paradox?
 

I don't know if I should put this somewhere else, the math isn't prime related, and while it's puzzling, I'm not sure it's a puzzle, so to the lounge I went.

Just for fun, I was thinking of a strange little function and playing with its properties, and I ran into a paradox.

Define Bn(x) to be the number obtained writing the base-n expansion of x in base-(n+1), for example, B10(23) = 23 base 11 = 25, B10(1.5) = 1 + 5/11

for the sake of clarity (because I don't know how to type subscripts here) let's say B(x) is the common form of the function, so B(x) = B10(x)


so here's where I came into the paradox

0.999... = 1

B(1) = 1
B(.999...) = 9/11 + 9/11^2 + 9/11^3 + ... = .9

but since they are equal, you shouldn't be able to put them into a function and get 2 different values

if it is because B(x) isn't a well defined function, why isn't it well-defined?

I believe it's because

[quote]0.999... = 1 [/quote]

isn't true.

0.9999.... can be said to be [i]approaching[/i] 1, but it is not equal to 1.

[quote="ColdFury"]I believe it's because

[quote]0.999... = 1 [/quote]

isn't true.

0.9999.... can be said to be [i]approaching[/i] 1, but it is not equal to 1.[/quote]

[url=http://db.uwaterloo.ca/~alopez-o/math-faq/node41.html#SECTION00531000000000000000]This[/url] is what the sci.math faq says about it

Re: paradox?
 

[quote="TravisT"]Define Bn(x) to be the number obtained writing the base-n expansion of x in base-(n+1), for example, B10(23) = 23 base 11 = 25, B10(1.5) = 1 + 5/11

for the sake of clarity (because I don't know how to type subscripts here) let's say B(x) is the common form of the function, so B(x) = B10(x)[/quote]

[quote]if it is because B(x) isn't a well defined function, why isn't it well-defined?[/quote]
Perhaps the definition needs to be specified in more detail in order to avoid using two different interpretations.

I think this is what you intended for Bn(x) (with a couple of details I'm throwing in):

The value of Bn(x), for n an integer and n > 1, is the base-n representation of the value of x, where the argument x is specified in base-(n+1) representation.

B10(23) = the base-10 representation of the number (23 in base-11) = 2 * 11 + 3 = 2 * 10 + 5 = 25.

[i]The function argument inside the parentheses is a base-11 number, not a base-10 number.[/i]

With this, B(1) = 0 * 11 + 1 = 0 * 10 + 1 = 1

and B(.999...) = 9/11 + 9/11^2 + 9/11^3 + ... < 1

There's no paradox, because [b]in base 11, the limit of .999... does not equal 1[/b].

Let "A" = the base-11 representation of the sum of 9 + 1.

Then in base-11, 1 equals the limit of .AAAAAA..., not the limit of .999999...

Re: paradox?
 

[quote="cheesehead"]I think this is what you intended for Bn(x) (with a couple of details I'm throwing in):

The value of Bn(x), for n an integer and n > 1, is the base-n representation of the value of x, where the argument x is specified in base-(n+1) representation.
[/quote]

Actually, I was looking at it as a real-valued function (Hmm .. can 0.999.. be considered an integer? can it not be considered an integer?)

[quote="cheesehead"]B10(23) = the base-10 representation of the number (23 in base-11) = 2 * 11 + 3 = 2 * 10 + 5 = 25.

The function argument inside the parentheses is a base-11 number, not a base-10 number.[/quote]

but if the 23 in B(23) is really 25, then the function is just B(x) = x, whereas I was thinking of a function where the value of 23 maps to the value of 25

I think the problem here is analogous to making a function f(x) that is equal to the number of letters in the written out name of the number, without the distinction of which language the numbers are in. Since both "1" and "0.999..." are valid decimal expansions, B(x) where x=1 is actually passing 2 values to the function, so 2 values are returned

Re: paradox?
 

[quote="TravisT"][quote="cheesehead"]I think this is what you intended for Bn(x) (with a couple of details I'm throwing in):

The value of Bn(x), for n an integer and n > 1, is the base-n representation of the value of x, where the argument x is specified in base-(n+1) representation.
[/quote]

Actually, I was looking at it as a real-valued function[/quote]
So was I.

I wrote that [b]n[/b] was an integer, not that [b]Bn(x)[/b] or [b]x[/b] were integers. So, yes, the latter two can be real-valued.

In fact, the specification that [b]n[/b] is an integer is just one of the things I threw in, intended to simplify the discussion. There's actually no compelling reason to bar nonintegral n.

[quote](Hmm .. can 0.999.. be considered an integer?[/quote]
Not if you want to avoid confusion and disagreements. It's an expression, or the limit of an infinite series, whose value equals the integer 1, but it's not an integer itself, for most purposes.

[quote][quote="cheesehead"]B10(23) = the base-10 representation of the number (23 in base-11) = 2 * 11 + 3 = 2 * 10 + 5 = 25.

The function argument inside the parentheses is a base-11 number, not a base-10 number.[/quote]

but if the 23 in B(23) is really 25,[/quote]
It's not.

The "23" is a base-11 representation of a number, whose base-10 representation is "25".

Actually, both "23" and "25" are abbreviations. If we had subscripts available, we'd have to have an "11" subscript on "23" and a "10" subscript on "25" to indicate that they're written in different bases. Or write "23 base-11" and "25 base-10" or something like that.

[quote]then the function is just B(x) = x,[/quote]
Only if you're willing to consider the numbers in the abstract without considering their representations in any base. But your own original definition rules that out!

[quote] whereas I was thinking of a function where the value of 23 maps to the value of 25[/quote]
That's "23 base-11 maps to the value of 25 base-10", of course.

Your function isn't really mapping numbers to numbers. It's mapping representations of numbers in one base to their corresponding representations in a different base.

[quote]I think the problem here is analogous to making a function f(x) that is equal to the number of letters in the written out name of the number, without the distinction of which language the numbers are in.[/quote]
Well, sort of ...

[quote]Since both "1" and "0.999..." are valid decimal expansions,[/quote]
Careful, careful, careful!

That's the sort of statment that can lead to confusion.

It's true, but both "1" and "0.999..." are also valid base-11 expansions, and valid base-12 expansions, and so on. See what I said above about abbreviations. We ordinary assume that all written numbers are in base-10, but since your function explicitly violates that assumption, we have to keep track of what base we're using to represent each number in this discussion.

[quote]B(x) where x=1 is actually passing 2 values to the function, so 2 values are returned[/quote]
No, just one. Only your confusion about base representations has led to thinking there are two values.

I noticed someone said that 0.999... isn't 1, but it can be approached by it.

Actually, the proof that 0.999... actually IS 1, is very simple :

let x=0.999...
then 10x=9.999...

10x - x = 9x = 9.99... - 0.999 = 9

9x = 9 so x=1

Since TravisT has stated that he is defining a mapping, I don't think that his function is defined over the reals. It is defined only on a subset of the rationals (where the denominator is a power of the base). Further, it is defined by first mapping each member of the set onto the sum of elements from a basis of the form i*b^n where the i are integers in the range 0 to b-1.

Viewed from the perspective of set theory, the trans-base mapping is well defined, [b]but it is not distributive[/b].

E.g., M(5+5) != M(5) + M(5) when b=5+5.

However, TravisT is expecting other properties of a group which is defined with distributive operators to hold here also.

[quote="Wackerbarth"]Since TravisT has stated that he is defining a mapping, I don't think that his function is defined over the reals.[/quote]
Why can't we have a mapping from an irrational number to its representation in a specified base?

(Note that in the following, the function argument is specified as the symbol "pi".)

B5(pi) = {base-5 representation of pi} = 3.032...

Bpi(pi) = 1

[quote]It is defined only on a subset of the rationals (where the denominator is a power of the base).[/quote]
... if, that is, one insists on mapping only from numbers with _finite_ representations in the appropriate base, right?

[quote="cheesehead"]Why can't we have a mapping from an irrational number to its representation in a specified base?
[/quote]

Because you cannot [i]represent[/i] an irrational by a series of "digits". You can only approximate it.

Can't an infinite representation exist even if it's not possible to physically record all its digits in finite time/space? In a finite time, one can produce any finite portion of the base-whatever representation of pi, just not the whole thing.