# Simple numbers question... Or maybe not so simple.

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Tufted Titmouse

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05 Jun 2013, 8:36 pm

Spiderpig wrote:
I don’t know if you can “efficiently” get rid of the radix point (that’s how you call the character separating the integer and fractional parts without referring to a specific base), because, in a positional system where digits are weighted with all integral powers of the base, you need a way to mark the place whose weight is 1 (i.e. whose exponent is 0). There may be ways to work around this, but they’re likely to involve more complexity than a humble dot or comma

maybe we can add a radix point on any place, so we have numbers written as ...7.4.9...8.21..8

The point is to make representation more efficient. Not necessarily to get rid of the "."

It would also be convenient (but not required), that numbers without radix would have the same representation that is conventional.

I suspect that it would also make calculations easier.

Tufted Titmouse

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05 Jun 2013, 9:06 pm

ruveyn wrote:
Ancalagon wrote:
But the system efficiency is ruined when we have decimals, or negative numbers. With these we have 2 extra symbols, the "." and the "-".

The "-" can only be placed once, and only on one position. Meanwhile the "." can be on any place, but can be present only once.

That's wasteful.

That isn't really wasteful. To represent numbers that are negative or fractional, we need something like the negative sign and the decimal sign. In any representation, something must take their place or we must give up representing that type of number. We even have reasonable rules for omitting them in most common cases, so they only take up space when they give us information.

If you just want a more compact representation, look into non-base 10 number systems.

You bet. the minus sign (-) effectively doubles the range of the integers and the decimal point gets us from the integers to all the rational numbers. Infinite non-repeating decimals gets us to ALL the real numbers.

ruveyn
Is wasteful in the sense that we introduce a new symbol, but we can't use any combination of the symbols to represent an unique number.

with 10 numerical symbols, and n places, we can represent 10^n different numbers. If we add an extra symbol, like "-", we should be able to represent 11^n different numbers, but since we can only put the "-" on one place, we only get 11x10^(n-1) different numbers. 11x10^(n-1) is lower than 11^n, so it is wasteful.

As was shown, the minus sign isn't even required if negative numbers are allowed for the symbols.

Using a decimal point is similarly wasteful, as is using "i" for complex numbers.

Ancalagon
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06 Jun 2013, 3:57 am

Is wasteful in the sense that we introduce a new symbol, but we can't use any combination of the symbols to represent an unique number.

What is being wasted? You might as well say base 10 is wasteful because it isn't base 11. Adding a bunch of new symbols to represent negatives avoids the "problem" of not being able to place the negative sign randomly, but it doubles the number of symbols needed. That isn't really a problem, but if you consider an extra symbol wasteful, then doubling the number of symbols is worse than adding one symbol.

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Tufted Titmouse

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06 Jun 2013, 7:22 am

Ancalagon wrote:
What is being wasted?

Unused combinations of the symbols are wasted.

For example, the traditional decimal system for integers have 11 symbols, which are "0","1","2","3","4","5","6","7","8","9","-", but not any combination of those symbols is utilized. 34--43--- means nothing.

Ancalagon wrote:
You might as well say base 10 is wasteful because it isn't base 11.

No. That's completely wrong. For example, the symbols 4,3,2,1,0,1,2,3,4,5 is base 10, and more efficient than "0","1","2","3","4","5","6","7","8","9","-", which uses 11 symbols

Ancalagon wrote:
Adding a bunch of new symbols to represent negatives avoids the "problem" of not being able to place the negative sign randomly, but it doubles the number of symbols needed.

No. That's a baseless claim. The base 10 symbols 4,3,2,1,0,1,2,3,4,5 does not doubles the number of symbols needed.

Ancalagon wrote:
That isn't really a problem, but if you consider an extra symbol wasteful, then doubling the number of symbols is worse than adding one symbol.

It does not matter how many symbols are used. What matters is how efficiently are these symbols used.

A base 3 of the symbols "0", "1" and "-" is more wasteful than a base 100 which uses all the possible combination of symbols.

Ancalagon
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06 Jun 2013, 9:14 am

It does not matter how many symbols are used. What matters is how efficiently are these symbols used.

Why would that matter?

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PsychoSarah
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06 Jun 2013, 9:25 am

This entire thread was based on an incorrect and not well presented assumption.

Tufted Titmouse

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06 Jun 2013, 11:03 am

PsychoSarah wrote:
This entire thread was based on an incorrect and not well presented assumption.

Why?

Spiderpig gave good answers to the problem presented on the first post. His answer to the problem of sign was very good.

PsychoSarah
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06 Jun 2013, 11:07 am

Literature and language are more my specialty. I have trouble communicating math ideas to people.

Spiderpig
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06 Jun 2013, 12:14 pm

For example, if base is -4, -3, -2, -1, 0, 1, 2, 3, 4, 5,
which we can represent with the symbols B, C, D, A, 0, 1, 2, 3, 4, 5 or for clarity: 4, 3, 2, 1, 0, 1, 2, 3, 4, 5

There’s a problem with an even base like 10, though: it makes the system unbalanced, and this does cause arithmetic to be more cumbersome. The opposite of a number can’t be obtained anymore by switching the sign of each individual digit, because “5” has no counterpart, so, while -3042 is still 3042, -2513 = 3513.

I think that it is just reordering the digits.

Of course. Once you get rid of the radix point, how do you encode the information about where the place whose weight is 1 is, other than with the digit order?

but that would not be more efficient, because it just sorts traditional representation, and adds lots of zeros.

Yes, it is more “efficient” according to the criterion you gave in your first post. Getting rid of both the minus sign—as part of the representation of the number itself—and the radix point, you’re left with only digits, and this enables you to represent b^n different numbers with n symbols (from a repertory of b). You can’t do it if you include the radix point among the symbols, as you already said. There was no mention in your first post of any extra requirement that some of those b^n have shorter representations. The fact that, in the standard system, you can leave out zeroes on the left before the radix point, or on the right after it, doesn’t allow you to represent more numbers with a given number of places, precisely because the extra zeroes carry no meaning.

maybe we can add a radix point on any place, so we have numbers written as ...7.4.9...8.21..8

That’s no longer a radix point. How would that system work?

If you come up with a system meeting both your initial requirements and your new ones, I’d like to know it.

[…] similarly wasteful, as is using "i" for complex numbers.

Heh! You can extend the approach of signed digits to have digits with complex values. With the ternary system, you need nine symbols, whose values are -1-i, -1, -1+i, -i, 0, i, 1-i, 1 and 1+i. You could, for example, use arrows as digits, pointing in the relevant direction in the complex plane, except for “0”, which works just fine. The weights are still powers of 3. Thus, 25.333… - 82.666… i (“3” and “6” recurring, respectively) is, in the balanced ternary system, 101.1 + 001.1 i; therefore, with complex digits, it’s represented as ↓→0↙↗.↗ .

With base 15, the one I suggested before, you’d need 225 digits, so this approach wouldn’t be practical, but you could still make double digits, combining the real signed digits (from to 7) in pairs, and clearly delimiting these, so the first member of each pair represents the real part of the value of the double digit, and the second, the imaginary part. The weights are still powers of 15.

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Last edited by Spiderpig on 06 Jun 2013, 12:36 pm, edited 1 time in total.

Tufted Titmouse

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06 Jun 2013, 12:19 pm

Ancalagon wrote:
It does not matter how many symbols are used. What matters is how efficiently are these symbols used.

Why would that matter?

-It matters because is the point of the thread.
-It matters because has important applications, like optimizing computer resources in programming.
-It matters because it may be (maybe not) easier to be taught, and save time and money on schools.
-It matters because it may have important applications, by generalizing the solutions to more complex problems.

Just try to solve arithmetic problems using a less efficient system, like roman numerals.

And it matters to me because my intuition says that the traditional system is wrong.

Look at this: a number like -Z is the same as 0 - Z, being 0 the number that added to any other doesn't change it.

Then, we have multiplication and division. The number that does not change the product is 1. So, we could use the same trick, and assume that /Z is the same than 1 / Z. But we do not use the same pattern, or convention; "-" is an operator, and also a sign, but "/" is not used as a sign. That's arbitrary.

If we start with positive integers, and only addition (or subtraction), we can only get integers, and subtraction leaded to the discovery of negative numbers.

Product is a repetition of addition, and his inverse, leaded to the discovery of real numbers.
I think that is natural to use "/" as a sign, but we, humans, also introduced another symbol, the radix point, or decimal separator "."

Here /4 = 0.25

That doesn't look good to me. Now there are 2 new symbols, "/" and ".".

If we repeat products, we get powers, and his inverse functions, logarithms and roots, which lead to the discovery of complex numbers.

For representation of complex numbers we need to introduce another symbol, which frequently is "i".
"i" may be though as a sign, and also as a separator. iZ = i x Z. It separates the real from the complex part of a number. Example: 2+i1 could be written as 2i1.
The "i" is sometimes omitted because complex numbers are thought as pair of numbers: 2+i1=(2,1)

But real numbers also may be though as a par of numbers: 2+.1=2.1
They may be thought as a par of numbers because the "." is present only once, so any real number have one integer and one fractional part (both may be zero).

So, we could write a number like 3-i0.25 as 3i/-4
Doesn't it looks like a mess? It doesn't looks good to me.

My experience taught me that introducing arbitrary complexities generally makes problems (even more) unnecessarily complex.

I think that the root of all the problems is the inefficiency of the symbols introduced. I can't justify it, so, I'm looking for an efficient way to represent real numbers, then the complex ones.

Last edited by Buxcador on 06 Jun 2013, 1:12 pm, edited 1 time in total.

Tufted Titmouse

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06 Jun 2013, 12:51 pm

PsychoSarah wrote:
Literature and language are more my specialty. I have trouble communicating math ideas to people.

I'm also have problems communicating. And I don't even speak English.

I know that I did not presented a strictly coded problem, but some interpretations of my post looks to me illogical and arbitrary.

What I would like to know is what you did interpreted and why.

naturalplastic
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06 Jun 2013, 4:54 pm

Lets get to the nub of the issue.
Basically you want to get rid of negative numbers.
Is that a fair summary of what youre after?

What you want is to have all of all negative integers replaced by nine newfangled digits that dont yet exist.

For the sake of arguement lets use the first nine letters of the alphabet as stand ins for these numerals.

A = -1, B=-2, C= -3, and so on.

That way we could eliminate subtraction all together.

"Nine plus G = 2" would replace both" 9-7=2", and "9+ -7=2" as the way we express subtraction. There would be no more subtraction-just adding these new newfangled numbers- which would have negative values.

Is that what this is all about?

Tufted Titmouse

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06 Jun 2013, 7:45 pm

Spiderpig wrote:
For example, if base is -4, -3, -2, -1, 0, 1, 2, 3, 4, 5,
which we can represent with the symbols B, C, D, A, 0, 1, 2, 3, 4, 5 or for clarity: 4, 3, 2, 1, 0, 1, 2, 3, 4, 5

There’s a problem with an even base like 10, though: it makes the system unbalanced, and this does cause arithmetic to be more cumbersome. The opposite of a number can’t be obtained anymore by switching the sign of each individual digit, because “5” has no counterpart, so, while -3042 is still 3042, -2513 = 3513.
Yes, you are right. The base 10 (-4..5) just helps to show how it works because it is alike to the standard base 10 (0..5), but it haves an asymmetry that makes it cumbersome and inconvenient.

A good system should be symmetric. Is unfortunate that the common system is base 10 instead of an odd base.

Spiderpig wrote:
Heh! You can extend the approach of signed digits to have digits with complex values. With the ternary system, you need nine symbols, whose values are -1-i, -1, -1+i, -i, 0, i, 1-i, 1 and 1+i. You could, for example, use arrows as digits, pointing in the relevant direction in the complex plane, except for “0”, which works just fine. The weights are still powers of 3. Thus, 25.333… - 82.666… i (“3” and “6” recurring, respectively) is, in the balanced ternary system, 101.1 + 001.1 i; therefore, with complex digits, it’s represented as ↓→0↙↗.↗ .
Since real numbers also are pair of numbers, the same trick could be used to get rid of the radix point.

>=→=1
<=←=-1
^=0.1
˅=^=-0.1
┐=1=1.1
└=1=-1.1
┘=1-0.1=0.9
┌=-1+0.1=-0.9

Tufted Titmouse

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06 Jun 2013, 10:05 pm

Since real numbers also are pair of numbers, the same trick could be used to get rid of the radix point.

>=→=1
<=←=-1
^=0.1
˅=^=-0.1
┐=1=1.1
└=1=-1.1
┘=1-0.1=0.9
┌=-1+0.1=-0.9

Hmm, no, I found that that method doesn't works, because it still needs a radix point.

eric76
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07 Jun 2013, 3:18 pm

We have 10 numbers, 0-9, precisely because we count in base 10.

If we counted in base 19, we would have 19 numbers.

When doing hexadecimal, we have 16 numbers: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, a, b, c, d, e, and f.

When doing octal, we have 8 numbers: 0, 1, 2, 3, 4, 5, 6, and 7.

When doing binary, we only have 2 numbers: 0 and 1.

eric76
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07 Jun 2013, 3:25 pm

If you want to have fun, do arithmetic in Egyptian Fractions.

In Egyptian Fractions, you write fractions as the sum of a finite series of fractions with a numerator equal to 1 and such that each fraction is less than the one in front.

For example, you would express 15/16 as 1/2 + 1/4 + 1/8 + 1/16.