Integers in base Pi

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Atran
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?

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

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

Atran
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 $Z$.

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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
Atran
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 $(Z_5,+)$.

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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?

Atran
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...

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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.

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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?

Atran
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.

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How is it written in the notation you are using?

Atran
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 $Z_\pi$.

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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.

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...

Atran
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.

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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.

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

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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)

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

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

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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, $4375_{\pi} = 4{\pi^3}+3{\pi^2}+7{\pi}+5$.

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 infinite number of digits.

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But ##5(\pi^0) + 5(\pi^0) \ne 1(\pi^1) + 0 (\pi^0)##.
No, the OP explicitly stated
5+5 equals π in $Z_{\pi}$.
If this thread is locked then the OP will not be able to come back and deal with the inconsistencies, but until they do we are just wasting each others time.

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In the meantime, anyone interested in the concept of an irrational radix could look at this example in base ## \tau ##, also known as ## \varphi ## or ## \phi = \dfrac {1 + \sqrt 5}2 ##.

No, the OP explicitly stated
##5 + 5 = \pi## in ##\mathbb{Z}_\pi ##

I don't understand what your "No" refers to.

I stated that post #1 intends to implement addition by an algorithm that implies ( notation base ##\pi##) that ##5 + 5 = \pi##. The point is that this operation is different that the addition operation in ##\mathbb{R}##.

pbuk
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I don't understand what your "No" refers to.
It refers to the fact that your statement was inconsistent with the information presented in the OP.

I stated that post #1 intends to implement addition by an algorithm that implies ( notation base π) that 5+5=π. The point is that this operation is different that the addition operation in R.

Addition and substraction are defined similarly as they are for (Z,+).

Homework Helper
My understanding is that the two different systems (base ##\pi## and base 10) share the exact same set of decimal strings. They share the same addition operation when it is understood to act on decimal strings as addends and deliver a decimal string as the sum. However, they map onto a different set of objects on the standard real number line.

For the base 10 representation, the notation scheme represents the standard set of integers on the real number line: 1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, 9.0, 10.0, 11.0, ...

For the base ##\pi## representation, the notation scheme represents a different set of numbers, most of which are not integers on the standard real number line: 1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, 9.0, 3.14..., 4.14..., ...

If the addition operation is interpreted as acting on real numbers to produce real numbers, this means that we have two distinct addition operations. One of them (base 10) respects the standard ordering of the real number line. One (base ##\pi##) does not.

Since essentially no properties of the real number line are respected, the mapping to the real number line seems completely superfluous. Which means that we do not have two systems of numeration. We have only one. This "base ##\pi##" is nothing of the sort. It is just base 10 re-labeled.

Last edited:
Stephen Tashi and pbuk
If the addition operation is interpreted as acting on real numbers to produce real numbers, this means that we have two distinct addition operations.

Yes, if , in the following passage, the term "addition" taken to mean an operation implemented by a particular algorithm peformed on digits and "similarly" is taken to mean "identical".

Addition and substraction are defined similarly as they are for $(Z,+)$.

Computing the result of 5+5 in base ##\pi## notation using the identical algorithm for base 10 digits results in ## 5 + 5 = 10 ## and the "##10##" represents ##\pi ##

jbriggs444
Jarvis323
I think maybe you want the 1's place the be the ##\frac{\pi}{10}##'s place?

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I think maybe you want the 1's place the be the ##\frac{\pi}{10}##'s place?
If that's true then the 100s place should be ##10\pi##, not ##\pi^2##. But then the set is just the integers scaled by ##\pi/10## which again is not interesting.

jbriggs444
Staff Emeritus
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: $0,1,2,3,4,5,6,7,8,9,{\pi},{\pi}+1,{\pi}+2,...,2{\pi},2{\pi}+1,...5{\pi},5{\pi}+1,5{\pi}+2,...,9{\pi}+8,9{\pi}+9,{\pi}^2,...$.