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

[swampwiz]

A commenter at Quora said that is possible for there be something like

12_π

which would evidently be ( π + 2 ). This seems ridiculous.

https://www.quora.com/What-mathemat...-non-mathematicians/answer/Peter-James-Thomas

 

[fresh_42]

A commenter at Quora said that is possible for there be something like

12_π

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.

 

[swampwiz]

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.

 

[fresh_42]

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

 

[fresh_42]

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 ...

 

[jbriggs444]

I guess negative numbers won't work, but I haven't elaborated it

Negative radix works fine and eliminates the need for unary negation.

 

[fresh_42]

Negative radix works fine and eliminates the need for unary negation.

So $27_{10}=300_{-3}$ and $-27_{10}=1000_{-3}\,?$

 

[jbriggs444]

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]

 

[fresh_42]

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 $2^{9}_{-2}\,?$

 

[jbriggs444]

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.

 

[fresh_42]

$1000_{-3}$ 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]

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}.

 

[jbriggs444]

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.

 

[fresh_42]

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]

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

 

[fresh_42]

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: $-9_{10}=1200_{-3}=30_{-3}$.

 

[jbriggs444]

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.

 

[fresh_42]

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]

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.

 

[Mark44]

By the way,

A commenter at Quora said that is possible for there be something like
12_π

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 $1 \cdot \pi + 2 \cdot 1 \ne 12_{10}$.

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).

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-$\pi$?

 

[jbriggs444]

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.

 

[fresh_42]

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.

 

[jbriggs444]

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.

 

[Mark44]

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.