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Irrational numbers

  1. Apr 16, 2012 #1
    Can anyone explain what is wrong with my reasoning?

    Suppose [itex]x = \frac{p}{q}[/itex] and let [itex]x = \sqrt 2 + \sqrt 3 [/itex]. Also, let [itex]a,b,c \in {\Bbb Z}[/itex] and assume [itex]a < xc < b[/itex]. If I show that xc must be an integer, and I know there does not exist c such that [itex]\sqrt 2 c[/itex], or [itex]\sqrt 3 c[/itex] is an integer. Then, [itex]\left( {\sqrt 2 + \sqrt 3 } \right)c[/itex] cannot be an integer, a contradiction.

    p and q are integers, where q > 1. I am supposing that [itex]\sqrt 2 + \sqrt 3 = p/q[/itex].
     
    Last edited: Apr 16, 2012
  2. jcsd
  3. Apr 16, 2012 #2
    What are you trying to prove exactly?

    If x is a rational number, p and q would have to be integers and therefore x could not be [itex]\sqrt{2}[/itex] + [itex]\sqrt{3}[/itex]. That's the contradiction. That means xc could never be rational for any integer c
     
  4. Apr 16, 2012 #3
    Suppose, beforehand we show that xc is an integer, and then show it is not. By the way, is it true that [itex]\sqrt 2 c + \sqrt 3 c[/itex] is never an integer? I think it is true. Then we have arrived at a contradiction.
     
  5. Apr 16, 2012 #4
  6. Apr 16, 2012 #5
    For instance, you could do a proof of this form:

    1. Assume x = [itex]\sqrt{2}[/itex] + [itex]\sqrt{3}[/itex] is rational.
    2. Define a < xc < b, where a, b, and c are integers.
    3. Prove that [itex]\sqrt{2}[/itex] and [itex]\sqrt{3}[/itex] are both irrational (by the contradiction proof I posted)
    4. Since integers are a subset of rational numbers, and x is not in the set of rational numbers, then xc cannot be an integer for any integer c.
     
  7. Apr 16, 2012 #6
    Yes, I know. I just gave an example. Suppose [itex]x = \sqrt 2 + \pi[/itex]. Would the same argument work? I know I left some detail out, but essentially I show that xc has to be an integer. Using the same reasoning would I arrive at a contradiction?
     
    Last edited: Apr 16, 2012
  8. Apr 16, 2012 #7
    Are you sure this would work? I think the only problem is part 2. You cannot just define a < xc < b. You have to show that is the case. Correct? If I suppose [itex]x = \sqrt 2 + \pi[/itex]. Would the same argument work? We know that both [itex]\sqrt 2[/itex] and [itex]\pi[/itex] are irrational. If I I show that part 2 must hold, etc. I would arrive at a contradiction. Correct?
     
    Last edited: Apr 16, 2012
  9. Apr 16, 2012 #8
    I'm a little confused because it's rather vague. a and b are arbitrary, correct? You originally just assumed a < xc < b just as i did. So I don't get what you're trying to prove with that. a and b are just boundaries that imply |a/c| < x < |b/c|
     
  10. Apr 16, 2012 #9
    For example, let [itex]x = \sqrt 2 + \pi[/itex] and assume [itex]x = p/q[/itex]. Suppose we can show that xc, where c is an integer, must be an integer between two integers, namely a and b i.e. a < xc < b. If I prove that xc cannot be an integer, would it be reasonable to infer that we have a contradiction? I know it may sound confusing. I hope it makes sense. Essentially, if we show that xc must be an integer between a and b, assuming x = p/q, and then show xc cannot be an integer, will we have a contradiction (assuming we can actually show xc must be between a and b)? Can we then infer that x is irrational?
     
    Last edited: Apr 16, 2012
  11. Apr 16, 2012 #10
    You can absolutely prove that x is irrational and therefore xc is irrational and not an integer.

    1. Show x is irrational by contradiction.
    2. Assume a < xc < b for integers a, b, c, where xc is an integer.
    3. Prove that since x is irrational, then xc is also irrational and therefore not an integer.

    What's the problem? You could pick a and b to be anything outside of xc. It seems like that part is pointless. If the heart of the proof is to prove that xc cannot be an integer between some integers a and b, then just prove xc cannot be an integer (no matter what xc is, there will be integers that it lies between).
     
  12. Apr 16, 2012 #11
    It seems like if we let [itex]x = \sqrt 2 + \pi[/itex] it does not work.
     
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