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Transcendental vs. SUPER transcendental?

  1. Jan 14, 2012 #1
    Is there any differentiation between transcendentals that can be expressed algebraically (through finite operations) and those that can't?

    IE, the golden ratio is transcendental but is also [1 + sqrt(5)]/2

    pi or e cannot be defined this way.

    So are there different classes of transcendentals, or do we view these types of numbers equally?
  2. jcsd
  3. Jan 14, 2012 #2


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    Gold Member

    The golden ratio is not transcendental. It is the solution to x2-x-1=0.
  4. Jan 14, 2012 #3
    Yes, I knew that. Then it is clear that my understanding of what constitutes a transcendental number is wrong.

    I always thought it was an irrational number that could not be brought to a rational through an operation on the number, but I can see that's incorrect. (what I mean is that, sqrt(2) is irrational, but squaring that gives me 2, therefore sqrt(2) is not transcendental.)

  5. Jan 14, 2012 #4
    Clearly the g.r. is algebraic, being a root of x^2 - x - 1.

    I agree that the g.r. has the property that most people think it's transcendental. In that respect it's a bit like a Grothendieck prime.

    One striking characteristic of Grothendieck’s
    mode of thinking is that it seemed to rely so little
    on examples. This can be seen in the legend of the
    so-called “Grothendieck prime”. In a mathematical
    conversation, someone suggested to Grothendieck
    that they should consider a particular prime number. “You mean an actual number?” Grothendieck
    asked. The other person replied, yes, an actual
    prime number. Grothendieck suggested, “All right,
    take 57.”

  6. Jan 14, 2012 #5


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    The golden ratio is an irrational number, as are pi and e.

    Irrational numbers can be subdivided into algebraic irrational numbers and transcendental numbers. The former (algebraic irrational numbers) can be expressed as the solution of a polynomial equation with integral coefficients, e.g. like (x^2 - 2 = 0 for √2) and (x^2 - x - 1 = 0 for the golden ratio). The latter (transcendental numbers) cannot be so expressed, but can be expressed as infinite sums or products of algebraic expressions with integer coefficients, or as limits where infinity is involved in some way. Neither sort of irrational number (algebraic or transcendental) has any periodicity in its decimal expansion (or digit expansion in any integer base), so a decimal representation of the number can never be written out in full.

    You've made a misstep in calling the golden ratio transcendental. Nevertheless, your question, slightly amended, has value. How many transcendental numbers can we actually express? I'm using express in the sense that we can write down a mathematically true expression with a finite span that equals the number. There are only three ways this can be done:

    1) To find a finite expression for the number in terms of a formula (involving infinite sums or limits) using elementary operations such as addition, subtraction, multiplication, division and exponentiation, with integer coefficients. The "famous" ones like [itex]\pi[/itex] and [itex]e[/itex] are covered here.

    2) To "construct" the decimal expansion of a number with a predictable but non-periodic pattern. Liouville's constant and the Champernowne constant fall under this category.

    3) To "construct" a transcendental number by algebraically transforming a known one. Note that all transcendental numbers formed by such an operation would be algebraically dependent with the original transcendental. As an example, we can form the number [itex]\pi + 1[/itex], which is transcendental, distinct from pi, but still algebraically related to pi. In fact, we can form a whole infinite class of such numbers defined by [itex]\pi + 10^{-n}, n \in \mathbb{Z}^+[/itex], which would each differ in a finite number of decimal places from pi.

    Using any of those methods, can we express or even enumerate all the transcendentals? The answer is an emphatic NO. You see, Georg Cantor did some great work in this area. The number of transcendental numbers is uncountably infinite, which means it is impossible to list them out. Even the last set of numbers I constructed using pi has only a countably infinite number of elements. This still falls far shorts of the uncountably infinite number of transcendentals that exist.

    The punchline is that there's no need for "SUPER" transcendentals - transcendentals alone are already "super" enough. One can never get a full grasp of how many there really are. Even if one can conceive of all the known transcendental numbers and think of all the algebraic manipulations that one can apply to them, one would still only have enumerated an infinitesimal (using it loosely here) fraction of the transcendental numbers that exist. There will always be an uncountable number of "unsung heroes" among the transcendental numbers. :biggrin:
  7. Jan 15, 2012 #6
    To further expound upon the distinction which I think may have motivated your post. There exist solutions to polynomial equations over the integers which cannot be expressed as some finite combination (sum/product/difference/quotient) of integers, plus the extraction of roots. These algebraic irrational numbers are those associated with polynomials which do not have solvable galois groups.

    Concisely put, there are roots of polynomials which are only explicitly expressible as some infinite series or continued fraction.
    Last edited: Jan 15, 2012
  8. Jan 15, 2012 #7
    What about a number such as 2^(sqrt(2))?
  9. Jan 15, 2012 #8
    Well yes that is transcendental, but I wouldn't consider it in the class of numbers capable of being written as some finite combination (sum/product/difference/quotient) of integers, plus the extraction of roots.
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