Proving the Rationality or Irrationality of Numbers with Non-Natural Bases

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In summary, determining whether a real number is rational or irrational depends on its form and representation. If the number can be written as a fraction of integers or has a terminating or repeating decimal expansion, it is rational. If the number has a non-terminating and non-repeating decimal expansion or is a root that cannot be written as a fraction, it is irrational. Natural numbers greater than one are typically used as bases, but it is possible to use rational non-natural bases as well.
  • #1
eljose
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How can you prove in general that given a real number "a" this is rational or irrational?..:confused: :confused: :confused:
 
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  • #2
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
[tex]\sum_{n=1}{\intfty}\frac{1}{10^{n^2}}[/tex]
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.
 
  • #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.
 
  • #4
BSMSMSTMSPHD said:
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.

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.
 
  • #5
NateTG said:
What is the "standard form" for a real number?

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.
 
  • #6
BSMSMSTMSPHD said:
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.

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.
 
  • #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.
 
  • #8
Curious3141 said:
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.

What about base pi?
 
  • #9
Office_Shredder said:
What about base pi?

Generally, only natural numbers greater than one are used as bases.
 
  • #10
Curious3141 said:
Generally, only natural numbers greater than one are used as bases.
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' //
 
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