = 10^2 + 2*10 + 7. After substitution, it becomes $\pi^2+2\pi+7$. Note that the latter is no rational number but it is a member of $Z_\pi$.

pbuk
Gold Member
That is not an answer to my question. Let me answer it for you: 5 + 5 is not a member of your set so it is not a group.

Mark44
Mentor
This is how I thought to set up the group, though I did not go through it thoroughly.

0 in $Z_\pi$ corresponds to 0 in $Z$.
$\pi$ in $Z_\pi$ corresponds to 10 in $Z$.
$2\pi$ in $Z_\pi$ corresponds to 20 in $Z$.
$3\pi$ in $Z_\pi$ corresponds to 30 in $Z$.
...
And so on...
The only one of the above that makes any sense is the first one. ##\pi, 2\pi##, etc. are numbers in ##\mathbb R##, not in ##Z_\pi##, whatever that's supposed to mean.

In addition, the 2nd, 3rd, and 4th lines above have things in the opposite order.
##\pi## in ##\mathbb R## corresponds to 10 in ##Z_\pi##
##2\pi## in ##\mathbb R## corresponds to 20 in ##Z_\pi##
##3\pi## in ##\mathbb R## corresponds to 30 in ##Z_\pi##
##\pi^2## in ##\mathbb R## corresponds to 100 in ##Z_\pi##
and so on...

That is not an answer to my question. Let me answer it for you: 5 + 5 is not a member of your set so it is not a group.
5+5 equals π in $Z_{\pi}$.

An isomorphism that takes a member of $Z$ to give a member of $Z_{\pi}$ could be: $f(b) = f(b_n{10}^n + b_{n-1}{10}^{n-1} + ... + b_1{10} + b_0) = a_n\pi^n + a_{n-1}\pi^{n-1} + ... + a_1\pi + a_0$, where $b$ is expanded to its decimal representation first. I don't worry about computing $f$ here.

Office_Shredder
Staff Emeritus
Gold Member
I'm going to reiterate post number 8 here.

Your addition is not addition on the real numbers, so it's not an interesting subgroup of R. The thing you've written down is just the normal integers, but in a confusingly written way.

pbuk
Gold Member
5 + 5 equals π in $Z_{\pi}$.
5 + 5 yields an integer. ## \pi ## is an irrational number. They cannot be equal.

An isomorphism that takes a member of $Z$ to give a member of $Z_{\pi}$ could be: $f(b) = f(b_n{10}^n + b_{n-1}{10}^{n-1} + ... + b_1{10} + b_0) = a_n\pi^n + a_{n-1}\pi^{n-1} + ... + a_1\pi + a_0$, where $b$ is expanded to its decimal representation first. I don't worry about computing $f$ here.
You should concentrate on the basics before considering isomorphisms.

jim mcnamara, Mark44, Vanadium 50 and 1 other person
Staff Emeritus
it's not an interesting subgroup of R
If π is the same π we all know and love, it's not even a subgroup, much less an interesting one.

If π = 10, we do have a group - the integers. We also have incredibly and unnecessarily misleading terminology. Furthermore, the statement that these are not rational is false.

Either way, there is no logical conclusion that can be drawn from these false premises.

pbuk
martinbn
It is already very unclear, but to me it looks like ##\mathbb Z[п]## i.e. polynomials in one variable over the integers.

Office_Shredder
Staff Emeritus
Gold Member
It is already very unclear, but to me it looks like ##\mathbb Z[п]## i.e. polynomials in one variable over the integers.
It's not, because they insist 5+5 is not equal to 10 (which doesn't even exist in this set)

martinbn
It's not, because they insist 5+5 is not equal to 10 (which doesn't even exist in this set)
Good point!

Stephen Tashi
I think there is a confusion about whether notation like "1" or "5" means an member of the integers or whether it refers to an element of the mathematical structure defined by @Atran

pbuk
Gold Member
There is certainly confusion somewhere, but the OP doesn't seem interested in clearing it up; there doesn't seem to be much point speculating until they return.

Having said that, it is clear that the symbols 3, 4, 5 and 7 represent integers in the first part of the post:
A proper number is expressed in $\pi$ in a similar way as a decimal integer is expressed in base 2. For example, $4375_{\pi} = 4{\pi^3}+3{\pi^2}+7{\pi}+5$.

Stephen Tashi
Addition and substraction are defined similarly as they are for ##(Z,+)##
Objections are being raised because that statement is interpreted to mean that the manipulations of digits used in computing the results of addition and subtraction when numbers are represented base 10 are to also be used when computing the results of those same operations base ##\pi##.

The algorithm to compute "5+5 = 10" is correct in base 10 because ##5(10^0) + 5(10^0) = 1(10^1) + 0(10^0)##. But ##5(\pi^0) + 5(\pi^0) \ne 1(\pi^1) + 0 (\pi^0)##.

It appears that implementing operations in ##\mathbb{R}## using a system of digits base ##\pi## requires introducing a different system of digits (perhaps more than a finite number) or using algorithms where an operation on two numbers with a finite number of digits may produce a number represented by an infinte number of digits.

pbuk
5+5 equals π in $Z_{\pi}$.