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How can you prove?

  1. Jul 28, 2006 #1
    How can you prove in general that given a real number "a" this is rational or irrational?..:confused: :confused: :confused:
  2. jcsd
  3. Jul 28, 2006 #2


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    It's not always easy.

    Obviously numbers like 1 or [itex]\frac{1}{2}[/itex] are rational

    Similiarly, it's pretty easy to see that
    is irrational.

    For algebraic numbers, like [itex]\sqrt{2}[/itex] the usual method is to assume that [itex]\sqrt{2}[/itex] is rational, so it can be expressed as a fraction [itex]\frac{a}{b}[/itex] and then show that [itex]2b^2=a^2[/itex] cannot be solved in the integers.
  4. Jul 28, 2006 #3
    Well, if I'm given a number - that is, any real number in its standard form (not a series or infinitely continued fraction, etc.) then it should be pretty easy.

    If the number is a fraction, or can be rewritten as a fraction (of integers) then it is rational. If its decimal expansion terminates or repeats a pattern of digits, then it is rational. There is no need to prove this, it simply meets the definition of a rational number.

    If, however, the decimal expansion does not terminate nor repeat (such as for pi or e) the number is irrational. If the number is a root, then it is irrational anytime it is not a perfect root. So, the square root of 2, 3, 5, 6, 7, 8, 10, etc. are all irrational as are the cube roots of 2, 3, 4, 5, 6, 7, 9, 10 etc.
  5. Jul 28, 2006 #4


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    What is the "standard form" for a real number?

    Consider, for example [itex]e^e[/itex]. It's relatively straightforward to calcuate the first few thousand digits, but determining whether it's irrational isn't exactly easy.
  6. Jul 28, 2006 #5
    I was thinking of numbers written in decimal or fraction form, using only the 10 digits and no symbols (such as pi or e). I realize that it was an elementary way of looking at things, and perhaps I didn't add much to the discussion. You're certainly right about e^e.
  7. Jul 29, 2006 #6


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    Any number that has either a terminating or a nonterminating periodic representation (in any base, including 10 of course) is rational. Irrational numbers are aperiodic in any base.
  8. Jul 29, 2006 #7
    If you can only "check" a finite number of decimals of a given number, then you can't decide whether it is rational or not.
  9. Jul 29, 2006 #8


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    What about base pi?
  10. Jul 29, 2006 #9


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    Generally, only natural numbers greater than one are used as bases.
  11. Jul 29, 2006 #10
    True; though, I have made a simple system for rational non-natural bases (greater than one of course!) that comply with the rules (coefficient selection included) for natural bases.
    Though, I do not think this is the standard way of dealing with non-natural rational bases:

    Given a positive rational base expressed as p/q (where p & q are naturals and p>q), I can express any rational number r/q (where r is natural) using powers of p/q with all coefficients being elements of {0,1/q, 2/q, ... , (p-1)/q}.

    Canceling the q's in the denominator of 'r' and each coefficient, this can be reduced to stating:
    [tex]\forall p,q,r \in \mathbb{N}\;{\text{where }}p > q,\;\exists \left( {x_0 , x_1 , \ldots ,x_n } \right) \in \left\{ {0,1, \ldots ,p - 1} \right\}^{n + 1} : r = \sum\limits_{k = 0}^n {x_k \left( {\frac{p}{q}} \right)^k }[/tex]
    ~For example,

    1) (Base 7/4) p=7, q=4, r=39. Thus, (x0, x1, ... , xn) is (4,6,1,4), as
    [tex]4 + 6\left( {\frac{7}{4}} \right) + 1\left( {\frac{7}{4}} \right)^2 + 4\left( {\frac{7}{4}} \right)^3 = 39 [/tex]

    2) (Base 13/10) p=13, q=10, r=29. Thus, (x0, x1, ... , xn) is (3,7,10), as
    [tex]3 + 7\left( {\frac{{13}}{{10}}} \right) + 10\left( {\frac{{13}}
    {{10}}} \right)^2 = 29 [/tex]

    3) (Base 17/11) p=17, q=11, r=94. Thus, the (x0, x1, ... , xn) is (9,4,16,11), as
    [tex]9 + 4\left( {\frac{{17}}{{11}}} \right) + 16\left( {\frac{{17}}{{11}}} \right)^2 + 11\left( {\frac{{17}}{{11}}} \right)^3 = 94 [/tex]

    ~For the special (also trivial) case q=1, (x0, x1, ... , xn) is just the base 'p' representation of 'r' //
    Last edited: Jul 29, 2006
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