Set of irrational numbers between 9 and 10 are countable

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Homework Statement


The set of irrational numbers between 9 and 10 is countable.


Homework Equations





The Attempt at a Solution


My belief is that I can prove by contradiction.

first, i must prove by contradiction using diagonalization that the real numbers between 9 and 10 are uncountable. (1)

second, i take set of rational numbers Q is countable, hence a subset Q (9,10) is also countable. (2)

third i can prove by contradiction stating I (9, 10) is countable
R(9,10) =
Q(9,10) [Countable per item (2) ] U I(9,10) [ Countable per statement]


This would imply that by closure properties R(9,10) is countable. which is a condtradiction of what we found in 1.

Is this logic sound? Can I prove (1) using the same diagonalization method used for (0,1)?
 

Answers and Replies

  • #2
jgens
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Your argument should work to prove that the set of irrational numbers between 9 and 10 is uncountable. However, if you know that the set of irrational numbers between 0 and 1 is uncountable already, why not just construct a bijection between the irrational numbers in (0,1) and the irrational numbers in (9,10)? This would show that they have the same cardinality and the set in question is thus uncountable.
 
  • #3
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Well, I agree, I should be able to do a bijection as visually and logically we would expect the numbers between 9 and 10 to have the same cardinality as between 0 and 1, but I am not sure on how I would prove this.... in class we taught the diagonaliation proof for reals between 0 and 1, and the union proof for irrationals being uncountable using R = I U Q, but we haven't attacked something such as mapping all the numbers in 0,1 to 9,10

I mean, i could prove bijection I *think* as follows:

for a function
F: R [tex]\rightarrow[/tex] R +9 : f(n) = n + 9

Prove it is a bijection
one-to-one:
If f(x_1) = f(x_2) [tex]\rightarrow[/tex] x_1 = x_2
f(x_1) = n_1 + 9 AND f(x_2) = n_2 + 9
If n_1 + 9 = n_2 + 9 [tex]\rightarrow[/tex] n_1 = n_2 [tex]\rightarrow[/tex] x_1=x_2

ONTO:
For all y [tex]\in[/tex] R + 9 [tex]\exists[/tex] x [tex]\in[/tex] R : f(x) = y
For all y = n + 9 [tex]\in[/tex] R + 9 [tex]\exists[/tex] x = n [tex]\in[/tex] R : f(n) = n + 9

Hence I proved any real number can be mapped to a real number +9, so I can map R(0,1) to R(9,10)

Is that right? Or I could do that for irrationals, i presume... but i figured getting the reals over would make my life easier...
 
  • #4
jgens
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There's nothing wrong with your initial argument, so if you're more comfortable with it, then by all means, stick with it; I just thought that constructing a bijection from the irrationals of (0,1) to the irrationals of (9,10) would be simpler. For example, you could have proven this result with the following argument:

Let [itex]A[/itex] be the set of irrational numbers in the open interval [itex](0,1)[/itex] and [itex]B[/itex] the set of irrational numbers in the open interval [itex](9,10)[/itex]. Consider the function [itex]f:A \to B[/itex] defined by [itex]f(x) = x+9[/itex]. Since [itex]0 < x < 1[/itex], it follows that [itex]9 < x+9 < 10[/itex]; moreover, because [itex]x[/itex] is irrational, [itex]x+9[/itex] is also irrational. Therefore, every member of [itex]A[/itex] maps to a unique member of [itex]B[/itex]. This proves that [itex]|A| \leq |B|[/itex], and thus [itex]B[/itex] is uncountable.
 

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