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Can one use an irrational number as a base?

  1. May 11, 2015 #1

    Grinkle

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    Is it sensible to consider a base pi number system? Can one make an irrational number rational by defining it as the unit of a counting system? I don't know what constitutes an mathematically consistent 'number line' - this question might not make sense. I'm just thinking that if I use pi as my unit, can write pi as a rational number, since 1 is a rational number. Then I guess most or maybe all other numbers on the number line except integral multiples of pi are irrational?
     
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  3. May 11, 2015 #2

    HallsofIvy

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    One can use any number as a base. For example, if we were to use [itex]\pi[/itex] as a base, then any number would be written as [itex]a+ b\pi+ c\pi^2+ d\pi^3+ ...[/itex], writing only the coefficients, a, b c in reversed order, That is [itex]2+ 3\pi+ \pi^2[/itex], which is about 21.29438 in base 10, would be written as 132 in base [itex]\pi[/itex].

    However, that would NOT make irrational numbers rational! All numbers, including integers, rational numbers, and irrational numbers, are defined independently of any numeration system.
     
  4. May 11, 2015 #3

    lavinia

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    You can also use negative bases. For instance any integer can be expressed as powers of -2.

    In base -2 , 1 is 1 , 10 is -2, 11 is -1, 100 is 4, 110 is 2.

    The rule of addition is 1 + 1 = 110. That is $$ 1+1 =(-2)^2 + (-2)^1 + 0 x (-2)^0$$.

    Interestingly 1 + 11 which is zero can not be computed by this rule without induction.

    Adding the 1's column gives 1 + 1 which equals 110. Put down the zero, carry the 11. Now one has 1+ 1 in the -2's column. Put down the zero carry the 11 and so forth.
     
    Last edited: May 17, 2015
  5. May 11, 2015 #4

    robphy

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    "Natural log" is "log base e".
     
  6. May 11, 2015 #5

    Grinkle

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    I kind of get that, but I am missing what a rational number is, I think.

    If I take and lay straight a rope of unit length in the base 1 number line, I can then cut a rope that is the length of the circle enclosing the unit length as its diameter (I make a circle of diameter 1). The length of the second rope I cannot express as a rational number using what I will call a base 1 number line. However, I can represent its length as '1' (I hope this is correct) using what will call a base pi number line. There is no existing number line, I claim, in which I can represent the lengths of both ropes as a fraction of two integers, I have to pick one or the other. I think this has something to do with what you mean by rationality being independent of the numeration system, but still it seems to me that I do have freedom to pick which numbers I want to be rational, I just can't have them all rational at once.

    Edit -

    Or maybe I am confusing the concept of a number with the concept of a base or numeration system.
     
  7. May 11, 2015 #6

    jbriggs444

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    Your terminology is sufficiently at odds with standard mathematical usage that no one here had properly understood what you are talking about.

    It appears that you are using "base pi number line" to describe a number line in which pi is essentially a unit of measure. The unit length is pi meters long. So every multiple of pi meters of length naturally corresponds to an integer.

    The wording "base pi" in standard mathematical language would denote a place value numbering system in which numerals (digit strings) are given values by multiplying the digit in each place by the "place value" of that position in the string. The place values are normally assigned by raising the "base" to an integer power corresponding to the position of the digit in the string (typically the position is measured relative to the decimal point). In base pi the digit string "10" would denote 1 times pi plus zero. i.e. pi.

    A rational number is a number which is the ratio of two integers.

    Numbers exist separate and apart from number lines. Your decision to represent the integer 1 as one meter, one foot, pi meters, pi feet or ##\sqrt{2}## furlongs does not change the nature of the integer 1.
     
  8. May 11, 2015 #7

    Grinkle

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    This feedback is not surprising.

    This is my biggest gap in understanding, I think.
     
  9. May 11, 2015 #8

    symbolipoint

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    Rational versus Irrational:
    A number which is a ratio of any two integers is a rational number.
    A number which is not the ratio of two integers is an IRRATIONAL number.
     
  10. May 12, 2015 #9
    I was wondering this too. In base Pi, 10 is Pi so since Pi is 10 in base Pi, well, 10 looks like an integer! Weird!
     
  11. May 12, 2015 #10

    Mark44

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    This brings up a question about what set to use for the coefficients a, b, c, and so on in the above.
    In computer studies there are a variety of bases in common use: base-2 (binary), base-8 (octal), base-16 (hexadecimal), and of course, base-10. Besides these, there are several other systems in use, such as base-32, base-36, base-58, and base-64. In each of these numbering schemes the number of digits used is equal to the number that represents the base.
    System Digits
    Binary - {0, 1}
    Octal - {0, 1, 2, 3, 4, 5, 6, 7}
    Decimal - {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
    Hex - {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F}
    Base 64 - various sets of characters and decimal digits are used.
     
  12. May 12, 2015 #11
    Any irrational number can be used as a base, but transcendental numbers like ##\pi## and ##e## have no intriguing properties as number bases, besides that no integer has a terminating expansion in the respective base systems. Nevertheless, I doubt anyone expected them to be very interesting--but I was surprised that no one mentioned base ##\varphi##. Every integer has a terminating expansion in "phinary":
    ##1_{10}=1_{\varphi}\\
    2_{10}=10.01_{\varphi}\\
    3_{10}=100.01_{\varphi}.##​
    You can go overboard, and take the base ##\zeta=\frac{1}{3}\Big(\sqrt[3]{\frac{27+3\sqrt{69}}{2}}+\sqrt[3]{\frac{27-3\sqrt{69}}{2}}\Big)##. Then we have
    ##1_{10}=1_{\zeta}\\
    2_{10}=1.011_{\zeta}\\
    3_{10}=11.00101_{\zeta},##​
    et cetera.These are a fews ways you can play with irrational (or even complex) base systems. In general, you can study algebraic number bases by examining the equivalent ring ##\mathbb{Z}[X,X^{-1}]/R(X)## where R(X) is the minimal polynomial of some (algebraic) number. The minimal polynomial of the number ##\zeta## I chose above is ##X^3-X-1##.
     
    Last edited: May 12, 2015
  13. May 14, 2015 #12

    Grinkle

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    That makes sense to me. I hope this is is well-phrased enough to answer - will the set of numbers with terminating expansions always have the same size, regardless of base? In my layman thinking, using very sloppy vernacular, does changing base change how many integers one can find? My question phrasing is at odds with your phrasing that integers are still integers even though they don't have a terminating expansion in every base, but I still hope my question makes enough sense to be answerable.
     
  14. May 14, 2015 #13

    jbriggs444

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    The set of numbers with terminating expansions will always be countable. That is, it will have (at most) the cardinality of the integers. That follows because two sets have the same cardinality if you can match up their respective members one to one. It is fairly simple to construct a one to one correspondence between digit strings that represent integers (i.e. finite digit strings without decimal points) and terminating digit strings that represent real numbers (i.e. finite digit strings with at most one decimal point).

    Since the set of digit strings in one base has the the same cardinality as the integers and the set of digit strings in another base also has the same cardinality as the integers, it follows that they have the same cardinality as each other. The cardinality of the set of numbers represented by terminating digit strings is always the same regardless of the base that is used.
     
  15. May 14, 2015 #14

    Grinkle

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    I thought the cardinality of integers is smaller than the cardinality of real numbers. I am reading the above to mean the two cardinalites are equal to each other. I am mis-interpreting something.

    edit:

    Maybe this - saying integers correspond to reals means that reals are at least as numerous as integers, and can be more numerous. So saying integers correspond to reals does imply that reals correspond to integers. Is that my error?
     
    Last edited: May 14, 2015
  16. May 14, 2015 #15

    Mark44

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    The cardinality of integers is less than the cardinality of the reals. What @jbriggs444 said was terminating digit strings, which are rational numbers. An example would be .423, or ##\frac{423}{1000}##. He didn't include nonterminating digit strings to the right of the decimal point, that have a repeating pattern, such as .444..., which is also rational and equal to ##\frac 4 9##.
    The rational number can be placed in a one-to-one correspondence with the integers, so both sets have the same cardinality.
     
  17. May 14, 2015 #16
    I get what you're asking--it actually took quite a bit of work to derive the "base ##\zeta##" expansion for the integer ##3## using only the digits ##1## and ##0## (otherwise the solution would be to trivially write the number ##3##!).
    The technique is to represent the numbers as their expansions, and then to add and subtract multiples of ##\zeta^3-\zeta-1##, which is zero-valued because ##\zeta## is a root of ##X^3-X-1##. To illustrate:
    ##1=1-\zeta^{-3}(\zeta^3-\zeta-1)=1-1+\zeta^{-2}+\zeta^{-3}=0.011_{\zeta}\\
    2=1+1=1+0.011_{\zeta}=1.011_{\zeta}\\##​
    The difficulty lies in finding an expansion with coefficients ##0\le c<\zeta##. For a suitable choice of polynomial this task is easy; if ##\zeta^n-1=0## then one can simply shift units left and right to cancel negative digits. I would imagine that the general case is not so simple.
    Finding the base-##\zeta## expansion of a number ##s##, where ##\zeta## is a root of an irreducible monic polynomial ##P(X)## with integer coefficients, is equivalent to finding polynomials ##Q(X), R(X)## with rational coefficients such that ##U(X)=\frac{P(X)Q(X)+sR(X)}{R(X)}## has all nonnegative integer coefficients less that ##\zeta##. Finding any such polynomial can be done efficiently using the extended GCD algorithm, but requiring that the coefficients be "principal"-valued complicates the problem.
    (The short answer is that I have no clue. :P)
     
    Last edited: May 14, 2015
  18. May 14, 2015 #17

    Grinkle

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    Suremarc - that is the most memorable response I have ever gotten in these forums! Awesome. And that's even if you made it up, which I can't Google because I can't make those Greek letters in the search box. :-) Anyway, thanks for a grin at the end of a pretty stressful day.
     
  19. May 14, 2015 #18

    DaveC426913

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    My favorite is
    Unary {|}

    Not computers, but used in card games a lot. :woot:

    (Dec) 2 = (Unary) ||
    (Dec) 6 = (Unary) ||||| |
    (Dec) 13 = (Unary) ||||| ||||| |||
     
  20. May 14, 2015 #19

    Mark44

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    I got in a loooonnnng discussion many years ago (on dialup Compuserve) with a guy who steadfastly maintained that 1 was a perfectly good number base. My argument that the only possible digit was 0 didn't seem to carry much weight with him.
     
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