A commenter at Quora said that is possible for there be something like
12_{π}
which would evidently be ( π + 2 ). This seems ridiculous.
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.A commenter at Quora said that is possible for there be something like
12_{π}
which would evidently be ( π + 2 ). This seems ridiculous.
So basically, it's a mathematical curiosity like the gamma function.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.
Negative radix works fine and eliminates the need for unary negation.I guess negative numbers won't work, but I haven't elaborated it
So ##27_{10}=300_{-3}## and ##-27_{10}=1000_{-3}\,?##Negative radix works fine and eliminates the need for unary negation.
Allow me to illustrate with base -2.So ##9_{10}=9_{-3}## and ##-9_{10}=100_{-3}\,?##
See my example for base -3.Allow me to illustrate with base -2.
##1000_{-3}## would denote -81 + 0 + 0 + 0.So ##27_{10}=300_{-3}## and ##-27_{10}=1000_{-3}\,?##
But ##1000_{-3}=1\cdot (-3)^3 + 0\cdot (-3)^2 + 0\cdot (-3)^1 + 0\cdot (-3)^0 =-27_{10}## not ##-81_{10}##.##1000_{-3}## would denote -81 + 0 + 0 + 0.
You would want ##1200_{-3}## to denote -81 + 54 + 0 + 0 = -27.
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}.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}\,?##
Sorry, I seem to have slipped a digit, you are correct.But ##1000_{-3}=1\cdot (-3)^3 + 0\cdot (-3)^2 + 0\cdot (-3)^1 + 0\cdot (-3)^0 =-27_{10}## not ##-81_{10}##.
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\,?##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}.
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]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\,?##
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}##.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
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.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}##.
Elegance is in the eye of the beholder, of course. I take the opposite view.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.
A commenter at Quora said that is possible for there be something like
12_{π}
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}##.By the way, ##12_\pi = (2+\pi)_{10}## could as well be ##12_\pi=C_{16}=12_{10}##
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:since we will never have on overflow, so all numbers are one digit numbers. Of course we would soon run out of symbols ...
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##?It is possible. And it is not ridiculous, just incredibly impractical.
Calling non-integer bases "ridiculous" is going a bit far.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##?
It is the base pi expansions which are infinite, not the integers. And it is not all integers. Almost all.all integers are infinitely long. This also can be called ridiculous.
That was my reason for not treating base/radix ##\pi## seriously.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.