# Is it possible for a non-natural number to be a number base?

• I
• swampwiz
In summary: You could go that way, but now you are no longer using the canonical representation and you have lost radix efficiency. Obviously there are lots of ways to slice and dice if one is building a convention from scratch.The support for non-canonical representations is one inelegance with non-integer radices.This conversation is about the impracticality of using a base -3 or base -2 in mathematical calculations. The commenter at Quora suggests that it is possible, but not practical, as it would require using negative numbers and would become confusing. Another commenter adds that using a negative radix eliminates the need for unary negation, but it would also result in a non-canonical representation and lose radix efficiency. Overall, using a negative base for mathematical
swampwiz said:
A commenter at Quora said that is possible for there be something like

12π

which would evidently be ( π + 2 ). This seems ridiculous.
It is possible. And it is not ridiculous, just incredibly impractical. And for ##b<1## or negative ##b## things become confusing, too. I guess negative numbers won't work, but I haven't elaborated it.

fresh_42 said:
It is possible. And it is not ridiculous, just incredibly impractical. And for ##b<1## or negative ##b## things become confusing, too. I guess negative numbers won't work, but I haven't elaborated it.
So basically, it's a mathematical curiosity like the gamma function.

It's mathematically useless, as well as practically useless. The gamma function is not.

swampwiz
By the way, ##12_\pi = (2+\pi)_{10}## could as well be ##12_\pi=C_{16}=12_{10}## since we will never have on overflow, so all numbers are one digit numbers. Of course we would soon run out of symbols ...

fresh_42 said:
I guess negative numbers won't work, but I haven't elaborated it
Negative radix works fine and eliminates the need for unary negation.

jbriggs444 said:
Negative radix works fine and eliminates the need for unary negation.
So ##27_{10}=300_{-3}## and ##-27_{10}=1000_{-3}\,?##

fresh_42 said:
So ##9_{10}=9_{-3}## and ##-9_{10}=100_{-3}\,?##
Allow me to illustrate with base -2.

The "one's" place has its usual meaning. Position 0, place value (-2)0 = 1.

So ##0_{-2} = 0## and ##1_{-2}## = 1.

The "two's" place is in position 1 with place value (-2)1 = -2.

So ##10_{-2} = -2## and ##11_{-2}## = -1.

The "four's" place is in position 2 with place value (-2)2 = 4.

So ##100_{-2}## = 4, ##101_{-2}## = 5, ##110_{-2}## = 2 and ##111_{-2}## = 3

[I am assuming a convention where each digit falls in the range from 0 to the ceiling of the absolute value of the radix minus one]

jbriggs444 said:
Allow me to illustrate with base -2.
See my example for base -3.
The positive values for odd powers of the base have then to be added lower even powers, which exceed the usual representation symbols 0,1,…,b−10,1,…,b−1.

For base -2 we would have 8−2=408−2=40, right? And the higher the odd power the more symbols we need. What is ##2^{9}_{-2}\,?##

fresh_42 said:
So ##27_{10}=300_{-3}## and ##-27_{10}=1000_{-3}\,?##
##1000_{-3}## would denote -81 + 0 + 0 + 0.
You would want ##1200_{-3}## to denote -81 + 54 + 0 + 0 = -27.

Edit: Blundered. ##1000_{-3}## is indeed -27 + 0 + 0 + 0.

jbriggs444 said:
##1000_{-3}## would denote -81 + 0 + 0 + 0.
You would want ##1200_{-3}## to denote -81 + 54 + 0 + 0 = -27.
But ##1000_{-3}=1\cdot (-3)^3 + 0\cdot (-3)^2 + 0\cdot (-3)^1 + 0\cdot (-3)^0 =-27_{10}## not ##-81_{10}##.

jbriggs444
fresh_42 said:
See my example for base -3.
The positive values for odd powers of the base have then to be added lower even powers, which exceed the usual representation symbols 0,1,…,b−10,1,…,b−1.

For base -2 we would have 8−2=408−2=40, right? And the higher the odd power the more symbols we need. What is ##2^{9}_{-2}\,?##
I am not sure why you are insisting on making this difficult. The alphabet for a radix -n digit-string representing a natural number is {0, 1, ..., n-1}.

fresh_42 said:
But ##1000_{-3}=1\cdot (-3)^3 + 0\cdot (-3)^2 + 0\cdot (-3)^1 + 0\cdot (-3)^0 =-27_{10}## not ##-81_{10}##.
Sorry, I seem to have slipped a digit, you are correct.

jbriggs444 said:
I am not sure why you are insisting on making this difficult. The alphabet for a radix -n digit-string representing a natural number is {0, 1, ..., n-1}.
You said a negative base makes a unary minus sign unnecessary. So what are ##-9_{10}## in base ##-3## and ##2^9_{10}## in base ##-2\,?##

We need to add copies of ##(-3)^1## up to three, so it is ##30_{-3}## and copies of ##(-2)^8##. The result are more ciphers than ##0,1,\ldots , b-1\,.## And I get ##+27_{10}=300_{-3}##.

O.k., I see, the minus sign results in an additional digit ##b##.

jbriggs444
fresh_42 said:
You said a negative base makes a unary minus sign unnecessary. So what are ##-9_{10}## in base ##-3## and ##2^9_{10}## in base ##-2\,?##
Ahhh, I had not understood the expected convention when you wrote ##2_{10}^9##. It seemed that you were intending ##2^9## as a single digit with value ##512_{10}##. But, in fact, you were trying to write ##2_{10}## raised to the ninth power. [With a single digit literal, the radix subscript adds nothing but confusion]So let's go down the place values.

0 = 1
1 = -2
2 = 4
3 = -8
4 = 16
5 = -32
6 = 64
7 = -128
8 = 256
9 = -512
10 = 1024

So to get ##-512_{10}## we need ##11000000000_2## = -1024 + 512 = -512.

To get ##-9_{10}## we need ##1200_3## = -27 + 18 + 0 + 0 = -9

jbriggs444 said:
Ahhh, I had not understood the expected convention when you wrote ##2_{10}^9##. It seemed that you were intending ##2^9## as a single digit with value ##512_{10}##. But, in fact, you were trying to write ##2_{10}## raised to the ninth power. [With a single digit literal, the radix subscript adds nothing but confusion]So let's go down the place values.

0 = 1
1 = -2
2 = 4
3 = -8
4 = 16
5 = -32
6 = 64
7 = -128
8 = 256
9 = -512
10 = 1024

So to get ##-512_10## we need ##11000000000_2## = -1024 + 512 = -512.

To get ##-9_10## we need ##1200_2## = -27 + 18 + 0 + 0 = -9
Yes, a negative overflow compensated by additional lower terms will do. But we could as well add ##b## to the set of digits and avoid the overflow: ##-9_{10}=1200_{-3}=30_{-3}##.

fresh_42 said:
Yes, a negative overflow compensated by additional lower terms will do. But we could as well add ##b## to the set of digits and avoid the overflow: ##-9_{10}=1200_{-3}=30_{-3}##.
You could go that way, but now you are no longer using the canonical representation and you have lost radix efficiency. Obviously there are lots of ways to slice and dice if one is building a convention from scratch.

The support for non-canonical representations is one inelegance with non-integer radices.

Depends on the preferred interpretation. If we have a negative base, then my thought was, that the gaps become longer between two useful powers, i.e. an additional sign ##b## is needed. With your classical view, the numbers will be longer. I find it a bit artificial to go knowingly in the wrong direction and compensate the mistake, rather than adding what actually contributes.

jbriggs444
fresh_42 said:
Depends on the preferred interpretation. If we have a negative base, then my thought was, that the gaps become longer between two useful powers, i.e. an additional sign ##b## is needed. With your classical view, the numbers will be longer. I find it a bit artificial to go knowingly in the wrong direction and compensate the mistake, rather than adding what actually contributes.
Elegance is in the eye of the beholder, of course. I take the opposite view.

fresh_42
By the way,
swampwiz said:
A commenter at Quora said that is possible for there be something like
12π

fresh_42 said:
By the way, ##12_\pi = (2+\pi)_{10}## could as well be ##12_\pi=C_{16}=12_{10}##
Nope. The hexadecimal numeral C, often written as 0xC, is equal to the decimal numeral 12, but ##12_\pi## means ##1 \cdot \pi + 2 \cdot 1 \ne 12_{10}##.
fresh_42 said:
since we will never have on overflow, so all numbers are one digit numbers. Of course we would soon run out of symbols ...
Right. In the programming world, a base b is a positive integer larger than 1, such as base-2, base-8, base-16, and base-64. All of these bases are in common use. For a given base, the digits are 0, 1, 2, ..., (b - 1).
fresh_42 said:
It is possible. And it is not ridiculous, just incredibly impractical.
I would even go so far as to say using ##\pi## as a base is ridiculous. In any number system that uses a positive integer base b, with b > 1, any integer is a finite sum of multiples of powers of b. How would you represent 253, for example, in base-##\pi##?

swampwiz
Mark44 said:
I would even go so far as to say using ##\pi## as a base is ridiculous. In any number system that uses a positive integer base b, with b > 1, any integer is a finite sum of multiples of powers of b. How would you represent 253, for example, in base-##\pi##?
Calling non-integer bases "ridiculous" is going a bit far.

Let us convert 253 to base pi. The log of 253 to base pi is 4.833... So we will need to start with five digits to the left of the radix point -- in the position with place value pi4 ~= 97.

So we divide 253 by pi4 and truncate to the next lower integer. The result is 2.597... which rounds down to 2.

So the first digit is 2.

We subtract 2pi4 from 253 yielding 58.181817931995...

Now it is just a matter of rinsing and repeating. We need to convert 58.181817931995... to base pi. We already know that the next digit will be in the pi3 place. So we divide by pi3 yielding 1.87... which rounds down to 1.

So the second digit is 1.

We subtract pi3 from 58.181817931995... yielding 27.17554125...

We repeat the process, dividing this by pi2 yielding 2.75... which rounds down to 2.

So the third digit is 2.

We subtract 2pi2 from 27.17554125... yielding 7.4363324495...

We repeat the process, dividing this by pi yielding 2.3670581356...

So the fourth digit is 2

We subtract 2pi from 7.4363324495... yielding 1.15314714233...

We repeat the process, dividing this by pi0 = 1 yielding 1.15314714233...

So the fifth digit is 1.

so ##253_{10}## ~= ##21221_{\pi}##.

We could continue the process indefinitely, populating further positions to the right of the radix point.

But there is no reason to calculate with powers of ##\pi##. Normally we write ##1,2,3,\ldots ,b-1## until ##b## marks the overflow from ##b^0## to ##b^1##. But with ##b=\pi## we will never reach an overflow by counting. Therefore we need for all numbers different from ##\pi^n## an extra sign. This is ridiculous. If we use your method instead, which is by the way the method you rejected for ##b < 0## since it accumulates regardless of an overflow, then all integers are infinitely long. This also can be called ridiculous.

I find we get funny notations for bases ##b=\frac{1}{n}##. I won't call them ridiculous, but a transcendental base is for some reason; even an irrational one is, since neither creates natural overflows.

swampwiz
fresh_42 said:
all integers are infinitely long. This also can be called ridiculous.
It is the base pi expansions which are infinite, not the integers. And it is not all integers. Almost all.

fresh_42 said:
But there is no reason to calculate with powers of ##\pi##. Normally we write ##1,2,3,\ldots ,b-1## until ##b## marks the overflow from ##b^0## to ##b^1##. But with ##b=\pi## we will never reach an overflow by counting. Therefore we need for all numbers different from ##\pi^n## an extra sign. This is ridiculous.
That was my reason for not treating base/radix ##\pi## seriously.

swampwiz

## 1. Can a non-natural number be used as a number base?

Yes, a non-natural number can be used as a number base. In fact, any number can be used as a base for a number system.

## 2. How does using a non-natural number as a base affect the number system?

Using a non-natural number as a base can create a non-standard number system, which may have different properties and operations compared to traditional number systems. For example, a number system with a base of pi would have a different set of rules for addition, subtraction, multiplication, and division.

## 3. Are there any practical applications for using a non-natural number as a base?

Yes, there are some practical applications for using a non-natural number as a base. For instance, in computer science, a number system with a base of 2pi (tau) has been proposed as an alternative to the traditional binary system, which could potentially improve efficiency in certain calculations.

## 4. What are some examples of non-natural number bases?

Some examples of non-natural number bases include pi, e (Euler's number), and the golden ratio. These numbers are commonly used in mathematics and have been proposed as bases for alternative number systems.

## 5. Are there any limitations to using a non-natural number as a base?

There are some limitations to using a non-natural number as a base. One limitation is that it may be more difficult to perform basic arithmetic operations, such as addition and multiplication, in a non-standard number system. Additionally, using a non-natural number as a base may not be practical or useful in all situations.

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